linear algebra
TRANSCRIPT
EXERCISE SET 1.1
1. (b) Not linear because of the term x1x3.
(d) Not linear because of the term x–21 .
(e) Not linear because of the term x3/51 .
6. (b) Substituting the given expressions for x and y into the equation x = 5 + 2y yields
Since this equation is valid for all values of t, the proposed solution is, indeed, thegeneral one.
Alternatively, we let x = t and solve for y, obtaining
7. Since each of the three given points must satisfy the equation of the curve, we have thesystem of equations
ax21 + bx1 + c = y1
ax22 + bx2 + c = y2
ax23 + bx3 + c = y3
If we consider this to be a system of equations in the three unknowns a, b, and c, theaugmented matrix is clearly the one given in the exercise.
8. If the system is consistent, then we can, for instance, subtract the first two equations fromthe last. This yields c – a – b = 0 or c = a + b. Unless this equation holds, the system cannothave a solution, and hence cannot be consistent.
y t=1
2–
5
2
t t t t= 5 + 21
2
5
2= 5 + 5 =− −
1
9. The solutions of x1 + kx2 = c are x1 = c – kt, x2 = t where t is any real number. If thesesatisfy x1 + x2 = d, then c – kt + t = d, or ( – k)t = d – c for all real numbers t. Inparticular, if t = 0, then d = c, and if t = 1, then = k.
10. If x – y = 3, then 2x – 2y = 6. Therefore, the equations are consistent if and only if k = 6;that is, there are no solutions if k ≠ 6. If k = 6, then the equations represent the same line,in which case, there are infinitely many solutions. Since this covers all of the possibilities,there is never a unique solution.
12. (a) If the system of equations fails to have a solution, then there are three possibilities:The three lines intersect in (i) distinct points, (ii) 2 distinct points, or (iii) not at all.
(i) (ii) (iii)
(b) If the system of equations has exactly one solution, then all of the lines must passthrough a common point. Moreover, at least two of the lines must be distinct.
2 Exercise Set 1.1
(c) If the system has in infinitely many solutions, then the three lines coincide; that is, thethree equations represent the same line.
14. If k = = m = 0, then all lines must intersect at the origin.
3
EXERCISE SET 1.2
1. (e) Not in reduced row-echelon form because Property 2 is not satisfied.
(f) Not in reduced row-echelon form because Property 3 is not satisfied.
(g) Not in reduced row-echelon form because Property 4 is not satisfied.
2. (c) Not in row-echelon form because Property 1 is not satisfieded.
(d) In row-echelon form.
(f) Not in row-echelon form because Property 1 is not satisfied.
4. (b) Let x4 = t. Then we can read off the solution
x1 = 8 + 7t
x2 = 2 – 3t
x3 = –5 –t
x4 = t
(c) Let x5 = t. Then x4 = 8 – 5t and x3 = 7 – 4t. Let x2 = s. Then x1 = –2 –3t + 6s.
5. (a) The solution is
x3 = 5
x2 = 2 – 2 x3 = –8
x1 = 7 – 4 x3 + 3x2 = –37
5. (b) Let x4 = t. Then x3 = 2 – t. Therefore
x2 = 3 + 9t – 4x3 = 3 + 9t – 4(2 – t) = –5 + 13t
x1 = 6 + 5t – 8x3 = 6 + 5t – 8(2 – t) = –10 + 13t
6. (a) The augmented matrix is
Replace Row 2 by Row 1 plus Row 2 and replace Row 3 by Row 3 plus –3 times Row 1.
Multiply Row 2 by –1.
Note that we resist the temptation to multiply Row 3 by –1/2. While this is sensible, itis not part of the formal elimination procedure. Replace Row 3 by Row 3 plus 10 timesRow 2.
Multiply Row 3 by –1/52.
1 1 2 8
0 1 5 9
0 0 52 104
− −− −
1 1 2 8
0 1 5 9
0 10 2 14
− −− − −
1 1 2 8
0 1 5 9
0 10 2 14
−− − −
1 1 2 8
1 2 3 1
3 7 4 10
– −−
6 Exercise Set 1.2
Now multiply Row 3 by 5 and add the result to Row 2.
Next multiply Row 3 by –2 and add the result to Row 1.
Finally, multiply Row 2 by –1 and add to Row 1.
Thus the solution is x1 = 3, x2 = 1, x3 = 2.
(c) The augmented matrix is
Replace Row 2 by Row 2 plus –2 times Row 1, replace Row 3 by Row 3 plus Row 1,and replace Row 4 by Row 4 plus –3 times Row 1.
1 1 2 1 1
2 1 2 2 2
1 2 4 1 1
3 0 0 3 3
− − −− − −
− −− −
1 0 0 3
0 1 0 1
0 0 1 2
1 1 0 4
0 1 0 1
0 0 1 2
1 1 2 8
0 1 0 1
0 0 1 2
1 1 2 8
0 1 5 9
0 0 1 2
− −
Exercise Set 1.2 7
Multiply Row 2 by 1/3 and then add –1 times the new Row 2 to Row 3 and add –3times the new Row 2 to Row 4.
Finally, replace Row 1 by Row 1 plus Row 2.
Thus if we let z = s and w = t, then we have x = –1 + w = –1 + t and y = 2z = 2s. Thesolution is therefore
x = –1 + t
y = 2s
z = s
w = t
7. (a) In Problem 6(a), we reduced the augmented matrix to the following row-echelonmatrix:
1 1 2 8
0 1 5 9
0 0 1 2
− −
1 0 0 1 1
0 1 2 0 0
0 0 0 0 0
0 0 0 0 0
− −−
1 1 2 1 1
0 1 2 0 0
0 0 0 0 0
0 0 0 0 0
− − −−
1 1 2 1 1
0 3 6 0 0
0 1 2 0 0
0 3 6 0 0
− − −−−−
8 Exercise Set 1.2
By Row 3, x3 = 2. Thus by Row 2, x2 = 5x3 – 9 = 1. Finally, Row 1 implies that x1 = –x2 – 2 x3 + 8 = 3. Hence the solution is
x1 = 3
x2 = 1
x3 = 2
(c) According to the solution to Problem 6(c), one row-echelon form of the augmentedmatrix is
Row 2 implies that y = 2z. Thus if we let z = s, we have y = 2s. Row 1 implies that x= –1 + y – 2z + w. Thus if we let w = t, then x = –1 + 2s – 2s + t or x = –1 + t. Hencethe solution is
x = –1 + t
y = 2s
z = s
w = t
8. (a) The augmented matrix is
Divide Row 1 by 2.
1 3 2 1
2 1 1
3 2 1
− −
2 3 2
2 1 1
3 2 1
− −
1 1 2 1 1
0 1 2 0 0
0 0 0 0 0
0 0 0 0 0
− − −−
Exercise Set 1.2 9
Replace Row 2 by Row 2 plus –2 times Row 1 and replace Row 3 by Row 3 plus –3times Row 1.
Divide Row 2 by 4. Then multiply the new Row 2 by –13/2 and add the result to Row3.
Thus, if we multiply Row 3 by –8/7, we obtain the following row-echelon matrix:
Finally, if we complete the reduction to the reduced row-echelon form, we obtain
The system of equations represented by this matrix is inconsistent because Row 3implies that 0x1 + 0x2 = 0 = 1.
8. (c) The augmented matrix is
Divide Row 1 by 4.
4 8 12
3 6 9
2 4 6
−−
− −
1 0 0
0 1 0
0 0 1
1 3 2 1
0 1 3 4
0 0 1
− −
1 3 2 1
0 1 3 4
0 0 7 8
− −
−
1 3 2 1
0 4 3
0 13 2 4
− −
/
/
10 Exercise Set 1.2
Replace Row 2 by Row 2 plus –3 times Row 1 and replace Row 3 by Row 3 plus 2times Row 1.
If we let x2 = t, then x1 = 3 + 2x2 = 3 + 2t. Hence, the solution is
x1 = 3 + 2t
x2 = t
9. (a) In Problem 8(a), we reduced the augmented matrix of this system to row-echelonform, obtaining the matrix
Row 3 again yields the equation 0 = 1 and hence the system is inconsistent.
(c) In Problem 8(c), we found that one row-echelon form of the augmented matrix is
Again if we let x2 = t, then x1 = 3 + 2x2 = 3 + 2t.
1 2 3
0 0 0
0 0 0
−
1 3 2 1
0 1 3 4
0 0 1
− −
/
/
1 2 3
0 0 0
0 0 0
−
1 2 3
3 6 9
2 4 6
−−
− −
Exercise Set 1.2 11
10. (a) The augmented matrix of the system is
Divide Row 1 by 5.
Replace Row 2 by Row 2 plus 2 times Row 1.
Multiply Row 2 by 5.
Replace Row 1 by Row 1 plus 2/5 times Row 2.
Thus, if we let x3 = t, then x2 = 5 – 27x3 = 5 – 27t and x1 = 2 – 12x3 = 2 – 12t.Therefore, the solution is
x1 = 2 – 12t
x2 = 5 – 27t
x3 = t
1 0 12 2
0 1 27 5
1 2 5 6 5 0
0 1 27 5
−
/ /
1 2 5 6 5 0
0 1 5 27 5 1
−
1 2 5 6 5 0
2 1 3 1
−−
5 2 6 0
2 1 3 1
−−
12 Exercise Set 1.2
(c) The augmented matrix of the system is
Interchange Rows 1 and 4 and multiply the new Row 1 by 1/2.
Interchange Rows 2 and 4. Then replace Row 3 by Row 3 plus –1 times the new Row 2.
Replace Row 4 by Row 4 plus –1 times Row 3.
Replace Row 1 by Row 1 plus –1/2 times Row 2.
1 2 0 5 2 1 2 3 2
0 0 1 2 1 4
0 0 0 1 1 3
0 0 0 0 0 0
−−
1 2 1 2 7 2 0 7 2
0 0 1 2 1 4
0 0 0 1 1 3
0 0 0 0 0 0
−−
1 2 1 2 7 2 0 7 2
0 0 1 2 1 4
0 0 0 1 1 3
0 0 0 1 1 3
−−−
1 2 1 2 7 2 0 7 2
0 0 0 1 1 3
0 0 1 3 2 7
0 0 1 2 1 4
−−−
0 0 1 2 1 4
0 0 0 1 1 3
0 0 1 3 2 7
2 4 1 7 0 7
−−−
Exercise Set 1.2 13
Replace Row 2 by Row 2 plus –2 times Row 3 and replace Row 1 by Row 1 plus –5/2times Row 3.
Thus, if we let y = t and v = s, we have
u = –6 – 2s – 3t
v = s
w = –2 – t
x = 3 + t
y = t
11. (a) From Problem 10(a), a row-echelon form of the augmented matrix is
If we let x3 = t, then Row 2 implies that x2 = 5 – 27t. Row 1 then implies that x1 =(–6/5)x3 + (2/5)x2 = 2 – 12t. Hence the solution is
x1 = 2 – 12t
x2 = 5 – 27t
x3 = t
(c) From Problem 10(c), a row-echelon form of the augmented matrix is
1 2 1 2 7 2 0 7 2
0 0 1 2 1 4
0 0 0 1 1 3
0 0 0 0 0 0
−−
1 2 5 6 5 0
0 1 27 5
−
1 2 0 0 3 6
0 0 1 0 1 2
0 0 0 1 1 3
0 0 0 0 0 0
−−
−
14 Exercise Set 1.2
If we let y = t, then Row 3 implies that x = 3 + t. Row 2 then implies that
w = 4 – 2x + t = –2 – t.
Now let v = s. By Row 1, u = 7/2 – 2s – (1/2)w – (7/2)x = –6 – 2s – 3t. Thus we havethe same solution which we obtained in Problem 10(c).
12. (a) There are more unknowns than equations. By Theorem 1.2.1, the system has infinitelymany solutions.
(d) If we let x2 = t, then x1 = 2–3
t. Thus the system has infinitely many solutions.
13. (b) The augmented matrix of the homogenous system is
This matrix may be reduced to
If we let x3 = 4s and x4 = t, then Row 2 implies that
4x2 = –4t – 4s or x2 = –t – s
Now Row 1 implies that
3x1 = –x2 – 4s – t = t + s – 4s – t = –3s or x1 = –s
Therefore the solution is
x1 = –s
x2 = –(t + s)
x3 = 4s
x4 = t
3 1 1 1 0
0 4 1 4 0
3 1 1 1 0
5 1 1 1 0− −
Exercise Set 1.2 15
14. (b) The augmented matrix of the homogenous system is
This matrix can be reduced to
If we let w = 2s and x = 2t, then the above matrix yields the solution
u = 7s – 5t
v = –6s + 4t
w = 2s
x = 2t
15. (a) The augmented matrix of this system is
2 1 3 4 9
1 0 2 7 11
3 3 1 5 8
2 1 4 4 10
−−
−
1 07
2
5
20
0 1 3 2 0
0 0 0 0 0
0 0 0 0 0
−
−
0 1 3 2 0
2 1 4 3 0
2 3 2 1 0
4 3 5 4 0
−−
−− − −
16 Exercise Set 1.2
Its reduced row-echelon form is
Hence the solution is
I1 = –1
I2 = 0
I3 = 1
I4 = 2
(b) The reduced row-echelon form of the augmented matrix is
If we let z2 = s and z5 = t, then we obtain the solution
Z1 = –s – t
Z2 = s
Z3 = –t
Z4 = 0
Z5 = t
16. (a) The augmented matrix of the system is
2 1
3 6
a
b
1 1 0 0 1 0
0 0 1 0 1 0
0 0 0 1 0 0
0 0 0 0 0 0
1 0 0 0 1
0 1 0 0 0
0 0 1 0 1
0 0 0 1 2
−
Exercise Set 1.2 17
Divide Row 1 by 2 and Row 2 by 3.
Subtract Row 1 from Row 2.
Multiply Row 2 by 2/3.
Thus
Row 1 then implies that
or
xa a b a
= −−
+
= −2
1
2 3
2
9
2
3
6
9
xa y
= −2 2
ya b
=−
+3
2
9
11
2 2
0 13
2
9
a
a b− +
11
2 2
03
2 2
2
3
a
a b− +
11
2 2
1 23
a
b
18 Exercise Set 1.2
17. The Gauss-Jordan process will reduce this system to the equations
x + 2y – 3z = 4
y – 2z = 10/7
(a2 – 16)z = a – 4
If a = 4, then the last equation becomes 0 = 0, and hence there will be infinitely manysolutions—for instance, z = t, y = 2 t + 10—
7 , x = –2 (2t + 10—
7 ) + 3t + 4. If a = – 4, then the last
equation becomes 0 = –8, and so the system will have no solutions. Any other value of a willyield a unique solution for z and hence also for y and x.
19. One possibility is
Another possibility is
20. Let x1 = sin α, x2 = cos β, and x3 = tan γ. If we solve the given system of equations for x1,x2, and x3, we find that x1 = 1, x2 = –1, and x3 = 0. Thus α = π/2, β = π, and γ = 0.
21. If we treat the given system as linear in the variables sin α, cos β, and tan γ, then theaugmented matrix is
This reduces to
1 2 3 0
2 5 3 0
1 5 5 0− −
1 3
2 7
2 7
1 3
1 7 2
1 3
1 7 2
→
→
→
11 1
1 3
2 7
1 3
0 1
→
RREF I= =
3
1 0 0
0 1 0
0 0 1
Exercise Set 1.2 19
so that the solution (for α, β, γ between 0 and 2 π) is
sin α = 0 ⇒ α = 0, π, 2π
cos β = 0 ⇒ β = π/2, 3π/2
tan γ = 0 ⇒ γ = 0, π, 2π
That is, there are 3•2•3 = 18 possible triples α, β, γ which satisfy the system of equations.
22. If (x, y) is a solution, then the first equation implies that y = –(λ – 3)x. If we thensubstitute this value for y into the second equation, we obtain [1 – (λ – 3)2]x = 0. But [1 –(λ – 3)2]x = 0 implies that either x = 0 or 1 – (λ – 3)2 = 0. Now if x = 0, then either of thegiven equations implies that y = 0; that is, if x = 0, then the system of equations has onlythe trivial solution. On the other hand, if 1 – (λ – 3)2 = 0, then either λ = 2 or λ = 4. Foreither of these values of λ, the two given equations are identical—that is, they representexactly the same line—and, hence, there are infinitely many solutions.
23. If λ = 2, the system becomes
– x2 = 0
2x1 – 3x2 + x3 = 0
–2x1 + 2x2 – x3 = 0
Thus x2 = 0 and the third equation becomes –1 times the second. If we let x1 = t, then x3= –2t.
24. We regard this as a system of linear equations in the variables 1/x, 1/y, and 1/z. Theaugmented matrix for this system of equations is therefore
1 2 4 1
2 3 8 0
1 9 10 5
−
−
1 0 0 0
0 1 0 0
0 0 1 0
20 Exercise Set 1.2
which reduces to
Hence, the solution is x = –13/7, y = 91/54, and z = –91/8.
25. Using the given points, we obtain the equations
d = 10
a + b + c + d = 7
27a + 9b + 3c + d = –11
64a + 16b + 4c + d = –14
If we solve this system, we find that a = 1, b = –6, c = 2, and d = 10.
26. Using the given points, we obtain the equations
41a – 4b + 5c + d = 0
53a – 2b + 7c + d = 0
25a + 4b – 3c + d = 0
This homogeneous system with more unknowns than equations will have infinitely manysolutions. Since we know that a ≠ 0 (otherwise we’d have a line and not a circle), we canexpect to solve for the other variables in terms of a. Since the augmented matrix reducesto
we have d = –29, c = –4a, and b = –2a. If we let a = 1, the equation of the circle becomesx2 + y2 – 2x – 4y – 29 = 0. Any other choice of a will yield a multiple of this equation.
29 0 0 1 0
4 0 1 0 0
2 1 0 0 0
1 0 0 7 13
0 1 0 54 91
0 0 1 8 91
−
−
Exercise Set 1.2 21
27. (a) If a = 0, then the reduction can be accomplished as follows:
If a = 0, then b ≠ 0 and c ≠ 0, so the reduction can be carried out as follows:
Where did you use the fact that ad – bc ≠ 0? (This proof uses it twice.)
29. There are eight possibilities. They are
(a)1 0 0
0 1 0
0 0 1
1 0
0 1
0 0 0
, ,
p
q
11 0
0 0 1
0 0 0
0 1 0
0 0 1
0 0 0
1
p
,
,
pp q p
0 0 0
0 0 0
0 1
0 0 0
0 0 0
0 0
, ,
11
0 0 0
0 0 0
∈, , ,where R any real np q uumbers.
and
0 0 0
0 0 0
0 0 0
0
0
1
0
1
0 1
b
c d
c d
b
d
c
b
d
c
→
→
→
→
1 0
0 1
a b
c d
b
a
c d
b
a
ad bc
a
→
→−
11
0
→
→
1
0 1
1 0
0 1
b
a
22 Exercise Set 1.2
(b)
30. (a) If the system of equations has only the trivial solution, then the three lines all passthrough (0,0) and at least two of them are distinct.
(b) Nontrivial solutions can exist only if the three lines coincide; that is, the threeequations represent the same line and that line passes through (0,0).
31. (a) False. The reduced row-echelon form of a matrix is unique, as stated in the remark inthis section.
(b) True. The row-echelon form of a matrix is not unique, as shown in the followingexample:
but
1 2
1 3
1 3
1 2
1 3
0 1
1 3
0 1
→
→
−
→
1 2
1 3
1 2
0 1
→
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 0 0
0 1 0
0 0 1
0
,
p
q
r
00 0 0
1 0 0
0 1 0
0 0 0 1
0 0 0 0
,
p
q
,
1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
0 1 0 0
0 0 1 0
p
,00 0 0 1
0 0 0 0
1 0
0 1
0 0 0 0
0 0 0 0
,
p q
r s
,
,
1 0
0 0 1
0 0 0 0
0 0 0 0
1p q
r
p q 00
0 0 0 1
0 0 0 0
0 0 0 0
0 1 0
0 0 1
0 0 0 0
0 0 0
,
p
q
00
0 1 0
0 0 0 1
0 0 0 0
0 0 0 0
,
p
,, ,
0 0 1 0
0 0 0 1
0 0 0 0
0 0 0 0
1
0 0 0 0
0 0
p q r
00 0
0 0 0 0
0 1
0 0 0 0
0 0 0 0
0 0 0 0
,
p q
, ,
0 0 1
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1
0
p
00 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
and
Exercise Set 1.2 23
(c) False. If the reduced row-echelon form of the augmented matrix for a system of 3equations in 2 unknowns is
then the system has a unique solution. If the augmented matrix of a system of 3equations in 3 unknowns reduces to
then the system has no solutions.
(d) False. The system can have a solution only if the 3 lines meet in at least one pointwhich is common to all 3.
32. (a) False. Let 2 of the equations be x1 + x2 + x3 + x4 + x5 = 0 and x1 + x2 + x3 + x4 + x5= 1 and the system cannot be consistent.
(b) False. Let all 5 equations be the same.
(c) False. The nth row might be n zeros followed by a 1. In that case, the system wouldhave no solution.
(d) False. If one equation is a multiple of another, those 2 equations represent the samesolution set. Thus, 1 of the 2 equations can be disregarded and we will have a systemof n – 1 equations in n unknowns. Such a system need not be inconsistent.
1 1 1 0
0 0 0 1
0 0 0 0
1 0
0 1
0 0 0
a
b
24 Exercise Set 1.2
EXERCISE SET 1.3
1. (c) The matrix AE is 4 × 4. Since B is 4 × 5, AE + B is not defined.
(e) The matrix A + B is 4 × 5. Since E is 5 × 4, E (A + B)is 5 × 5.
(h) Since AT is 5 × 4 and E is 5 × 4, their sum is also 5 × 4. Thus (AT + E)D is 5 × 2.
2. Since two matrices are equal if and only if their corresponding entries are equal, we havethe system of equations
a – b = 8
b + c = 1
c + 3d = 7
2a –4d = 6
The augmented matrix of this system may be reduced to
Hence, a = 5, b = –3, c = 4, and d = 1.
3. (e) Since 2B is a 2 × 2 matrix and C is a 2 × 3 matrix, 2B – C is not defined.
1 0 0 0 5
0 1 0 0 3
0 0 1 0 4
0 0 0 1 1
−
25
3. (g) We have
(j) We have tr(D – 3E) = (1 – 3(6)) + (0 – 3(1)) + (4 – 3(3)) = –25.
4. (b) We have
(d) Since BT is a 2 × 2 matrix and 5CT is a 3 × 2 matrix, the indicated addition isimpossible.
(h) We have
5. (b) Since B is a 2 × 2 matrix and A is a 3 × 2 matrix, BA is not defined (although AB is).
(d) We have
AB =−
−
12 3
4 5
4 1
=−
− −−
=9 1 1
13 2 4
0 1 6
9 −−
− − −
13 0
1 2 1
1 4 6
( – )2 3
12 2 8
2 2 2
6 4 6
3 3
E DT T T =
−
−− 99
15 0 6
6 3 12
T
D ET T– =
−
−−1 1 3
5 0 2
2 1 4
6 1 4
1 1 1
3 2 33
5 0 1
4 1 1
1 1 1
=− −
−− −
= –3
13 7 8
3 2 5
11 4 10
39 21 24
9 6−
=− − −
– −−− − −
15
33 12 30
– ( ) –3 2 3
1 5 2
1 0 1
3 2 4
D E+ = −
+112 2 6
2 2 4
8 2 6
−
26 Exercise Set 1.3
Hence
(e) We have
(f) We have
(j) We have tr(4ET – D) = tr(4E – D) = (4(6) – 1) + (4(1) – 0) + (4(3) – 4) = 35.
6. (a) We have
( – )2
2 2 6
10 0 4
4 2 8
6 1 3
1 1 2D E AT =
−
− −44 1 3
3 0
1 2
1 1
−
=
−− −−
−
4 3 3
11 1 2
0 1 5
3 0
1 2
1 1
− −
=
6 3
36 0
4 7
CCT =
1 4 2
3 1 5
1 3
4 1
2 5
21 17
=117 35
A BC( ) = −
=
3 0
1 2
1 1
1 15 3
6 2 10
33 45 9
11 11 17
7 17 13
−
( )AB C = −
3 45 9
11 11 17
7 17 13
Exercise Set 1.3 27
6. (b) Since (4B)C is a 2 × 3 matrix and 2B is a 2 × 2 matrix, then (4B)C + 2B is undefined.
(d)
Thus,
(f) Each of the matrices DTET and (ED)T is equal to
Thus their difference is the 3 × 3 matrix with all zero entries.
7. (a) The first row of A is
A1 = [3 -2 7]
Thus, the first row of AB is
A B1 3 2 7 6 2 4
0
= [67 41 41]
= −[ – ]
11 3
7 7 5
14 4 12
36 1 26
25 7 21
−
(BA CT T– )2
10 6
14 2
1 8
=−
−− −
BA CT =– 2
4 1
0 2
3 1 1
0 2 1
−
−
− 2
1 4 2
3 1 5
12 6 3
0 4 2
2 8 4
6 2 10
=−
−
=− −
− −
10 14 1
6 2 8
28 Exercise Set 1.3
(c) The second column of B is
Thus, the second column of AB is
(e) The third row of A is
A3 = [0 4 9]
Thus, the third row of AA is
8.
BA =−
6 6 70
6 17 31
63 41 122
6
6
63
=
+−
3
6
0
7
6
2
1
7
+
−
= −
0
4
3
5
6
17
41
2
6
0
7
+−
+
6
2
1
7
4
4
3
5
=
+70
31
122
7
6
0
7
4
−−
+
2
1
7
9
4
3
5
A A3 0 4 9 3 2 7
6 5 4
0 4 9
= −
[ ]
== [ ]24 56 97
AB2
3 2 7
6 5 4
0 4 9
2
1
7
=−
−
=41
21
67
B2
2
1
7
=−
Exercise Set 1.3 29
9. (a) The product yA is the matrix
[y1a11 + y2a21 + … + ym
am
] y1a12 + y2a22 + … + ym
am2
…
y1a1n+ y2a2n
+ … + ym
amn
]
We can rewrite this matrix in the form
y1 [a11 a12… a1n
] + y2 [a21 a22… a2n
] + … + ym
[am1 am2 …
amn]
which is, indeed, a linear combination of the row matrices of A with the scalarcoefficients of y.
(b) Let y = [y1, y2, …, ym
]
Taking transposes of both sides, we have
(yA)T = ATyT = (A1 | A2 | … | Am
)
= (y1A1 | y2A2 | … | ymAm=
y A
y A
y Am m
1 1
2 2
T
y1
ym
by 9a, =yA y A
y A
y Am m
1 1
2 2
and = be the m rowsA A
A
Am
1
2
oof .A
30 Exercise Set 1.3
10. (a) By the result of Exercise 9, the rows of AB are the matrices
3[6 –2 4] – 2[0 1 3] + 7[7 7 5]
6[6 –2 4] + 5[0 1 3] + 4[7 7 5]
0[6 –2 4] + 4[0 1 3] + 9[7 7 5]
11. Let fij
denote the entry in the ith row and jth column of C(DE). We are asked to find f23. Inorder to compute f23, we must calculate the elements in the second row of C and the thirdcolumn of DE. According to Equation (3), we can find the elements in the third column ofDE by computing DE3 where E3 is the third column of E. That is,
12. (a) Suppose that A is m × n and B is r × s. If AB is defined, then n = r. That is, B is n ×s. On the other hand, if BA is defined, then s = m. That is, B is n × m. Hence AB is m× m and BA is n × n, so that both are square matrices.
(b) Suppose that A is m × n and B is r × s. If BA is defined, then s = m and BA is r × n.If A(BA) is defined, then n = r. Thus B is an n × m matrix.
f23 3 1 5
1 5 2
1 0 1
3 2 4
3
= −
[ ] 22
3
19
0
25
= [3 1 5]
= 182
Exercise Set 1.3 31
15. (a) By block multiplication,
16. (b) Call the matrices A and B. Then
17. (a) The partitioning of A and B makes them each effectively 2 × 2 matrices, so blockmultiplication might be possible. However, if
then the products A11B11, A12B21, A11B12, A12B22, A21B11, A22B21, A21B12, and A22B22 areall undefined. If even one of these is undefined, block multiplication is impossible.
AA A
A AB
B B
B=
=11 12
21 22
11 12
21and
BB22
=
− − −
4 7 19 43
2 2 18 17
0 5 25 35
2 3 23 24
AB =
−
−
−2 5
1 3
0 5
2 1 3
0 1 5
2 5
1 3
0 55
4
7
1 42 1 3
0 1 51
−
−
44
4
7
−
AB =
−−
−
+
1 2
0 3
2 1
3 5
1 5
4 2
7 1
0 3
1 2
0 3
4
2
1 5
4 2
−
−−
+
−
−
+
5
3
1 52 1
3 566 1
7 1
0 31 5
4
26 1
−
+
5
3
8 9
9 15
−
=
−−
+
7 14
28 2
0
6
10
14
13 26
−
+
−
− + − +
=
42 3 14 27
− −−
1 23 10
37 13 8
29 23 41
32 Exercise Set 1.3
18. (a) The entry in the ith row and jth column of AB is
ai1b1j
+ ai2b2j
+ … + ain
bnj
Thus, if the ith row of A consists entirely of zeros, i.e., if aik
= 0 for i fixed and for k =1, 2, …, n, then the entry in the ith row and the jth column of AB is zero for fixed i andfor j = 1, 2, …, n. That is, if the ith row of A consists entirely of zeros, then so does theith row of AB.
20. The element in the ith row and jth column of I is
The element in the ith row and jth column of AI is
ai1δ1j
+ ai2δ2j
+ … + ain
δnj
But the above sum equals aij
because δkj
= 0 if k ≠ j and δjj
= 1. This proves that AI = A.
The proof that IA = A is similar.
21. (b) If i > j, then the entry aij
has row number larger than column number; that is, it liesbelow the matrix diagonal. Thus [a
ij] has all zero elements below the diagonal.
(d) If |i – j| > 1, then either i – j > 1 or i – j < –1; that is, either i > j + 1 or j > i + 1. Thefirst of these inequalities says that the entry a
ijlies below the diagonal and also below
the “subdiagonal“ consisting of all entries immediately below the diagonal ones. Thesecond inequality says that the entry a
ijlies above the diagonal and also above the
entries immediately above the diagonal ones. Thus we have
[a
a a
a a a
a a a
ij ] =
11 12
21 22 23
32 33 34
0 0 0 0
0 0 0
0 0 0
00 0 0
0 0 0
0 0 0 0
43 44 45
54 55 56
65 66
a a a
a a a
a a
δij
i j
i j=
0
1
if
if
≠=
Exercise Set 1.3 33
22. (b) For example, a11 = 11–1 = 10 = 1, a13 = 13–1 = 12 = 1, a23 = 23–1 = 22 = 4, and a32 = 32–1
= 31 = 3. Thus the matrix is
23.x f (x)
x
x
x
f (x) =
f (x)
f (x)
) =2
1
f x
) =2
0
f x
) =7
4
f x
) =0
-2
f x
fx
x
x x
x
1
2
1
2
= 2
+
x =
1
1
x =
2
0
x =
4
3
x =−
2
2
1 1 1 1
1 2 4 8
1 3 9 27
1 4 16 64
34 Exercise Set 1.3
(a)
(b)
(c)
(d)
24. (a)
27. The only solution to this system of equations is, by inspection,
29. (a) Let B = Then B2 = A implies that
(*)a2 + bc = 2 ab + bd = 2
ac + cd = 2 bc + d2 = 2
One might note that a = b = c = d = 1 and a = b = c = d = –1 satisfy (*). Solving thefirst and last of the above equations simultaneously yields a2 = d2. Thus a = ±d. Solvingthe remaining 2 equations yields c(a + d) = b(a + d) = 2. Therefore a ≠ –d and a andd cannot both be zero. Hence we have a = d ≠ 0, so that ac = ab = 1, or b = c = 1/a.The first equation in (*) then becomes a2 + 1/a2 = 2 or a4 – 2a2 + 1 = 0. Thus a = ±1.That is,
and
are the only square roots of A.
− −− −
1 1
1 1
1 1
1 1
a b
c d
A = −
1 1 0
1 1 0
0 0 0
f w + z( )α β
α
=
+f
w
w
wm
1
2 β
α
z
z
z
f
w
m
1
2
=
11 1
2 2
++
+
βα β
α β
z
w z
w zm m
=
+
α βf
w
w
f
z
zm m
1 1
= ( ) ( )α βf w + f z
Exercise Set 1.3 35
(b) Using the reasoning and the notation of Part (a), show that either a = –d or b = c = 0.If a = –d, then a2 + bc = 5 and bc + a2 = 9. This is impossible, so we have b = c = 0.This implies that a2 = 5 and d2 = 9. Thus
are the 4 square roots of A.
Note that if A were , say, then B = would be a square root of A for
every nonzero real number r and there would be infinitely many other square roots as well.
(c) By an argument similar to the above, show that if, for instance,
A = and B =
where BB = A, then either a = –d or b = c = 0. Each of these alternatives leads to acontradiction. Why?
31. (a) True. If A is an m × n matrix, then At is n × m. Thus AAT is m × m and AT A is n × n.Since the trace is defined for every square matrix, the result follows.
(b) True. Partition A into its row matrices, so that
A = and AT =
Then
AA
r r r r r r
r r r r r r
r
T
T TmT
T TmT
m
=
1 1 1 2 1
2 1 2 2 2
rr r r r rT
mT
m mT
1 2
r r rT T
mT
1 2
r
r
rm
1
2
a b
c d
−
1 0
0 1
1
4 1
r
r −
5 0
0 5
5 0
0 3
5 0
0 3
5 0
0 3
−
−
−−
5 0
0 3
36 Exercise Set 1.3
Since each of the rows ri
is a 1 × n matrix, each rTinferior/superior alignedi
is an n× 1 matrix, and therefore each matrix r
irTinferior/superior aligned
jis a 1 × 1 matrix.
Hence
tr(AAT) = r1 rT1 + r2 rT
2 + … + rm
rTm
Note that since ri
rTinferior/superior alignedi
is just the sum of the squares of theentries in the ith row of A, r1 rTinferior/superior aligned1 + r2 rTinferior/superioraligned2 + … + r
mrTinferior/superior aligned
mis the sum of the squares of all of the
entries of A.
A similar argument works for ATA, and since the sum of the squares of the entries of AT
is the same as the sum of the squares of the entries of A, the result follows.
31. (c) False. For instance, let A = and B = .
(d) True. Every entry in the first row of AB is the matrix product of the first row of A witha column of B. If the first row of A has all zeros, then this product is zero.
32. (a) True. If the rows of A are r1, … , rn
and the columns are c1, … , cn, then the ith row
of AA is given by ric1
… ric
n. Thus if r
i= r
jfor some i ≠ j, then the ith row of AA will
equal its jth row.
(b) False. For instance if
A = then A2 =
(c) True. If every entry of B is a positive even integer, we can write B = [bij] = [2c
ij] where
each entry cij
is a positive integer. If A = [aij] where each a
ijis also a positive integer,
then each entry of AB and of BA is the sum of numbers of the form 2cijakl. Thus each
will be an even positive integer.
(d) True. If A is m × n and B is r × s, then for AB to be defined we must have n = r andfor BA to be defined, we must have s = m. Thus AB is m × m and BA is n × n. But forAB + BA to be defined, we must have m = n. Thus A and B must both be square m ×m matrices.
1 −
1 0
0 0 0
0 0 0
1 1 0
0 0
0 0 0
−
0
1 1
1 1
0 1
0 1
Exercise Set 1.3 37
EXERCISE SET 1.4
1. (a) We have
Hence,
On the other hand,
B C+ =− −
−
8 5 2
1 8 6
7 2 15
=−
−
10 6 1
1 12 11
5 1 19
( )A B C+ + =− −
−
+10 4
0 5
2 6
2
7
10
0 −−
2 3
1 7 4
3 5 9
A + B =
10 4 2
0 5 7
2 6 10
− −
−
39
Hence,
1. (c) Since a + b = –3, we have
Also
2. (a) We have
On the other hand
( )aB C =− −
−
32 12 20
0 4 8
16 28 24
−
=− − −0 2 3
1 7 4
3 5 9
72 248 132
288 68 88
44 108 152−
=− − −
−
( )4
18 62 33
7 17 22
11 27 38
=− − −
−
72 248 132
28 68 88
44 108 1522
a BC( ) ( )=− −
−
−4
8 3 5
0 1 2
4 7 6
0 2 3
1 7 4
3 55 9
aC bC+ =−
+0 8 12
4 28 16
12 20 36
0 14 −−− − −− − −
=−
− −21
7 49 28
21 35 63
0 6 9
3 21 −−− − −
12
9 15 27
( ) ( )a b C+ = −−
3
2 3
7 4
5 9
0
1
3
==−
− − −− − −
0 6 9
3 21 12
9 15 27
A B C+ + =−
−
+( )
2 1
0 4
2 1
3
5
4
88 5 2
1 8 6
7 2 15
10 6 1
1 12 1
− −
−
=−
11
5 1 19−
40 Exercise Set 1.4
Finally,
(c) Since
and
the two matrices are equal.
3. (b) Since
and
the two matrices are equal.
A BT T+ =
−−
−
+2 0 2
1 4 1
3 5 4
88 0 4
3 1 7
5 2 6
10 0 2
4 5 6− −−
= − −−−
2 7 10
( )A BT
T
+ =− −
−
=10 4 2
0 5 7
2 6 10
10 0 2
−− −−
4 5 6
2 7 10
BA CA
–
–
–
+ =− −
+26 25– 11
4 6 13
4 26 1
66 2
6 31 54
12 26 70
10 37
– – –5 20 30 9
–
–
–
= 667
16 0 71–
( )
–
B C A+ =−
8 5
2
2– 2
1 8 6
7 15
–– – –1
0
20 30 93
4 5
2 1 4
1
–
–
= 00 37
16 0–
67
71
B aC( ) =−
8 3 5
7
0 8– –
0 1 2
4 6
12
4 288 16
12 20 36
28 6
=– – –72 248 132
88 88
44 108 152−
Exercise Set 1.4 41
3. (d) Since
and
the two matrices are equal.
4. For A, we have ad – bc = 1. Hence,
For B, we have ad – bc = 20. Hence,
For C, we have ad – bc = 2. Hence,
5. (b)
( )BT
T
− =−
=1 1
20
4 3
4 2
1
20
4 3
4 2
1
20
4 4
3 2−
=
−
T
( )BT −
−
=−
=
−
11
2 4
3 4
1
20
4 4
3 2
C− =
− −
1 1
2
1 4
2 6
B− =
−
1 1
20
4 3
4 2
A− =
−−
1 2 1
5 3
B AT T = − −
−
−8 0 4
3 1 7
5 2 6
2 0 2
−−
= −1 4 1
3 5 4
28 20 0
228 31 21
6 38 36
− −
( )ABT
T28 28 6
20 31 38
0 21 36
−−−
=28 20 0
28 31 21
6 38 36
− − −
42 Exercise Set 1.4
6. (a)
7. (b)
Thus,
7. (d)
8. To compute A–3, we can either compute A3 and then take its inverse or else we can find A–1
and then find (A–1)3. We choose the first alternative. Thus
A3 2 0
4 1
2 0
4 1
2 0
4 1=
=
=
4 0
12 1
2 0
4 1
8 0
28 1
2
5
13
2
134
13
1
13
1 0
0 1A =
−
−
=
−
−
,
18
13
2
134
13
12
13
so that A .=−
−
9
13
1
132
13
6
13
If then( ) ,I A I+ =−
+−2
1 2
4 51 22
1 2
4 5
5
13
2
134
13
1
13
1
A =−
=
−
−
. Hence
A =
2 7 1
1 7 3 7
7 73 7
1 2
2 7
1 31 1
1
A A(( ) )= =−
−
=
− −−
We are given that Th( ) .73 7
1 21
A− =
−−
eerefore
B A− − =
−
−−
1 1 1
20
4 3
4 2
2 1
5 3 =−
−
1
20
7 5
18 10
( )AB−
−
=−−
=
−−
11
10 5
18 7
1
20
7 5
18 10
Exercise Set 1.4 43
so that
9. (b) We have
10. (b) Let A be any square matrix. Then
p2(A)p3(A) = (A + 3)(A – 3) = A2 + 3A – 3A – 9I = A2 – 9 = p1(A)
11. Call the matrix A. By Theorem 1.4.5,
since cos2 θ + sin2 θ = 1.
=−
cos sin
sin cos
θ θθ θ
A− =
+
−
12 2
1
cos sin
cos sin
sin cosθ θθ θθ θ
=
20 7
14 6
=
−
+
22 8
16 6
3 1
2 1
1 0
0 1
=
−
+
2
11 4
8 3
3 1
2 1
1 0
0 1
p A( ) =
−
+2
3 1
2 1
3 1
2 11
1 0
0 1
2
A− =
−
3
1
80
7
21
44 Exercise Set 1.4
12. Since the determinant is
12(ex + e–x)1
2(ex + e–x) – 12(ex – e–x)12(ex – e–x)
= 14(e2x + e–2x + 2) – 14(e2x + e–2x – 2)
= 1 ≠ 0
Theorem 1.4.5 applies.
13. If a11a22… a
nn≠ 0, then a
ii≠ 0, and hence 1/a
iiis defined for i = 1,2, . . ., n. It is now easy
to verify that
14. We are given that 3A – A2 = I. This implies that A(3I – A) = (3I – A)A = I; hence A–1 = 3I – A.
15. Let A denote a matrix which has an entire row or an entire column of zeros. Then if B is anymatrix, either AB has an entire row of zeros or BA has an entire column of zeros,respectively. (See Exercise 18, Section 1. 3.) Hence, neither AB nor BA can be the identitymatrix; therefore, A cannot have an inverse.
16. Not necessarily; for example, both I and –I are invertible, but their sum is not.
17. Suppose that AB = 0 and A is invertible. Then A–1(AB) = A–10 or IB = 0. Hence, B = 0.
A
a
a
ann
− =
1
11
22
1 0 0
0 1 0
0 0 1
Exercise Set 1.4 45
18. By Exercise 15 of Section 1.3, we are looking for a matrix C such that
= I4
That is, BA–1 + AC = O. Thus
AC = –BA–1
and hence
C = –A–1BA–1
19. (a) Using the notation of Exercise 18, let
Then
so that
C = −−
−1
4
1 1
1 1
1 1
1 1
1 1
1 1
= −
=−
1
4
0 0
4 0
0 0
1 0
A− =
−
1 1
2
1 1
1 1
A B=−
=
1 1
1 1
1 1
1 1and
=+
−
I O
BA AC I
21
2
A O
B A
A O
C A
AA O O O
BA
=+ +−
−
−1
1
1
−− −+ +
1 1
AC O AA
46 Exercise Set 1.4
Thus the inverse of the given matrix is
20. (a) One such example is
In general, any matrix of the form
will work.
(b) One such example is
In general, any matrix of the form
will work.
0
0
0
b c
b f
c f
−− −
A =−−
0 0 1
0 0 2
1 2 0
a d e
d b f
e f c
A =−−
− −
1 0 1
0 2 2
1 2 3
1
2
1
20 0
1
2
1
20 0
0 01
2
1
2
1 01
2
1
−
−
−22
Exercise Set 1.4 47
21. We use Theorem 1.4.9.
(a) If A = BBT, then
AT = (BBT)T = (BT)TBT = BBT = A
Thus A is symmetric. On the other hand, if A = B + BT, then
AT = (B + BT)T = BT + (BT)T = BT + B = B + BT = A
Thus A is symmetric.
(b) If A = B – BT, then
AT = (B – BT)T = [B + (–1)BT]T = BT + [(–1)BT]T
= BT + (–1)(BT)T = BT + (–1)B = BT – B = –A
Thus A is skew-symmetric.
22. By Theorem 1.4.9, (An)T = (A(An–1))T = (An–1)T AT. If we repeat this process n – 1 times, weget that (An)T = (AT)n which justifies the equality.
23. Let
Then
AA
x x x
x x− =
111 12 13
21 2
1 0 1
1 1 0
0 1 122 23
31 32 33
11 31 12 32 1
x
x x x
x x x x x
=+ + 33 33
11 21 12 22 13 23
21 31 22 32 2
++ + ++ +
x
x x x x x x
x x x x x 33 33+
x
A
x x x
x x x
x x x
− =
111 12 13
21 22 23
31 32 33
48 Exercise Set 1.4
Since AA–1 = I, we equate corresponding entries to obtain the system of equations
The solution to this system of equations gives
24. (a) Let ij
denote the ijth entry of the matrix on the left and rij
be the corresponding entryfor the matrix on the right. Then
ij
= aij
+ (bij
+ cij)
Since the associative rule holds for real numbers, this implies that
ij
= (aij
+ bij) + c
ij= r
ij
(c) Suppose that B is m × n and C is n × p. Then
[a(BC)]ij
= a(bi1c1j
+ bi2c2j
+ … + bin
cnj
)
= (abi1)c1j
+ (abi2)c2j
+ … + (abin
)cnj
= [(aB)C]ij
= bi1 (ac
ij) + b
i2(ac2j) + … + b
in(ac
nj) = [B(aC)]
ij
A− =
−−
−
11 2 1 2 1 2
1 2 1 2 1 2
1 2 1 2 1 2
x x
x x
x x
x x
x x
11 31
12 32
13 33
11 21
12 22
1
0
0
0
+ =+ =
+ =+ =
+ ==+ =
+ =+ =
+ =
1
0
0
0
1
13 23
21 31
22 32
23 33
x x
x x
x x
x x
Exercise Set 1.4 49
25. We wish to show that A(B – C ) = AB – AC. By Part (d) of Theorem 1.4.1, we have A(B – C) = A(B + (–C)) = AB + A(–C). Finally by Part (m), we have A(–C) = –AC and the desired result can be obtained by substituting this result in the above equation.
26. Let aij
denote the ijth entry of the m × n matrix A. All the entries of 0 are 0. In the proofsbelow, we need only show that the ijth entries of all the matrices involved are equal.
(a) aij
+ 0 = 0 + aij
= aij
(b) aij
– aij
= 0
26. (c) 0 – aij
= – aij
(d) ai10 + a
i20 + … + ain
0 = 0 and 0 a1j+ 0a2j
+ … + amj
= 0
27. (a) We have
On the other hand,
( ) ) ( ) (A AA A AA A AA Ar s
r r r
= ( )
factors ffactors aactors
factors
factor
s
rs
AA A
=
ss
A A AA A AA Ar s
r s
= ( (
factors
)
factors
))
= =+
+AA A A
r s
r s
factors
50 Exercise Set 1.4
(b) Suppose that r < 0 and s < 0; let ρ = –r and δ = –s, so that
Ar As = A–ρ A–δ
= (A–1)ρ (A–1)ρ (by the definition)
= (A–1)ρ+δ (by Part (a))
= A–(ρ+δ) (by the definition)
= A–ρ–δ
= Ar+s
Also
(Ar)s = (A–ρ)–δ
= [(A–1)ρ]–δ (by the definition)
= ([(A–1)ρ]–1)δ (by the definition)
= ([(A–1)–1)]ρ)δ (by Theorem 1.4.8b)
= ([A]ρ)δ (by Theorem 1.4.8a)
= Aρδ (by Part (a))
= A (–ρ)(–δ)
= Ars
29. (a) If AB = AC, then
A–1(AB) = A–1(AC)
or
(A–1 A)B = (A–1 A)C
or
B = C
(b) The matrix A in Example 3 is not invertible.
Exercise Set 1.4 51
30. Following the hint, let ij
= [A(BC)]ij
and rij
= [(AB)C]ij
and suppose that A is m × n, B is n × p, and C is p × q. Then
i j
= ai1 [BC]1j
+ ai2[BC]2j
+ … + ain
[BC]nj
= ai1
(b11c1j+ … + b1p
cpi
) + ai2(b
21c
1j+ … + b
2pcpj
) + … + ain
(bn1
c1j
+ … + bnp
cpj
)
= (ai1b11 + a
i2b21 + … + ain
bn1)c1j
+ … + (ai1b1p
+ ai2b2p
+ … + ain
bnp)cpj
= [AB]i1c1j
+ … + [AB]ip
cpj
= rij
For anyone familiar with sigma notation, this proof can be simplified as follows:
31. (a) Any pair of matrices that do not commute will work. For example, if we let
then
whereas
A AB B2 22
1 3
0 1+ + =
( )A B+ =
=
22
1 1
0 1
1 2
0 1
A B=
1 0
0 0==
0 1
0 1
l a BC a b cij is sj is st tjt
p
s
n
= ===∑∑
11ss
n
is sts
n
tj ita b c AB
=
=
∑
∑=
=
1
1
cc
r
tjt
p
t
p
ij
==∑∑
=
11
52 Exercise Set 1.4
(b) In general,
(A + B)2 = (A + B)(A + B) = A2 + AB + BA + B2
32. (a) As in Exercise 31, let A and B be any appropriate matrices which do not commute.
33. If
Thus, A2 = I if and only if a211 = a 2
22 = a233 = 1, or a11 = ±1, a22 = ±1, and a33 = ±1. There are
exactly eight possibilities:
34. (a) The logical contrapositive is “If A is invertible, then AT is invertible.”
(b) The statement is true by Theorem 1.4.10.
35. (b) The statement is true, since (A – B)2 = (–(B – A))2 = (B – A)2.
(c) The statement is true only if A–1 and B–1 exist, in which case
(AB–1)(BA–1) = A(B–1B)A–1 = AInA–1 = AA–1 = I
n
36. D = (CT)–1 B–1 BT C–1 (AT)–1 B–1
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
1 0
−
00
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
1
−
−−
− 00 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
−
−
−−
−1 0 0
0 1 0
0 0 1
11 0 0
0 1 0
0 0 1
−−
A
a
a
a
A
a
=
=11
22
33
20 0
0 0
0 0
then
1112
222
332
0 0
0 0
0 0
a
a
Exercise Set 1.4 53
EXERCISE SET 1.5
1. (a) The matrix may be obtained from I2 by adding –5 times Row 1 to Row 2. Thus, it iselementary.
(c) The matrix may be obtained from I2 by multiplying Row 2 of I2 by √3. Thus it iselementary.
(e) This is not an elementary matrix because it is not invertible.
(g) The matrix may be obtained from I4 only by performing two elementary row operationssuch as replacing Row 1 of I4 by Row 1 plus Row 4, and then multiplying Row 1 by 2.Thus it is not an elementary matrix.
3. (a) If we interchange Rows 1 and of A, then we obtain B. Therefore, E1 must be the matrixobtained from I3 by interchanging Rows 1 and 3 of I3, i.e.,
(c) If we multiply Row 1 of A by –2 and add it to Row 3, then we obtain C. Therefore, E3must be the matrix obtained from I3 by replacing its third row by –2 times Row 1 plusRow 3, i.e.,
5. (a) R1 ↔ R2 Row 1 and Row 2 are swapped
(b) R1 → 2R1
R2 → –3R2
(c) R2 → –2R1 + R2
E3
1 0 0
0 1 0
2 0 1
=−
E1
0 0 1
0 1 0
1 0 0
=
55
6. (a)
Therefore
(c)
Thus, the given matrix is not invertible because we have obtained a row of zeros on theleft side.
6 4 1 0
3 2 0 1
1 2 3 1 6 0
3 2 0 1
−−
−−
−
1 2 3 1 6 0
0 0 1 2 1
1 4
2 7
7 4
2 1
1
=
−−
−
1 4 1 0
2 7 0 1
1 4 1 0
0 1 3 1
1 0 7 4
0 1 2 1
− −
−−
56 Exercise Set 1.5
Add –2 times Row1 to Row 2.
Multiply Row 2 by –1;then multiply Row 2 by –4and add to Row 1.
Divide Row 1 by 6.
Multiply Row 1 by 3and add to Row 2.
7. (a)
Thus, the desired inverse is
3
2
11
10
6
51 1 1
1
2
7
10
2
5
− −
−
−
3 4 1 1 0 0
1 0 3 0 1 1
2 5 4 0 0 1
1 0 3 0 1 0
3 4 1
−
−
−− 11 0 0
2 5 4 0 0 1
1 0 3 0 1 0
0 4 10 1 3 0
0
−
− −55 10 0 2 1
1 0 3 0 1 0
0 4 10 1
− −
− −33 0
0 1 0 1 1 1
1 0 3 0 1 0
0 1
−
0 1 1 1
0 0 10 5 7 4
1 0 03
2
−− − −
−− −
−
−
11
10
6
50 1 0 1 1 1
0 0 11
2
7
10
2
5
Exercise Set 1.5 57
Interchange Rows1 and 2.
Add –3 times Row 1to Row 2 and –2 timesRow 1 to 3.
Add –4 times Row 3to Row 2 and inter-change Rows 2 and 3.
Multiply Row 3 by–1/10. Then add –3times Row 3 to Row 1.
Add –1 times Row 2to Row 3.
7. (c)
Thus
(e)
1 0 1
0 1 1
1 1 0
1
2
1
2
1
21
2
1
2
1
1
=
−
−
−
221
2
1
2
1
2−
1 0 1 1 0 0
0 1 1 0 1 0
1 1 0 0 0 1
1 0 1 1 0 0
0 1 1
0 1 0
0 1 1 1 0 1
1 0 1 1 0 0
0 1 1 0 1 0
0 0 11
2
1
2
1
− −
22
1 0 01
2
1
2
1
2
0 1 01
2
1
2
−
−1
2
1 1 01
2
1
2
1
2−
58 Exercise Set 1.5
Subtract Row 1from Row 3.
Subtract Row 2 fromRow 3 and multiplyRow 3 by -1/2.
Subtract Row 3 from Rows 1 and 2.
Thus
1 0 1
1 1 1
0 1 0
1
2
1
2
1
20 0
1
−
=
−−
11
1
2
1
2
1
2−
1 0 1 1 0 0
1 1 1 0 1 0
0 1 0 0 0 1
1 0 1 1
−
00 0
0 1 2 1 1 0
0 0 2 1 1 1
1 0 1 1 0 0
0
− − −
11 0 0 0 1
0 0 11
2
1
2
1
2
1 0 01
2
1
2
1
20
−
−
11 0 0 0 1
0 0 11
2
1
2
1
2−
Exercise Set 1.5 59
Add Row 1 to Row 2and subtract the newRow 2 from Row 3.
Add Row 3 to Row 2and then multiplyRow 3 by -1/2.
Subtract Row 3 from Row 1.
8. (a)
1
5
1
5
2
51 0 0
1
5
1
5
1
100 1 0
1
5
4
5
1
100 0 1
−
−
−
−
1 1 2 5 0 0
1 11
20 5 0
1 41
220 0 5
1 1 2 5 0 0
0 05
25
−
− 55 0
1 55
25 0 5
1 1 2 5 0 0
0
− −
−00 1 2 2 0
0 11
21 0 1
1 1 0
−
− −
11 4 0
0 0 1 2 2 0
0 1 0 0 1 1
1 0 0 1 3 1
−−
00 1 0 0 1 1
0 0 1 2 2 0
−−
60 Exercise Set 1.5
Multiply all threeRows by 5.
Subtract Row 3 from both Row 2 and Row 3.
Multiply Row 2 by 2/5and Row 3 by -1/5.
Add 2 times Row 2 to Row 1 and 1/2times Row 2 to Row 3.
Subtract Row 3 from Row 1 and interchangeRows 2 and 3.
Thus the desired inverse is
(b) Hint: First multiply Rows 1 and 2 by 1/√2.
(c)
1 0 0 0 1 0 0 0
1 3 0 0 0 1 0 0
1 3 5 0 0 0 1 0
1 3 5 7 0 0 0 1
−−−
1 0 0 0 1 0 0 0
0 3 0 0 1 1 0 0
0 3 5 0 1 0 1 0
0 3 5 7 1 0 0 1
−
1 0 0 0 1 0 0 0
0 1 0 01
3
1
30 0
0 0 5 0 0 1 1 0
0 0 5 7 0 1 0 1
1 0 0 0 1
−−
0 0 0
0 1 0 01
3
1
30 0
0 0 1 0 01
5
1
50
0 0 0 1
−
−
0 01
7
1
7−
1 3 1
0 1 1
2 2 0
−−
Exercise Set 1.5 61
Subtract Row 1 from Rows 2-4.
Subtract Row 2 fromRows 3 and 4 andmultiply Row 2 by 1/3.
Subtract Row 3 from Row 4 andmultiply Row 3 by 1/5and Row 4 by 1/7.
Thus the desired inverse is
8. (d) Since the third rowconsists entirely of zeros, this matrix is not invertible.
9. (b) Multiplying Row i of
by 1/ki
for i = 1, 2, 3, 4 and then reversing the order of the rows yields I4 on the leftand the desired inverse
on the right.
(c) To reduce
we multiply Row i by 1/k and then subtract Row i from Row (i + 1) for i = 1, 2, 3.Then multiply Row 4 by 1/k. This produces I4 on the left and the inverse,
k
k
k
k
0 0 0 1 0 0 0
1 0 0 0 1 0 0
0 1 0 0 0 1 0
0 0 1 0 0 0 1
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
4
3
2
1
k
k
k
k
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
1
2
3
4
k
k
k
k
1 0 0 0
1
3
1
30 0
01
5
1
50
0 01
7
1
7
−
−
−
62 Exercise Set 1.5
on the right.
10. (a) It takes 2 elementary rowoperations to convert A to I: first multiply Row 1 by 5 andadd to Row 2, then multiply Row 2 by 1/2. If we apply each of these operations to I, weobtain the 2 elementary matrices
From Theorem 1.5.1, we conclude that
E2E1A = I
(b) Since the inverse of a matrix is unique, the above equation guarantees that (with E1and E2 as defined above)
A–1 = E2E1
(c) From the equation E2E1A = I, we conclude that
E2–1E2E1A = E2
–1I
or
E1A = E2–1
or
E1–1E1A = E1
–1E2–1
or
A = E 1–1E 2
–1
E E121 0
5 1
1 0
0 1 2=
=
and
1 0 0 0
1 1 0 0
1 1 1 0
1 1 1 1
2
3 2
4 3 2
/ k
k k
k k k
k k k k
−
−
−
Exercise Set 1.5 63
so that
12.
13. (a) E3E2E1A =
(b) A = (E3E2E1)–1 = E–1
–1E2–1E3
–1
14. We first reduce A to row-echelon form, keeping track as we do so of the elementary rowoperations which are required. Here
1. If we interchange Rows 1 and 2, we obtain the matrix
1 3 3 8
0 1 7 8
2 5 1 8− − −
A =− − −
0 1 7 8
1 3 3 8
2 5 1 8
=−
1 0 2
0 1 0
0 0 1
1 0 0
0 1 3
0 0 1
1 0 0
0 4 0
0 0 1
1 0 0
0 1 4 0
0 0 1
1 0 0
0 1 3
0 0 1
−
−
1 0 2
0 1 0
0 0 1
1 0 2
0 4 3
0 0 1
= I3
3 2
3 1
1 0
1 1
1 2
0 1
3 0
0 1
−−
=
−
A =−
1 0
5 1
1 0
0 2
64 Exercise Set 1.5
2. If we add twice Row 1 to Row 3, we obtain the matrix
3. If we subtract Row 2 from Row 3, we obtain the matrix
Note that the above matrix R is in row-echelon form. Thus, Theorem 1.5.1 implies thatE3E2E1A = R, where
Thus, A = E1–1E2
–1E 3–1R. We need only let
and
G E= =
−3
11 0 0
0 1 0
0 1 1
F E=−
−2
11 0 0
0 1 0
2 0 1
E E1
11
0 1 0
1 0 0
0 0 1
=
−
E E1 2
0 1 0
1 0 0
0 0 1
1 0 0
0 1 0
2 0 1
,=
=
=−
, and E3
1 0 0
0 1 0
0 1 1
R =
1 3 3 8
0 1 7 8
0 0 0 0
1 3 3 8
0 1 7 8
0 1 7 8
Exercise Set 1.5 65
15. If A is an elementary matrix, then it can be obtained from the identity matrix I by a singleelementary rowoperation. If we start with I and multiply Row 3 by a nonzero constant, thena = b = 0. If we interchange Row 1 or Row 2 with Row 3, then c = 0. If we add a nonzeromultiple of Row 1 or Row 2 to Row 3, then either b = 0 or a = 0. Finally, if we operate onlyon the first two rows, then a = b = 0. Thus at least one entry in Row 3 must equal zero.
17. Every m × n matrix A can be transformed into reduced row-echelon form B by a sequenceof rowoperations. From Theorem 1.5.1,
B = EkE
k–1… E1A
where E1, E2, …, Ek
are the elementary matrices corresponding to the row operations. If wetake C = E
kE
k–1… E1, then C is invertible by Theorem 1.5.2 and the rule following Theorem
1.4.6.
18. The hypothesis that B is rowequivalent to A can be expressed as follows:
B = EkE
k–1… E1A
where E1, E2, …, Ek
are elementary matrices. Since A is invertible, then (by the rule followingTheorem 1.5.3) B is invertible and
B–1 = A–1E1–1E2
–1 … Ek–1
19. (a) First suppose that A and B are row equivalent. Then there are elementary matrices E1, . . ., E
psuch that A = E1
… EpB. There are also elementary matrices E
p+1, . . ., Ep+q
such that Ep+1
… Ep+q
A is in reduced row-echelon form. Therefore, the matrix Ep+1
…E
p+qE1
… EpB is also in (the same) reduced row-echelon form. Hence we have found,
via elementary matrices, a sequence of elementary row operations which will put B inthe same reduced row-echelon form as A.
Now suppose that A and B have the same reduced row-echelon form. Then there are elementary matrices E1, . . ., E
pand E
p+1, . . ., Ep+q
such that E1… E
pA =
Ep+1
… Ep+q
B. Since elementary matrices are invertible, this equation implies that A = E
p–1 … E–1
1Ep+1… E
p+qB. Since the inverse of an elementary matrix is also an
elementary matrix, we have that A and B are row equivalent.
20. There are kinds of elementary row operations, and therefore basic kinds of elementarymatrices. First, suppose that we multiply row k by the constant c ≠ 0. This results in the elementary matrix E with e
kk= c and all other entries identical to those of I
m. Now
consider the product EA. Since
[EA]ij
= ei1a1j
+ ei2a2j
+ … + eim
amj
66 Exercise Set 1.5
where eij
= 0 unless i = j, we have that [EA]ij
= aij
unless i = k, in which case [EA]kj
= cakj
.That is, EA is just A with the kth row multiplied by c.
Next, suppose that we interchange rows k and . This results in the elementary matrix E in which e
k = ek = 1, all other entries in rows k and are zero, and otherwise
E coincides with Im
. Thus [EA]ij
= aij
unless i = k or i = , in which case [EA]kj
= ajand
[EA]j = akj. That is, EA is just A with rows k and interchanged.
Finally, suppose that we multiply row k by c and add the result to row where k ≠ .This results in the elementary matrix E in which ek
= c and all other entries coincide withthose of I
m. Thus [EA]
ij= a
ijunless i = , in which case [EA]j
= cakj
+ aj, so that EA is just
A with c times row k added to row .
21. The matrix A, by hypothesis, can be reduced to the identity matrix via a sequence ofelementary row operations. We can therefore find elementary matrices E1, E2, . . . E
ksuch
that
Ek
… E2• E1
• A = In
Since every elementary matrix is invertible, it follows that
A = E1–1E2
–1 … Ek–1I
n
22. (a) False. A square matrix need not be invertible, but the product of elementary matricesis. Thus the zero matrix provides a counterexample.
(b) False. The product of two elementary matrices is a matrix which represents twosuccessive elementary row operations. These operations can usually not beaccomplished by a single elementary row operation. For instance, let
Then
which is clearly not an elementary matrix since it cannot be obtained from I2 via asingle elementary row operation.
(c) True. If A is invertible and we apply an elementary row operation to A, then we have,in effect, multiplied A by an elementary matrix E. The resulting matrix EA is invertibleas the product of invertible matrices.
E E1 22 0
0 2=
E E1 22 0
0 1
1 0
0 2=
=
and
Exercise Set 1.5 67
23. (a) True. Suppose we reduce A to its reduced row-echelon form via a sequence ofelementary row operations. The resulting matrix must have at least one row of zeros,since otherwise we would obtain the identity matrix and A would be invertible. Thusat least one of the variables in x must be arbitrary and the system of equations willhave infinitely many solutions.
(b) See Part (a).
(d) False. If B = EA for any elementary matrix E, then A = E–1B. Thus, if B wereinvertible, then A would also be invertible, contrary to hypothesis.
24. There is not. For instance, let b = 1 and a = c = d = 0. Then there is no matrix A which willsatisfy the equation.
68 Exercise Set 1.5
EXERCISE SET 1.6
1. This system of equations is of the form Ax = b, where
By Theorem 1.4.5,
Thus
That is,
x1 = 3 and x2 = –1
3. This system is of the form Ax = b, where
By direct computation we obtain
A
x
x
x
=
=
1 3 1
2 2 1
2 3 1
1
2
3
xx
= −
and bb
4
1
3
xx bb= =−
−
=
−−
A1 6 1
5 1
2
9
3
11
A− =
−−
1 6 1
5 1
Ax
x=
1 1
5 61
2xx = and b =
2
9
69
so that
That is,
x1 = –1, x2 = 4, and x3 = –7
5. The system is of the form Ax = b, where
By direct computation, we obtain
Thus,
That is, x1 = 1, x2 = 5, and x3 = –1.
xx bb= =−
−A
11
5
1
A− =
−
−
1 1
5
1 0 1
3 1 1
1 1 0
A
x
x
x
= −−
1 1 1
1 1 4
4 1 1
1
2
3
xx =
=
and bb
5
10
0
xx bb= =−
−
−A
11
4
7
A− =
−−
−
11 0 1
0 1 1
2 3 4
70 Exercise Set 1.6
7. The system is of the form Ax = b where
By Theorem 1.4.5, we have
Thus
That is,
x1 = 2b1 – 5b2 and x2 = –b1 + 3b2
9. The system is of the form Ax = b, where
We compute
so that
xx bb=
− ( ) + ( ) +
( ) −−A
b b b
b1
1 2 3
1
1 3 1 3
1 3 1 33
2 3 1 3
2
1 2 3
( )( ) + ( ) −
b
b b b
A− =
−−
−
11 3 1 3 1
1 3 1 3 0
2 3 1 3 1
A
x
x
x
= −
=1 2 1
1 1 1
1 1 0
1
2
3
xx
=
and bb
b
b
b
1
2
3
xx bb= =−
− +
−A
b b
b b
1 1 2
1 2
2 5
3
A− =
−−
1 2 5
1 3
Ax
x=
=
3 5
1 20 1
2and bb =
b
b
1
2
Exercise Set 1.6 71
9. (a) In this case, we let
Then
That is, x1 = 16/3, x2 = –4/3, and x3 = –11/3.
9. (c) In this case, we let
Then
That is, x1 = 3, x2 = 0, and x3 = –4.
10. We apply the method of Example 2 to the two systems (a) and (c) solved above. Thecoefficient matrix augmented by the two b matrices yields
1 2 1 1 1
1 1 1 3 1
1 1 0 4 3
− −− −
xx bb=−
−A
13
0
4
bb =−−
1
1
3
xx bb= −−
−A
116 3
4 3
11 3
bb =−
1
3
4
72 Exercise Set 1.6
The reduced row-echelon form may be obtained as follows:
This, fortunately, yields the same results as in Exercise 9.
11. The coefficient matrix, augmented by the two b matrices, yields
This reduces to
or
Thus the solution to Part (a) is x1 = 22/17, x2 = 1/17, and to Part (b) is x1 = 21/17, x2 = 11/17.
1 0
0 1
22 17 21 17
1 17 11 17
1 5 1 2
0 17 1 11
− −
1 5 1 2
3 2 4 5
− −
1 2 1 1 1
0 3 0 4 0
0 2 1 1 4
1 0 2 2
− −−
−
− −55
0 1 0 4 3 0
0 0 1 2 11 6 2
1 0 0 16 3 3
0
−−
/
11 0 4 3 0
0 0 1 11 3 4
−− −
Exercise Set 1.6 73
Add –3 times Row 1to Row 2.
Add –1 times Row 2 toRow 3 and –1 times Row 1to Row 2.
Add –1 times Row 3 toRow 2, divide Row 2 by–3 and ROw 3 by 2, add–1 times Row 2 to Row 3
Multiplt Row 3 by –2 andthen add –2 times Row 3to Row 1.
Divide Row 2 by 17 and add5 times Row 2 to Row 1.
15. As above, we set up the matrix
This reduces to
or
or
Thus if we let x3 = t, we have for Part (a) x1 = –12 – 3t and x2 = –5 – t, while for Part (b) x1 = 7 – 3t and x2 = 3 – t.
16. The augmented matrix for this sytem is
which, when reduced to row-echelon form, becomes
1 2 3 3
0 0 22
1 2
−−
b
b b
6 4
3 21
2
−−
b
b
1 0 3 12 7
0 1 1 5 3
0 0 0 0 0
−−
1 2 1 2 1
0 1 1 5 3
0 0 0 0 0
− −−
1 2 1 2 1
0 1 1 5 3
0 1 1 5 3
− −− − −− − −
1 2 1 2 1
2 5 1 1 1
3 7 2 1 0
− −− −− −
74 Exercise Set 1.6
Add twice Row 2to Row 3.
Add appropriatemultiples of Row 1to Rows 2 and 3.
Add –1 times Row 2 toRow 3 and multiplyRow 2 by –1.
The system is consistent if and only if b1 – 2b2 = 0 or b1 = 2b2. Thus Ax = b is consistent ifand only if b has the form
17. The augmented matrix for this system of equations is
If we reduce this matrix to row-echelon form, we obtain
The third row implies that b3 = b1 – b2. Thus, Ax = b is consistent if and only if b has theform
20. (a) In general, Ax = x if and only if Ax – x = 0 if and only if (A – I)x = 0. In this case,
and hence,
A = −
2 1 2
2 2 2
3 1 1
bb =−
b
b
b b
1
2
1 2
1 2 5
0 1 41
34
0 0 0
1
2 1
1 2 3
−
− −( )− + +
b
b b
b b b
1 2 5
4 5 8
3 3 3
1
2
3
−−
− −
b
b
b
bb =
2 2
2
b
b
Exercise Set 1.6 75
The reduced row-echelon form of A–I is
Therefore, the system of equations (A – I)x = 0 has only the trivial solution x1 = x2 = x3 = 0; that is, x = 0 is the only solution to Ax = x.
22. (a) It is clear that x = 0 is the only solution to the equation Ax = 0. We may then invoke Theorem 1.6.4 to conclude that A is invertible.
23. Since Ax = has only x = 0 as a solution, Theorem 1.6.4 guarantees that A is invertible. ByTheorem 1.4.8 (b), Ak is also invertible. In fact,
(Ak)–1 = (A–1)k
Since the proof of Theorem 1.4.8 (b) was omitted, we note that
Because Ak is invertible, Theorem 1.6.4 allows us to conclude that Akx = 0 has only thetrivial solution.
24. First, assume that Ax = 0 has only the trivial solution. It then follows from Theorem 1.6.4that A is invertible. Now Q is invertible by assumption. Thus, QA is invertible because theproduct of two invertible matrices is invertible. If we apply Theorem 1.6.4 again, we see that(QA)x = 0 has only the trivial solution.
Conversely, assume that (QA)x = 0 has only the trivial solution; thus, QA is invertibleby Theorem 1.6.4. But if QA is invertible, then so is A because A = Q–1(QA) is the productof invertible matrices. But if A is invertible, then Ax = 0 has only the trivial solution.
A A A A A A
k k
− − − =1 1 1
factors factors
II
A I− =
1 0 0
0 1 0
0 0 1
A I− =
1 0 0
0 1 0
0 0 1
76 Exercise Set 1.6
25. Suppose that x1 is a fixed matrix which satisfies the equation Ax1 = b. Further, let x be anymatrix whatsoever which satisfies the equation Ax = b. We must then show that there is amatrix x0 which satisfies both of the equations x = x1 + x0 and Ax0 = 0.
Clearly, the first equation implies that
x0 = x – x1
This candidate for x0 will satisfy the second equation because
Ax0 = A(x – x1) = Ax – Ax1 = b – b = 0
We must also show that if both Ax1 = b and Ax0 = 0, then A(x1 + x0) = b. But
A(x1 + x0) = Ax1 + Ax0 = b + 0 = b
26. We are given that AB = I. By Part (a) of Theorem 1.6.3, A must be the inverse of B. But then
A = B–1 ⇒ A–1 = (B–1)–1 = B
That is, A–1 = B.
27. (a) x ≠ 0 and x ≠ y
(b) x ≠ 0 and y ≠ 0
(c) x ≠ y and x ≠ –y
Gaussian elimination has to be performed on (A I) to find A–1. Then the productA–1B is performed, to find x. Instead, use Gaussian elimination on (A B) to find x. Thereare fewer steps in the Gaussian elimination, since (A B) is a m × (n+1) matrix in general,or n × (n+1) where A is square (n × n). Compare this with (A I) which is n × (2n) in theinversion approach. Also, the inversion approach only works for A n × n and invertible.
28. (a) For the equation x = Ax + b to have a unique solution for x, the equation (I – A)x = bmust also have a unique solution. We are assured of this if the matrix I – A is invertible.
29. No. The system of equations Ax = x is equivalent to the system (A – I)x = 0. For thissystem to have a unique solution, A – I must be invertible. If, for instance, A = I, then anyvector x will be a solution to the system of equations Ax = x.
Note that if x ≠ 0 is a solution to the equation Ax = x, then so is kx for any real number
Exercise Set 1.6 77
k. A unique solution can only exist if A – I is invertible, in which case, x = 0.
30. Yes. The matrices A and B need not be square. For example, if
then AB = I2 but
In fact, only square matrices can be invertible.
31. Let A and B be square matrices of the same size. If either A or B is singular, then AB issingular.
BA =
1 0
0 0 0
0 0 1
0
A =
=
1 0 0
0 0 1
1 0
0 0
0 1
and B
78 Exercise Set 1.6
EXERCISE SET 1.7
6. For A to be symmetric, the following equations must hold:
a – 2b + 2c = 3
2a + b + c = 0
a + c = –2
The solution to this system of equations is a = 11, b = –9, c = –13. This is the only set ofvalues for which A is symmetric.
7. The matrix A fails to be invertible if and only if a + b – 1 = 0 and the matrix B fails to beinvertible if and only if 2a – 3b – 7 = 0. For both of these conditions to hold, we must havea = 2 and b = –1.
8. (a) Since the product is not symmetric, the matrices do not commute.
(b) Since the product is symmetric, the matrices commute.
9. We know that A and B will commute if and only if
is symmetric. So 2b + d = a – 5b, from which it follows that a – d = 7b.
ABa b
b d
a b b d
a b b=
−
=
+ +−
2 1
1 5
2 2
5 −−
5d
79
10. (b) If
so that
Thus there are 8 possible matrices A which satisfy the given equation.
11. (b) Clearly
for any real number k = 0.
13. We verify the result for the matrix A by finding its inverse.
−
−
− − −
1 2 5 1 0 0
0 1 3 0 1 0
0 0 4 0 0 1
1 2 5 1 0 0
0 1 3 0 11 0
0 0 1 0 0 1 4−
A
ka ka ka
ka ka ka
ka ka ka
=
11 12 23
21 22 23
31 32 33
3 0 0
0 5 0
0 0 7
k
k
k
A =
±
±
±
1
30 0
01
20
0 0 1
A A− −=
2 19 0 0
0 4 0
0 0 1
then =±
±±
3 0 0
0 2 0
0 0 1
80 Exercise Set 1.7
Multiply Row 1 by –1and Row 3 by –1/4.
Thus A–1 is indeed upper triangular.
14. (a) Since A–1 = , the inverse is indeed symmetric.
15. (a) If A is symmetric, then AT = A. Then (A2)T = (AA)T = ATAT = A . A = A2, so A2
is symmetric.
(b) We have from part (a) that
(2A2 – 3A + I)T = 2(A2)T + 3AT + IT = 2A2 + 3A + I
16. (a) The statement is true for k = 1, since A is symmetric. It is true for k = 0, since A0 = Iis symmetric. Assume AK is symmetric. Show that AK+1 is also symmetric.
We have (AK+1)T = (AK • A)T = AT • (AK)T = A • AK = AK+1. So AK+1 is symmetric.
By induction, AK is symmetric for all K = 0,1,2,3, . . ., if A is symmetric.
(b) Yes—this follows from part (a), with use of the rules (A + B)T = AT + BT, and (kA)T = KAT. So, let p(x) = a0x
n + a, xn + … + an–1x + a
nbe any real polynomial in
x. Then P(A) = a0An + a, An–1 + … + a
n–1A + anI. Then [P(A)]T = (a0A
n + a, An–1 + …+ a
n–1 A + anI)T
= a0(An)T + a,(An–1)T … + an–1A
T + anIT
= a0 An + a, An–1 + … + an–1A + a
nI = P(A).
So P(A) is symmetric.
3 5 1 5
1 5 2 5
1 0 1 1 2 0
0 1 0 0 1 3 4
0 0 1 0 0 1 4
1 0 0 1 2 1 4
0 1 0
−
−
−00 1 3 4
0 0 1 0 0 1 4−
Exercise Set 1.7 81
Add –1 times Row3 to Row 1.
Add 2 times Row 2 toRow 1 and –3 timesRow 3 to Row 2.
17. From Theorem 1.7.1(b), we have if A is an n × n upper triangular matrix, so is A2. Byinduction, if A is an n × n upper triangular matrix, so is Ak, k = 1, 2, 3, . . . We note that theidentity matrix I
n= A0 is also upper triangular. Next, if A is n × n upper triangular, and k
is any (real) scalar, then KA is upper triangular. Also, if A and B are n × n upper triangularmatrices, then so is A+B. These facts allow us to conclude if p(x) is any (real) polynomial,and A is n × n upper triangular, then p(A) is an n × n upper triangular matrix.
18. If A = AT A, then AT = (AT A)T = AT (AT)T = AT A = A, so A is symmetric. This implies thatA = AT, so A = A2.
19. Let
Then if A2 – 3A – 4I = O, we have
This leads to the system of equations
x2 – 3x – 4 = 0
y2 – 3y – 4 = 0
z2 – 3z – 4 = 0
which has the solutions x = 4, –1, y = 4, –1, z = 4, –1. Hence, there are 8 possible choicesfor x, y, and z, respectively, namely (4, 4, 4), (4, 4, –1), (4, –1, 4), (4, –1, –1), (–1, 4, 4),(–1, 4, –1), (–1, –1, 4), and (–1, –1, –1).
20. (a) Since aij
= i2 + j2 = j2 + i2 = aji, A is symmetric.
(d) Since 2i2 + 2j3 need not equal 2j2 + 2i3, A is not symmetric. For instance a21 ≠ a12.In fact, A will be symmetric only in the trivial case where n = 1.
x
y
z
x
y
z
2
2
2
0 0
0 0
0 0
3
0 0
0 0
0 0
−
−
=4
1 0 0
0 1 0
0 0 1
O
A
x
y
z
=
0 0
0 0
0 0
82 Exercise Set 1.7
22. (a) If A is invertible and skew-symmetric, then by Theorem 1.4.10,
(A–1)T = (AT)–1 = (–A)–1 = –(A–1)
so A–1 is skew-symmetric.
(c) From the hint,
A = 12(A + AT) + 12(A – AT)
so we need only prove that 12 (A + AT) is symmetric and that 1
2 (A – AT) is skew-symmetric. To this end, note that
12(A + AT)T = 12(AT + (AT)T) = 12(A + AT)
and that
12(A – AT)T = 12(AT – (AT)T) = 12(AT – AT – A) = 12(A – AT)
23. The matrix
is skew-symmetric but
is not skew-symmetric. Therefore, the result does not hold.
In general, suppose that A and B are commuting skew-symmetric matrices. Then (AB)T = (BA)T = AT BT = (–A)(–B) = AB, so that AB is symmetric rather than skew-symmetric. [We note that if A and B are skew-symmetric and their product is symmetric,then AB = (AB)T = BT AT = (–B)(–A) = BA, so the matrices commute and thus skew-symmetric matrices, too, commute if and only if their product is symmetric.]
24. (a) Here
AA A= =−
−
2 1 0
0 1
A =−
0 1
1 0
Exercise Set 1.7 83
has the solution y1 = 1, y2 = 0, y3 = –2. Thus
so that x1 = 7/4, x2 = 1, x3 = –1/2.
25. Let
Then
Hence, x3 = 1 which implies that x = 1, and z3 = –8 which implies that z = –2. Therefore,3y = 30 and thus y = 10.
26. An n × n matrix has n2 entries, n of them along the diagonal and (n2 – n)/2 of them abovethe diagonal. Thus we could assign distinct values to n + (n2 – n)/2 = (n2 + n)/2 entries.The remaining (n2 – n)/2 entries below the diagonal would coincide with the correspondingentries above the diagonal.
27. To multiply two diagonal matrices, multiply their corresponding diagonal elements to obtaina new diagonal matrix. Thus, if D1 and D2 are diagonal matrices with diagonal elements d1, . . ., d
nand e1, . . ., e
nrespectively, then D1D2 is a diagonal matrix with diagonal elements
d1e1, . . ., dne
n. The proof follows directly from the definition of matrix multiplication.
Ax y x xz z
z
33 2 2
30
1 30
0 8=
+ +( )
=−
Ax y
z=
0
U
x
x
x
xx
2 1 3
0 1 2
0 0 4
1
2
3
−
=11
0
2−
= yy
L
y
y
y
yy
1 0 0
2 3 0
2 4 1
11
2
3
−
= −22
0
= b
84 Exercise Set 1.7
28. Suppose that D has diagonal elements d1, . . ., dn. Since A is a square matrix and AD = I,
Theorem 1.6.3 guarantees that A = D–1. Thus, A is a diagonal matrix with diagonal elements1/d1, . . ., 1/d
n. Note that since D is invertible, it follows that all of its diagonal elements
must be nonzero.
29. In general, let A = [aij]n×n
denote a lower triangular matrix with no zeros on or below thediagonal and let Ax = b denote the system of equations where b = [b1, b2, . . ., b
n]T. Since
A is lower triangular, the first row of A yields the equation a11x1 = b1. Since a11 ≠ 0, we cansolve for x1. Next, the second row of A yields the equation a21x1 + a22x2 = b2. Since weknow x1 and since a22 ≠ 0, we can solve for x2. Continuing in this way, we can solve forsuccessive values of x
iby back substituting all of the previously found values x1, x2, . . ., x
i–1.
30. (a) False. If A were invertible, then AT would also be invertible, and hence so would AAT.
(c) True. Use Theorems 1.4.10 and 1.5.3.
(d) False. Let A = . Then if A2 is symmetric, we must have ab + bd = ac + cd
so that either a = –d or b = c. If b = c, then A is symmetric. However, if b ≠ c but
a = –d, then A2 is symmetric, but A is not. For instance, let a = .1 0
1 1−
a b
c d
Exercise Set 1.7 85
SUPPLEMENTARY EXERCISES 1
1.
3
5
4
54
5
3
5
14
3
5
34
5
3
5
−
−
x
y
x
y
−
− +
−
14
3
5
3
05
3
4
3
1
x
x y
44
3
5
3
0 14
5
3
5
1 03
5
4
5
0 14
5
x
x y
x y
− +
+
− xx y+
3
5
87
Multiply Row 1 by 5/3.
Multiply Row 2 by 3/5.
Add –4/5 times Row 1to Row 2.
Add –4/3 times Row 2to Row 1.
Thus,x′ = 3–
5x + 4–
5y
y′ = – 4–5
x + 3–5
y
3. We denote the system of equations by
a11x1 + a12x2 + a13x3 + a14x4 = 0
a21x1 + a22x2 + a23x3 + a24x4 = 0
If we substitute both sets of values for x1, x2, x3, and x4 into the first equation, we obtain
a11 – a12 + a13 + 2a14 = 0
2a11 + 3a13 – 2a14 = 0
where a11, a12, a13, and a14 are variables. If we substitute both sets of values for x1, x2, x3,and x4 into the second equation, we obtain
a21 – a22 + a23 + 2a24 = 0
2a21 + 3a23 – a24 = 0
where a21, a22, a23, and a24 are again variables. The two systems above both yield the matrix
which reduces to
This implies that
a11 = –(3/2)a13 + (1/2)a14
a12 = –(1/2)a13 + (5/2)a14
1 0 3 2 1 2 0
0 1 1 2 5 2 0
−−
1 1 1 2 0
2 0 3 1 0
−−
88 Supplemental Exercises 1
and similarly,
a21
= (–3/2)a23
+ (1/2)a24
a22 = (–1/2)a23 + (5/2)a24
As long as our choice of values for the numbers aij
is consistent with the above, then the system will have a solution. For simplicity, and to insure that neither equation is amultiple of the other, we let a13 = a14 = –1 and a23 = 0, a24 = 2. This means that a11 = 1, a12 = –2, a21 = 1, and a22 = 5, so that the system becomes
x1 – 2x2 – x3 – x4 = 0
x1 + 5x2 + 2x4 = 0
Of course, this is just one of infinitely many possibilities.
4. Suppose that the box contains x pennies, y nickels, and z dimes. We know that
x + 5y + 10z = 83
x + y + z = 13
The augmented matrix of this system is
which can be reduced to the matrix
by Gauss-Jordan elimination. Thus,
x = 5–4
z – 9–2
z and y = 35—2
Since x, y, and z must all be integers between 0 and 13, then z must be either 2, 6, or 10because those are the only values of z for which x and y are both integers. However, z = 2implies that x < 0 while z = 10 implies that y < 0. Therefore, the only solution is x = 3, y = 4, and z = 6.
1 05
4
9
2
0 19
4
35
2
− −
1 5 10 83
1 1 1 13
Supplemental Exercises 1 89
5. As in Exercise 4, we reduce the system to the equations
Since x, y, and z must all be positive integers, we have z > 0 and 35 – 9z > 0 or 4 > z. Thuswe need only check the three values z = 1, 2, 3 to see whether or not they produce integersolutions for x and y. This yields the unique solution x = 4, y = 2, z = 3.
6. The augmented matrix of this system is
which can be reduced to the matrix
Row 3 of the above matrix implies that the system of equations will have a solution if andonly if 2a2 – a – 6 = 0; that is, if and only if a = 2 or a = –3/2. For either of those values ofa, the solution is x1 = 2 – t, x2 = t, and x3 = 2. Hence, the system has infinitely manysolutions provided a = 2 or a = –3/2. For all other values of a, there will be no solution.
8. Let s = xy, t = √y, and u = zy. Then we have a system of linear equations in s, t, and uwith augmented matrix
1 2 3 8
2 3 2 7
1 1 2 4
−−
−
1 1 0 2
0 0 1 2
0 0 0 2 62a a− −
1 1 1 4
0 0 1 2
0 0 4 22a a− −
x
yz–
=+
=
1 5
4
35 9
4
z
90 Supplemental Exercises 1
This reduces to
so thatxy = 5
√y = 3
zy = 3
Thus from the second equation, y = 9, and from the first and third equations, x = 5/9 andz = 3/9 = 1/3.
9. Note that K must be a 2 × 2 matrix. Let
Then
or
or
2 8 4 4
4 6 2 3 2 3
2 4 2
a c b d b d
a c b d b d
a c b d b
+ + − −− + − + −
− − − + 22
8 6 6
6 1 1
4 0 0d
=−
−−
1 4
2 3
1 2
2
2
8 6 6
6−−
−−
=
−a b b
c d d−−
−
1 1
4 0 0
1 4
2 3
1 2
2 0 0
0 1 1−
−
−
a b
c d==
−−
−
8 6 6
6 1 1
4 0 0
Ka b
c d=
1 0 0 5
0 1 0 3
0 0 1 3
Supplemental Exercises 1 91
Thus2a + 8c = 8
b + 4d = 6
– 4a + 6c = 6
– 2b + 3d = –1
2a – 4c = –4
b – 2d = 0
Note that we have omitted the 3 equations obtained by equating elements of the lastcolumns of these matrices because the information so obtained would be just a repeat ofthat gained by equating elements of the second columns. The augmented matrix of theabove system is
The reduced row-echelon form of this matrix is
Thus a = 0, b = 2, c = 1, and d = 1.
1 0 0 0 0
0 1 0 0 2
0 0 1 0 1
0 0 0 1 1
0 0 0 0 0
0 0 0 0 0
2 0 8 0 8
0 1 0 4 6
4 0 6 0 6
0 2 0 3 1
2 0 4 0 4
0 1 0 2 0
−− −
− −−
92 Supplemental Exercises 1
10. If we substitute the given values for x, y, and z into the system of equations, we obtain
a – b – 6 = –3
–2 + b + 2c = –1
a – 3 – 2c = –3
or
a – b = 3
b + 2c = 1
a – 2c = 0
This system of equations in a, b, and c yields the matrix
which reduces to
Hence, a = 2, b = –1, and c = 1.
11. The matrix X in Part (a) must be 2 × 3 for the operations to make sense. The matrices inParts (b) and (c) must be 2 × 2.
(b) Let X = . Then
If we equate matrix entries, this gives us the equations
Xx y x x y
z w z z w
1 1 2
3 0 1
3 2
3 2
−
=
+ − ++ − +
x y
z w
1 0 0 2
0 1 0 1
0 0 1 1
−
1 1 0 3
0 1 2 1
1 0 2 0
−
−
Supplemental Exercises 1 93
x + 3y = –5 x + 3w = 6
– x = –1 – z = –3
2x + y = 0 2z + w = 7
Thus x = 1 and z = 3, so that the top two equations give y = –2 and w = 1. Sincethese values are consistent with the bottom two equations, we have that
11. (c) As above, let X = , so that the matrix equation becomes
This yields the system of equations
2x – 2y + z = 2
–4x + 3y + w = –2
–x + z – 2w = 5
–y – 4z + 2w = 4
with matrix
which reduces to
Hence, x = –113/37, y = –160/37, z = –20/37, and w = –46/37.
1 0 0 0 113 37
0 1 0 0 160 37
0 0 1 0 20 37
0 0 0 1 46 37
−−
−−
2 2 1 0 2
4 3 0 1 2
1 0 1 2 5
0 1 4 2 4
−− −− −
− −
3 3
2 2
2 4
2 4
x z y w
x z y w
x y x
z w z
+ +
− + − +
−+
+
+−
2 2
5 4
x y
z w
X =−
1 2
3 1
94 Supplemental Exercises 1
12. (a) By inspection
and
Thus,
14. (a) If A4 = 0, then
(I – A)(I + A + A2 + A3) = I + A + A2 + A3 – A – A2 – A3 – 0 = I
Therefore, (I – A)–1 = I + A + A2 + A3 by Theorem 1.6.3.
(b) If An+1 = 0, then
(I – A)(I + A + A2 + … + An) = I – An + 1 = I + 0 = I
Thus Theorem 1.6.3 again supplies the desired result.
15. Since the coordinates of the given points must satisfy the polynomial, we have
p(1) = 2 ⇒ a + b + c = 2
p(–1) = 6 ⇒ a – b + c = 6
p(2) = 3 ⇒ 4a + 2b + c = 3
=− −
−
1 7 11
14 10 26
1
2
3
x
x
x
z
z
1
2
4 1 1
3 5 1
1 1 1
3 1 4
2 2 3
=
−− −
−−
− −
x
x
x
1
2
3
z
z
y
y
y
1
2
1
2
3
4 1 1
3 5 1
=
−− −
y
y
y
1
2
3
1 1 1
3 1 4
2 2 3
=−
−− −
xx
x
x
1
2
3
Supplemental Exercises 1 95
The reduced row-echelon form of the augmented matrix of this system of equations is
Thus, a = 1, b = – 2, and c = 3.
16. If p(x) = ax2 + bx + c, then p′(x) = 2ax + b. Hence,
p(–1) = 0 ⇒ a – b + c = 0
p(2) = –9 ⇒ 4a + 2b + c = –9
p′(2) = 0 ⇒ 4a + b = 0
The augmented matrix of this system reduces to
so that a = 1, b = –4, and c = –5.
17. We must show that (I – Jn) (I –
n–11 J
n) = I or that (I –
n–11 J
n) (I – J
n) = I. (By virtue of
Theorem 1.6.3, we need only demonstrate one of these equalities.) We have
But Jn2 = nJ
n(think about actually squaring J
n), so that the right-hand side of the above
equation is just I, as desired.
18. If A3 + 4A2 – 2A + 7I = 0, then (A3 + 4A2 – 2A + 7I)T = 0 as well. Now repeated applicationsof Theorem 1.4.9 will finish the problem.
I J In
J In
IJ J In n n n−( ) −−
= −−
− +1
1
1
1
12
nnJ
In
nJ
nJ
n
n n
−
= −−
+−
1
1
1
1
2
2
1 0 0 1
0 1 0 4
0 0 1 5
−−
1 0 0 1
0 1 0 2
0 0 1 3
−
96 Supplemental Exercises 1
19. First suppose that AB–1 = B–1 A. Note that all matrices must be square and of the samesize. Therefore
(AB–1)B = (B–1 A)B
or
A = B–1 AB
so that
BA = B(B–1 AB) = (BB–1)(AB) = AB
It remains to show that if AB = BA then AB–1 = B–1 A. An argument similar to the one givenabove will serve, and we leave the details to you.
20. Suppose that A and A + B are both invertible. Then there is a matrix C such that (A + B)C = I. We must show that I + BA–1 is also invertible. Consider the matrix
(I + BA–1)(AC) = AC + (BA–1)(AC)
= AC + BIC
= AC + BC
= (A + B)C
= I
Thus AC is the inverse of I + BA–1.
Now suppose that A is invertible but A + B is not. We must show that (I + BA–1) is also not invertible. Suppose that (I + BA–1)D = I for some matrix D. Then
(A + B)A–1 D = AA–1 D + BA–1 D
= D + BA–1 D
= (I + BA–1)D
= I
This implies that A + B is invertible. Since this is contrary to our hypothesis, I + BA–1 is notinvertible.
Supplemental Exercises 1 97
21. (b) Let the ijth entry of A be aij. Then tr(A) = a11 + a22 + … + a
nn, so that
tr(kA) = ka11 + ka22 + … + kann
= k (a11 + a22 + … + ann
)
= ktr(A)
(d) Let the ijth entries of A and B be aij
and bij, respectively. Then
tr(AB) = a11b11 + a12b21 + … + a1nb
n1
+ a21b12 + a22b22 + … + a2nb
n2
+ …
+ an1b1n
+ an2b2n
+ … + ann
bnn
and
tr(BA) = b11a11 + b12a21 + … + b1na
n1
+ b21a12 + b22a22 + … + b2na
n2
+ …
+ bn1a1n
+ bn2a2n
+ … + bnn
ann
If we rewrite each of the terms bija
jiin the above expression as a
jib
ijand list the terms
in the order indicated by the arrows below,
tr(BA) = a11b11 + a21b12 + … + an1b1n
+ a12b21 + a22b22 + … + an2b2n
+ …
+ a1nb
n1 + a2nb
n2 + … + ann
bnn
then we have tr(AB) = tr(BA).
22. Suppose that A and B are square matrices such that AB – BA = I. Then
tr(AB – BA) = tr(AB) – tr(BA) = tr(I).
But tr(I) = n for any n × n identity matrix I and tr(AB) – tr(BA) = 0 by Part (d) of Problem 21.
98 Supplemental Exercises 1
24. (c) Let A = [aij(x)] be an m × n matrix and let B = [b
ij(x)] be an n × s matrix. Then the
ijth entry of AB is
ai1(x)b1j
(x) + … + ain
(x)bnj
(x)
Thus the ijth entry of d—dx
(AB) is
d—dx
(ai1(x)b1 j
(x) + … + ain
(x)bnj
(x))
= a′i1(x)b1j
(x) + ai1(x)b′1 j
(x) + … + a′in
(x)bnj
(x) + ain
(x)b′nj
(x)
= [a′i1(x)b1j
(x) + … + a′in
(x)bnj
(x)] + [ai1(x)b′1j(x) + … + a
in(x)b′nj(x)]
This is just the ijth entry of dA—dx
B + AdB—dx
.
25. Suppose that A is a square matrix whose entries are differentiable functions of x. Supposealso that A has an inverse, A–1. Then we shall show that A–1 also has entries which aredifferentiable functions of x and that
dA–1
——dx
= –A–1 dA—dx
A–1
Since we can find A–1 by the method used in Chapter 1, its entries are functions of x whichare obtained from the entries of A by using only addition together with multiplication anddivision by constants or entries of A. Since sums, products, and quotients of differentiablefunctions are differentiable wherever they are defined, the resulting entries in the inversewill be differentiable functions except, perhaps, for values of x where their denominatorsare zero. (Note that we never have to divide by a function which is identically zero.) Thatis, the entries of A–1 are differentiable wherever they are defined. But since we are assumingthat A–1 is defined, its entries must be differentiable. Moreover,
or
dA
dxA A
dA
dx
−−
+ =11
0
d
dxAA
d
dxI( ) ( )− = =1
0
Supplemental Exercises 1 99
Therefore
so that
26. Following the hint, we obtain the equation
x2 + x – 2 = A(x2 + 1) + (Bx + C)(3x – 1)
or
x2 + x – 2 = (A + 3B)x2 + (3C – B)x + (A – C)
If we equate coefficients of like powers of x, we obtain the system of equations
A + 3B = 1
– B + 3C = 1
A – C = –2
the solution to this system is
A = –7/5, B = 4/5, C = 3/5
27. (b) Let H be a Householder matrix, so that H = I – 2PPT where P is an n × 1 matrix. Thenusing Theorem 1.4.9,
HT = (I – 2PPT)T
= IT – (2PPT)T
= I – 2(PT)T PT
= I – 2 PPT
= H
––
– –dA
dxA
dA
dxA
11 1=
––A
dA
dx
dA
dxA
11=
100 Supplemental Exercises 1
and (using Theorem 1.4.1)
HT H = H2 (by the above result)
= (I – 2PPT)2
= I2 – 2PPT – 2PPT + (–2PPT )2
= I – 4PPT + 4PPT PPT
= I – 4PPT + 4PPT (because PT P = I)
= I
28. (a) We are asked to show that (C–1 + D–1)–1 = C(C + D)–1D. Since all the matrices in sightare invertible and the inverse of a matrix is unique, it will suffice to show that theinverses of each side are equal, or that (C(C + D)–1D)–1 = C–1 + D–1. We have
(C(C + D)–1 D)–1 = D–1(C + D)C–1
= D–1CC–1 + D–1 DC–1
= D–1 + C–1
= C–1 + D–1
(b) Since we don’t know whether or not C is invertible, we can’t use the technique of Part (a). Instead, we observe that
(*) (I + CD)C = C(I + DC)
since each side of (*) is just C + CDC. If we multiply both sides of (*) on the left by(I + CD)–1 and on the right by (I + DC)–1, we obtain the desired result.
(c) Start with D + DDT C–1 D, write it in two different ways, and proceed more or less asin Part (b).
29. (b) A bit of experimenting and an application of Part (a) indicates that
An
n
n
n
a
b
d c
=
0 0
0 0
0
Supplemental Exercises 1 101
where
d = an–1 + an–2 c + … + acn–2 + cn–1 = if a ≠ c
If a = c, then d = nan–1. We prove this by induction. Observe that the result holdswhen n = 1. Suppose that it holds when n = N. Then
Here
Thus the result holds when n = N + 1 and so must hold for all values of n.
a cda c
a c
a c
a a c a c c
a cN
NN N N N N N
++
−−
=− + −
−=
+ +
=
1 1aa c
a ca c
a a Na N a a
N N
N N N
+ +
−
−−
≠
+ ( ) = +( ) =
1 1
1 1
if
if cc
A AA A
a
b
d c
a
N N
N
N
N
+ = =
=1
0 0
0 0
0
NN
N
N cd N
b
a c
+
+
+ +
1
1
1
0 0
0 0
0
a c
a c
n n–
–
102 Supplemental Exercises 1
EXERCISE SET 2.1
1. (a) The number of inversions in (4,1,3,5,2) is 3 + 0 + 1 + 1 = 5.
(d) The number of inversions in (5,4,3,2,1) is 4 + 3 + 2 + 1 = 10.
2. (a) The permutation is odd because 5 is odd.
(d) The permutation is even because 10 is even.
3.
5.
7.
9.
−− = − − + − + + = −
2 1 4
3 5 7
1 6 2
20 7 72 20 84 6 65( ) ( )
a
aa a a a
−− −
= − − − − = − +3 5
3 23 2 3 5 5 212( )( ) ( )( )
−− −
= − − − − =5 6
7 25 2 7 6 52( )( ) ( )( )
3 5
2 412 10 22
−= − − =( )
103
11.
13. (a)
= λ2 + 2λ – 3 = (λ – 1)(λ + 3)
Hence,det(A) = 0 if and only if λ = 1 or λ = –3.
14. The following permutations are even:
(1,2,3,4) (1,3,4,2) (1,4,2,3)
(2,1,4,3) (2,3,1,4) (2,4,3,1)
(3,1,2,4) (3,2,4,1) (3,4,1,2)
(4,1,3,2) (4,2,1,3) (4,3,2,1)
The following permutations are odd:
(1,2,4,3) (1,3,2,4) (1,4,3,2)
(2,1,3,4) (2,3,4,1) (2,4,1,3)
(3,1,4,2) (3,2,1,4) (3,4,2,1)
(4,1,2,3) (4,2,3,1) (4,3,1,2)
15. If A is a 4 × 4 matrix, then
det(A) = ∑(–1)p a1i1a2i2
a3i3a4i4
where p = 1 if (i1, i2, i3, i4) is an odd permutation of 1,2,3,4 and p = 2 otherwise. Thereare 24 terms in this sum.
det ( ) ( )( )A =−
− += − + +
λλ
λ λ2 1
5 42 4 5
( ) ( )
3 0 0
2 1 5
1 9 4
12 0 0 0 135 0−−
= + + − + + == −123
104 Exercise Set 2.1
17. (a) The only nonzero product in the expansion of the determinant is
a15a24a33a42a51 = (–3)(–4)(–1)(2)(5) = –120
Since (5,4,3,2,1) is even, det(A) = –120.
(b) The only nonzero product in the expansion of the determinant is
a11a25a33a44a52 = (5)(–4)(3)(1)(–2) = 120
Since (1,5,3,4,2) is odd, det(A) = –120.
18. We have
and
Thus x2 – 2x = 3 + x – x2 or 2x2 – 3x – 3 = 0. By the quadratic formula, x = (3 ± 33)/4.
19. The value of the determinant is
sin2 θ – (–cos2 θ) = sin2 θ + cos2 θ = 1
The identity sin2 θ + cos2 θ = 1 holds for all values of θ.
20. We have
and
BAd e
f
a b
c
ad bd ce
cf=
=
+
0 0 0
ABa b
c
d e
f
ad ae bf
cf=
=
+
0 0 0
1 0 3
2 6
1 3 5
5 0 3 2 3 3
−−−
= − + + − − − +x
x
x x x( ) ( )( )( ) ( (−− + = −6 3 0 22)( ) ) x x
x
xx x x x
−−
= − − − = + −1
3 11 3 1 3 2( ) ( )
Exercise Set 2.1 105
Thus AB = BA if and only if ae + bf = bd + ce, which is just the condition that
21. Since the product of integers is always an integer, each elementary product is an integer.The result then follows from the fact that the sum of integers is always an integer.
22. The signed elementary products of an n × n matrix (n > 1) all of whose entries are 1 willall be either 1 or –1.There are n! such elementary products, half of which are 1 and theother half of which are –1. Thus the determinant of such a matrix will always be zero.
23. (a) Since each elementary product in the expansion of the determinant contains a factorfrom each row, each elementary product must contain a factor from the row of zeros.Thus, each signed elementary product is zero and det(A) = 0.
24. Suppose that A = [aij] denotes an n by n diagonal matrix. That is a
ij= 0 unless i = j. If any
elementary product a1j1a2j2
… anjn
contains a factor of the form aiji
with i ≠ ji, then that
elementary product is zero. But, the only elementary product where i = ji
for all i = 1, 2,…, n is a11a22
… ann
, that is, the product of elements from the diagonal of A. Since thecolumn indices for this elementary product are in natural order, the sign of this elementaryproduct is positive. Hence, the determinant of any diagonal matrix is the product of itsdiagonal elements.
25. Let U = [aij] be an n by n upper triangular matrix. That is, suppose that a
ij= 0 whenever
i > j. Now consider any elementary product a1j1a2j2
… anjn
. If k > jk
for any factor akjk
in thisproduct, then the product will be zero. But if k ≤ j
kfor all k = 1, 2, …, n, then k = j
kfor all
k because j1, j2, …, jn
is just a permutation of the integers 1, 2, …, n. Hence, a11a22… a
nn
is the only elementary product which is not guaranteed to be zero. Since the column indicesin this product are in natural order, the product appears with a plus sign. Thus, thedeterminant of U is the product of its diagonal elements. A similar argument works forlower triangular matrices.
b a e
e d fbd bf ae ce .
−−
= − − + = 0
106 Exercise Set 2.1
EXERCISE SET 2.2
1. (b) We have
det( )A =−
−=
− −
− −
=
2 1 3
1 2 4
5 3 6
0 5 5
1 2 4
0 13 14
(( )( )
( )( )
− −− −
= − −−
1 5
1 2 4
0 1 1
0 13 14
1 5
1 2 4
0 1 1
0 0 1
== − − − = −
= − − =
( )( )( )
det( )
1 5 1 5
2 1 5
1 2 3
3 4 6
0 5
AT
−−− −
−
= −− −
−
1
1 2 3
0 10 3
1
1 2 3
0 5 1
0 0
( )
−−
= − − − = −
1
1 1 5 1 5( )( )( )( )
107
By Theorem 2.2.2.
Add 13 timesRow 2 to Row 3.
Factor –5 from Row 1and interchange Row 1 and Row 2.
Add 2 times Row 2to Row 1 and 3 timesRow 2 to Row 3.
Add –2 times Row 1 toRow 3, and interchangeRow 1 and Row 2.
By Theorem 2.2.2.
Add –2 times Row 2 toRow 1 and –5 timesRow 2 to Row 3.
2. (a) Since the matrix is upper triangular, its determinant equals the product of the diagonalelements. Hence, if we call the matrix A, then det(A) = –30.
(c) The first and third rows of the given matrix are proportional—in fact, they are equal—and hence, by Theorem 2.2.5, det(A) = 0.
3. (b) Since this matrix is just I4 with Row 2 and Row 3 interchanged, its determinant is –1.
5.
If we factor –5/3 from Row 3 and apply Theorem 2.2.2 we fInd that
det(A) = –3(–5/3)(1) = 5
7.
det( )A =−
− − =−3 6 9
2 7 2
0 1 5
3
1 2 3
0 3 4
0 1 5
det( )A = = −( )0 3 1
1 1 2
3 2 4
1
1 1 2
0 3 1
3 22 4
1
1 1 2
0 3 1
0 1 2
1 3
1 1 2
0 1 1 3
0 1
( )
( )( )
= −− −
= −− −22
3
1 1 2
0 1 1 3
0 0 5 3
= −−
108 Exercise Set 2.2
InterchangeRow 1 andRow 2.
Add –3 times Row 1to Row 3.
Factor 3from Row 2.
Add Row 2to Row 3.
Factor 3 fromRow 1 and Addtwice Row 1 to Row 2.
9.
det( )A = = −( )2 1 3 1
1 0 1 1
0 2 1 0
0 1 2 3
1
1 0 1 1
2 1 3 1
0 2 1 0
0 1 2 33
1
1 0 1 1
0 1 1 1
0 2 1 0
0 1 2 3
1
1 0 1 1
0 1
= −−
= −
( )
( )11 1
0 0 1 2
0 0 1 4
1
1 0 1 1
0 1 1 1
0 0 1 2
0
−−
= −−
−( )
00 0 6
1 1 6 1 6= − − =( )( )( )( )
(= 3))( )3
1 2 3
0 1 4 3
0 0 11 3
911
3
1
−
=
−−
= =
2 3
0 1 4 3
0 0 1
9 11 3 1 33( )( )
Exercise Set 2.2 109
Factor 3 fromRow 2 andsubtract Row 2from Row 3.
Factor 11/3from Row 3.
InterchangeRow 1 andRow 2.
Add –1 times Row 2to Row 3.
Multiply Row 3 by–1/10. Then add –3times Row 3 to Row 1.
Add –1 times Row 2to Row 3.
11.
12. Let
We are given that det(A) = –6
A
a b c
d e f
g h i
=
det( )A =− − −1 3 1 5 3
2 7 0 4 2
0 0 1 0 1
0 0 2 1 11
0 0 0 1 1
1 3 1 5 3
0 1 2 6 8
0 0 1 0 1
0 0 0 1 1
0 0 0
=−
−11 1
1 3 1 5 3
0 1 2 6 8
0 0 1 0 1
0 0 0 1 1
0 0 0 0 2
=−
−
Hencee A,det( ) ( )( )( ) .= − = −1 2 1 2
110 Exercise Set 2.2
Add 2 times Row 1 toRow 2; add –2 timesRow 3 to Row 4.
Add –1 timesRow 4 to Row 5.
(a)
(b)
(c) Let
Then det(D) = det(A) = –6 because the matrix D may be obtained from A by replacingthe first row by the sum of the first and third rows.
D
a g b h c i
d e f
g h i
=+ + +
det( )
( )( )( )
C
a b c
d e f
g h i
a b c
d
= − − −
= −
3 3 3
4 4 4
3 1 4 ee f
g h i
A= − =12 72det( )
det( )
( )
B
d e f
g h i
a b c
d e f
a b c
g h i
a
= = −( )
= −
1
1 2bb c
d e f
g h i
A= = −det( ) 6
Exercise Set 2.2 111
InterchangeRow 2 andRow 3.
Interchange Row 1and Row 2.
Factor 3 from Row 1,–1 from Row 2, and 4 from Row 3.
12. (d) Let
If we factor –3 from Row 1, we obtain
But the above matrix can be obtained from A by adding –4 times Row 2 to Row 3.Hence det(E) = (–3)det(A) = 18.
13.
Since b2 – a2 = (b – a)(b + a), we add –(b + a) times Row 2 to Row 3 to obtain
= (b –a)[(c2 – a2) – (c – a)(b + a)]
= (b – a)(c –a)[(c + a) – (b + a)]
= (b – a)(c – a)(c – b)
det( )A b a c a
c a c a
= − −
−( ) − −
1 1 1
0
0 0 2 2 (( ) +( )b a
det( )A a b c
a b c
b a c a
b a c a
=
= − −
− −
1 1 1
1 1 1
0
0
2 2 2
2 2 2 22
det( ) ( )E
a b c
d e f
g d h e i f
= −− − −
3
4 4 4
E
a b c
d e f
g d h e i f
=− − −
− − −
3 3 3
4 4 4
112 Exercise Set 2.2
Add –a times Row 1 toRow 2; add –a2 timesRow 1 to Row 3.
14. (a) Consider the elementary product p = a1j1a2j2
a3j3. If j1 = 1 or j1 = 2, then p = 0 because
a11 = a12 = 0. If j1 = 3, then the only choices for j2 are j2 = 1 or j2 = 2. But a21 = 0, sothat p = 0 unless j2 = 2. Finally, if j1 = 3 and j2 = 2, then the only choice for j3 is j3 =1. Therefore, a13a22a31 is the only elementary product which can be nonzero. Sincethe permutation (3,2,1) is odd, then
det(A) = –a13a22a31
(b) By an argument which is completely analogous to that given in Part (a), we can arguethat a14a23a32a41 is the only nonzero elementary product. Since (4,3,2,1) is an evenpermutation, then
det(A) = a14a23a32a41
15. In each case, d will denote the determinant on the left and, as usual, det(A) =
∑ ±a1j1a2j2
a3j3, where ∑ denotes the sum of all such elementary products.
(a) d = ∑ ± (ka1j1
)a2j2
a3j3 = k ∑ ±
a1j1
a2j2
a3j3 = k det(A)
(b) d = ∑ ± a2j1a1j2
a3j3= ∑ ± a1j2
a2j1a3j3
= (a11 + ka21)(a22)(a33) + (a12 + ka22)(a23)(a31)
+ (a13 + ka23)(a21)(a32) – (a13 + ka23)(a22)(a31)
– (a12 + ka22)(a21)(a33) – (a11 + ka21)(a23)(a32)
= a11 a22 a33 + a12 a23 a31 + a13 a21 a32 – a13 a22 a31
– a12 a22 a33 – a11 a23 a32
+ ka21 a22 a33 + ka22 a23 a31 + ka23 a21 a32
– ka23 a22 a31 – ka22 a21 a33 – ka21 a23 a32
=a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a ka a ka a ka
a a a
a a a
11 21 12 22 13 23
21 22 23
31 32 3
+ + +
33
Exercise Set 2.2 113
16.
(4)
(5)
R3 → R3 – 3R2
= (–1)(–6 + 1) = 5
(6)
R3 → R3 – 2R2
= –(–10 + 27) = –17
(7)
R2 → R2 + 2R1
= = − =33 4
1 53 15 4 33( )
3 6 9
2 7 2
0 1 5
3
1 2 3
2 7 2
0 1 5
3
1 2 3
0 3 4
0 1 5
−− − =
−− − =
−
1 3 0
2 4 1
5 2 2
1 3 0
2 4 1
9 10 0
1 3
9 10
−−
−=
−−
−= −
−−
0 3 1
1 1 2
3 2 4
0 3 1
1 1 2
0 1 2
13 1
1 2=
− −= −( )
− −
3 6 9
0 0 2
2 1 5
23 6
2 12 3 12(
−−
−=
−= + )) = 30
114 Exercise Set 2.2
17. (8)
R2 → R2 – 2R1
R3 → R3 + 2R1
R4 → R4 – 2R1
R3 → R3 + R1 = 39
(9)
R1 → R1 – 2R2 R3 → R3 + 3R1
= −−
= −( ) −( ) =( )1
1 1 1
2 1 0
4 5 0
1 12 1
4 56
2 1 3 1
1 0 1 1
0 2 1 0
0 1 2 3
0 1 1 1
1 0 1 1
0 3 1 0
0 1 2 3
1
1 1
=
−
= −( )−−1
2 1 0
1 2 3
=−−
−=
−− =
−=3
3 5 1
1 2 0
0 12 1
3
3 5 1
1 2 0
3 7 0
31 2
3 739
1 2 3 1
5 9 6 3
1 2 6 2
2 8 6 1
1 2 3 1
3 5 0 1
1 2 0 0
0
−−
− − −=
−−−112 0 1−
Exercise Set 2.2 115
(10)
R1 → R1 – 2R2
R1 → R1 + 3R2
= –16 (2
3 + 13 ) = –16
(11)
R2 → R2 – R3 = –2
1 3 1 5 3
2 7 0 4 2
0 0 1 0 1
0 0 2 1 1
0 0 0 1 1
1 3 1 5 3
0 1 2 6 8
0
− − −=
−00 1 0 1
0 0 2 1 1
0 0 0 1 1
1 2 6 8
0 1 0 1
0 2 1 1
0 0 1 1
22 2
=
−
→ +R R R11
1
1 0 1
2 1 1
0 1 1
1
1 0 1
2 0 0
0 1 1
1 20 1
1 1= −( ) = −( ) = −( ) −( )
=
− −
−
=
−
=1
2
1 0 1
23
1
3
1
31
32
30
1
2
1 1 0
23
1
3
1
31
32
30
11
2
1
3
1 1
1
32
3
−
−
0 1 1 1
12
12
1 12
23
1
3
1
30
1
3
2
30 0
1 0 1 0
12
12
1 12
23
−
=
− −
11
3
1
30
1
3
2
30 0−
116 Exercise Set 2.2
18. (a) Since A can be obtained from I3 by interchanging the first and third rows, it followsthat det(A) = –1.
19. Since the given matrix is upper triangular, its determinant is the product of the diagonalelements. That is, the determinant is x(x + 1)(2x – 1). This product is zero if and only ifx = 0, x = – 1, or x = 1/2.
20. (a) If x = 1orif x = –3, then two rows of the given matrix will be the same and thedeterminant will be zero.
(b) Since the determinant is 12 – 8x – 4x2 and since a quadratic polynomial has exactlytwo roots, it follows that the values of x which we found in Exercise 18(a) are theonly possibilities.
Exercise Set 2.2 117
EXERCISE SET 2.3
1. (a) We have
and
2.
Thus det(A) = 10 and det(B) = –17. Now by direct computation,
det( ) ( )
/
/
det( )
A
B
= = =
=
2 1 0
3 4 0
0 0 2
2
1 1 2 0
0 5 2 0
0 0 2
10
11 1 3
7 1 2
5 0 1
14 1 0
3 1 0
5 0 2
17 0 0
3 1 0
5 0
−=
− −−
=−−
11
17= −
det( ) ( )( ) ( )( ) ( )22 4
6 82 8 4 6 40 2 102
A =−
= − − = − = −
det( )A =−
= − − = −1 2
3 44 6 10
119
If we add 9 times Column 2 to Column 1, we find that
4. (a)
Since the determinant of this matrix is not zero, the matrix is invertible.
(c) Since the first two rows of this matrix are proportional, its determinant is zero. Hence,it is not invertible.
5. (a) By Equation (1),
det(3A) = 33 det(A) = (27)(–7) = –189
(c) Again, by Equation (1), det(2A–1) = 23 det(A–1). By Theorem 2.3.5, we have
det(2A–1) = 8 det(A) = –8
7
1 0 1
9 1 4
8 9 1
1 0 0
9 1 13
8 9 7
1 0 0
9 1 0
8 9 124
−−
−= − = − = −124
det( ) ( )
(
AB =−
=−
=
0 1 8
40 1 17
10 0 2
10
0 1 8
4 1 17
1 0 2
10)) ( )
– det( )
0 1 8
0 1 9
1 0 2
10
1 0 2
0 1 8
0 0 17
170
−= −
= = A det( )B
AB =−
9 1 8
31 1 17
10 0 2
120 Exercise Set 2.3
(d) Again, by Equation (1), det(2A) = 23 det(A) = –56. By Theorem 2.3.5, we have
det[(2A)–1] = 1 det(2A) = – 1
56
(e)
7. If we replace Row 1 by Row 1 plus Row 2, we obtain
because the first and third rows are proportional.
8. If we add –1 times Column 1 to Column 3 and –1 times Column 2 to Column 3 in thedeterminant on the left, we obtain the determinant on the right.
10. If we add –t times Row 1 to Row 2 in the determinant on the left, we get
a b t a b t a b t
t b t b t b
c
1 1 2 2 3 32
12
22
31 1 1
+ + +
− − −( ) ( ) ( )
11 2 3c c
b c c a b a
a b c
f
a b c b c a c b a
a b c
f
+ + +=
+ + + + + +=
1 1 1 1
0
a g d
b h e
c i f
a d g
b e h
c f i
a b c
d e f
g h i
= −
= − .
= 7
Exercise Set 2.3 121
Interchange Columns2 and 3.
Take the transpose ofthe matrix.
or
Adding –t times Row 2 to Row 1 in the above determinant yields the desired result.
12. (a) We have
det(A) = (k – 3)(k – 2) – 4
= k2 – 5k + 2
From Theorem 2.3.3, A will be invertible if and only if det(A) ≠ 0. But det(A) = 0 if andonly if k = (5 ± 17)/2. Thus A is invertible unless k = (5 ± 17)/2.
13. By adding Row 1 to Row 2 and using the identity sin2 x + cos2 x = 1, we see that thedeterminant of the given matrix can be written as
But this is zero because two of its rows are identical. Therefore the matrix is not invertible.
14. (b) The system can be written as
or
λ1 0
0 1
2 3
4 31
2
–
x
x==
0
0
2 3
4 31
2
1
2
=
x
x
x
xλ
sin sin sin2 2 2
1 1 1
1 1 1
α β γ
( )11 1 2 2 3 3
1 2 3
1 2 3
−+ + +
t
a b t a b t a b t
b b b
c c c
2
122 Exercise Set 2.3
This expression can be simplified—
15. We work with the system from Part (b).
(i) Here
so the characteristic equation is λ2 – 5λ – 6 = 0.
(ii) The eigenvalues are just the solutions to this equation, or λ = 6 and λ = –1.
(iii) If λ = 6, then the corresponding eigenvectors are the nonzero solutionsto the equation
The solution to this system is x1 = (3/4)t, x2 = t, so is an eigenvectorwhenever t ≠ 0.
If λ = –1, then the corresponding eigenvectors are the nonzero solutions
to the equation
If we let x1 = t, then x2 = –t, so is an eigenvector whenever t ≠ 0.
It is easy to check that these eigenvalues and their corresponding eigenvectors satisfythe original system of equations by substituting for x1, x2, and λ. The solution is valid for allvalues of t.
x =−
t
t
− −− −
=
3 3
4 4
0
01
2
x
x
x =
x
x
1
2
x = ( )
3 4 t
t
6 2 3
4 6 3
4 3
4 31
2
− −− −
=
−−
x
x
x
x
1
2
0
0
=
x =
x
x
1
2
det( ) ( )( )λλ
λλ λ λI A− =
−− −
= − − − = −
2 3
4 32 3 12 2 55 6λ −
λλ
−− −
=
2 3
4 3
0
01
2
x
x
Exercise Set 2.3 123
16. Since A is invertible, det(A) ≠ 0. Thus
det(A–1 BA) = det(A–1) det(B) det(A)
= 1 det(A) det(B) det(A)
= det(B)
17. (a) We have, for instance,
The answer is clearly not unique.
18. We know that
A is invertible if and only if det(A) ≠ 0
if and only if det(AT) = 0 (since det(A) = det(AT)
if and only if det(AT) det(A) ≠ 0
if and only if det(AT A) ≠ 0
if and only if AT A is invertible
19. Let B be an n × n matrix and E be an n × n elementary matrix.
Case 2: Let E be obtained by interchanging two rows of In. Then det(E) = –1 and EA is
just A with (the same) two rows interchanged. By Theorem 2.2.3, det(EA) = –det(A) =det(E) det(A).
Case 3: Let E be obtained by adding a multiple of one row of In
to another. Then det(E)= 1 and det(EA) = det(A). Hence det(EA) = det(A) = det(E) det(A).
a b c d
a b c d
a b c d
a c
a1 1 1 1
2 2 2 2
1 1 1 1
2 2
1+ ++ +
=+ +
++ bb c d
b d
a c
a c
b d
a c
a c
b d
1 1 1
2 2
1 1
2 2
1 1
2 2
1 1
2 2
+
= + + +b d
b d
1 1
2 2
124 Exercise Set 2.3
20. No, since det(AB) = det(A) det(B) = det(B) det(A) = det(BA). Hence, the 2 determinantsare always equal.
21. If either A or B is singular, then either det(A) or det(B) is zero. Hence, det(AB) = det(A)det(B) = 0. Thus AB is also singular.
22. (a) False. If A is an n by n matrix, then det(2A) = 2n det(A). Hence, the statement holdsonly in the trivial cases when n = 1 or when det(A) = 0.
(c) False. For instance, let .
(d) True. By Theorem 1.6.1, any system of linear equations has no solution, one solution,or infinitely many solutions. Since A0 = 0, there is at least one solution. Since det(A)= 0, it follows from Theorem 2.3.6 that there cannot be just one solution. Hence, theremust be infinitely many. Also see the argument used in Exercise 20(a) of Section 1.5.
23. (a) False. If det(A) = 0, then A cannot be expressed as the product of elementarymatrices. If it could, then it would be invertible as the product of invertible matrices.
(b) True. The reduced row echelon form of A is the product of A and elementary matrices,all of which are invertible. Thus for the reduced row echelon form to have a row ofzeros and hence zero determinant, we must also have det(A) = 0.
(c) False. Consider the 2 × 2 identity matrix. In general, reversing the order of thecolumns may change the sign of the determinant.
(d) True. Since det(AAT) = det(A) det(AT) = [det(A)]2, det(AAT)cannot be negative.
A =
1 0
0 0
Exercise Set 2.3 125
EXERCISE SET 2.4
1. a) 5 b) 9 c) 6d) 10 e) 0 f) 2
2. a) odd b) odd c) evend) even e) even f) even
3. 22
4. 0
5. 52
6. –3 6)
7. a2 – 5a + 21
8. 0
9. –65
10. –4
11. –123
12. –c4 + c3 – 16c2 + 8c – 2
127
13. (a) λ = –3, λ = 1
(b) λ = 4, λ = 3, λ = –2
14. 1 2 3 4 even 2 1 3 4 odd1 2 4 3 odd 2 1 4 3 even1 3 4 2 even 2 3 1 4 even1 3 2 4 odd 2 3 4 1 odd1 4 2 3 even 2 4 1 3 even1 4 3 2 odd 2 4 1 2 odd
3 1 2 4 even 4 1 2 3 odd3 1 4 2 odd 4 1 3 2 even3 2 1 4 odd 4 2 1 3 even3 2 4 1 even 4 2 3 1 odd3 4 1 2 even 4 3 1 2 odd3 4 2 1 odd 4 3 2 1 even
15. (a) detA = a11a23a32a44 + a11a24a32a43 – a11a24a33a42
– a11a23a32a44 + a11a24a32a43 – a11a24a33a42
– a12a21a33a44 + a12a21a34a43 + a12a23a31a44
– a12a23a34a41 + a12a24a33a41 + a12a24a31a43
+ a13a21a32a44 + a13a21a34a42 + a13a22a32a41
+ a13a22a34a41 + a13a24a31a42 + a13a24a32a41
(b) There is no diagonal pattern
16. 275
17. (a) –120
(b) 120
128 Exercise Set 2.4
18.
19. det = sin2 θ + cos2 θ = 1
20.
So AB = BA ae + bf = bd + ce.
But = bd – bf – ea + ee
So if = 0, then ae + bf = bd + ce, and conversely, and so AB = BA.
21. detA would also be an integer, since detA is formed by products and sums of its entries, andthe integers are closed under multiplication and addition.
22. detA = 0, since all products are either ±1, half being +1 (corresponding to an evenpermutation), and half being –1 corresponding to an odd permutation.
23. (a) Every product in detA must have an element of any given row.
(b) Likewise, every product in detA must have an element of any given column, or becausedetAT = detA.
24. detA = a11a22a33… a
nn—the product of the diagonal entries.
25. detA = a11a22… a
nn—the product of the diagonal entries, in both cases.
b a c
e d f
−−
b a c
e d f
−−
ABa b
c
d e
f
ad ae bf
cf=
=
+
0 0 0
=
=
+
BA
d e
f
a b
c
ad bd ce
cf0 0 0
x =±3 33
4
Exercise Set 2.4 129
SUPPLEMENTARY EXERCISES 2
1.
3. The determinant of the coefficient matrix is
The system of equations has a nontrivial solution if and only if this determinant is zero;that is, if and only if α = β. (See Theorem 2.3.6.)
1 1 1 11 1
αβ
α β
αβ α
α ββ
α β1 1
1= 0 0 −
1= =–( – )a ––( – )( – )β βa a
′ =
−
−=
+
+=x
x
y x y
45
35
35
45
45
35
35
45
925
1625
3
5
4
5
3545
35
45
45
35
35
45
1
x y
y
x
y y x
+
′ =−
=−
= − +4
5
3
5x y
131
4. Recall that det(A) is the sum of 6 signed elementary products, 3 of which are positive and3 of which are negative. Since each of the factors in these products is either 0 or 1, then theonly values which the products can assume are 0, +1, or –1. If det(A) has 3 signedelementary products which are equal to +1, then the remaining 3 products must equal –1.(Why?) In this case, det(A) = 0. However, det(A) can have 2 signed elementary productswhich are equal to +1 and the remaining 4 products equal to 0. For instance, this is true of
Thus, the largest possible value for det(A) is 2.
5. (a) If the perpendicular from the vertex of angle α to side a meets side a between anglesβ and γ, then we have the following picture:
Thus cos β = 1st sfa1divisor2nd sfc and cos γ = 1st sfa2divisor2nd sfb andhence
a = a1 + a2 = c cos β + b cos γ
This is the first equation which you are asked to derive. If the perpendicular intersectsside a outside of the triangle, the argument must be modified slightly, but the sameresult holds. Since there is nothing sacred about starting at angle α, the sameargument starting at angles β and γ will yield the second and third equations.
Cramer’s Rule applied to this system of equations yields the following results:
a
a1 a2
c b
det
1 0 1
1 1 0
0 1 1
132 Supplementary Exercises 2
6. The system of equations
(1 – λ)x – 2y = 0
x – (1 + λ)y = 0
always has the trivial solution x = y = 0. There can only be a nontrivial solution in case thedeterminant of the coefficient matrix is zero. But this determinant is just
–(1 – λ)(1 + λ) + 2 = λ2 + 1
and λ2 + 1 can never equal zero for any real value of λ.
7. If A is invertible, then adj(A), or adj(A) = [det(A)]A–1. Thus
adj( )det( )
AA
AI= =
AA
–
det( )1 1
=
cos α (–= =
+ +
a c b
b a
c a
c b
c a
b a
a a b c
0
0
0
0
0
2 2 22
2 2
0
2
)
abc
b c a
bc
a b
c b a
b c a
abc
=+ +
=cos β ( )=
− +=
+ −b a b c
abc
a b c
ac
2 2 2 2 2 2
2 2
cos γ ( )= =
+ −=
0
0
2 2
2 2 2
c a
c b
b a c
abc
c a b c
abc
aa b c
ab
2 2 2
2
+ −
Supplementary Exercises 2 133
That is, adj(A) is invertible and
It remains only to prove that A = det(A)adj(A–1).This follows from Theorem 2.4.2 andTheorem 2.3.5 as shown:
8. From the proof of Theorem 2.4.2, we have
(*) A[adj(A)] = [det(A)]I
First, assume that A is invertible. Then det(A) ≠ 0. Notice that [det(A)]I is a diagonalmatrix, each of whose diagonal elements is det(A). Thus, if we take the determinant of thematrices on the right- and left-hand sides of (*), we obtain
det(A[adj(A)]) = det([det(A)]I)
or
det(A) • det(adj(A)) = [det(A)]n
and the result follows.
Now suppose that A is not invertible. Then det(A) = 0 and hence we must show thatdet(adj(A)) = 0. But if det(A) = 0, then, by (*), A[adj(A)] = 0. Certainly, if A = 0, thenadj(A) = 0; thus, det(adj(A)) = 0 and the result holds. On the other hand, if A ≠ 0, then Amust have a nonzero row x. Hence A[adj(A)] = 0 implies that xadj(A) = 0 or
[adj(A)]TxT = 0T
Since xT is a nontrivial solution to the above system of homogeneous equations, it followsthat the determinant of [adj(A)]T is zero. Hence, det(adj(A)) = 0 and the result is proved.
A AA
A A A= = =[ ]det( )
( ) det ( ) ( )– ––
– –1 11
1 11adj adj
[adj( )]det( )
–AA
A1 1=
134 Supplementary Exercises 2
9. We simply expand W. That is,
= ddx
(f1(x)g2(x) – f2(x)g
1(x))
= f′1(x)g2(x) + f1(x)g′2(x) – f′2(x)g1(x) – f2(x)g′1(x)
= [f′1(x)g2(x) – f′2(x)g1(x)] + [f1(x)g′2(x) – f2(x)g′1(x)]
10. (a) As suggested in the figure and the hints,
area ABC = area ADEC + area CEFB – area ADFB
= 12 (x3 – x1)(y1 + y3) + 12 (x2 – x3)(y3 + y2)
– 12 (x2 – x1)(y1 + y2)
= 12 [x3y1 – x1y1 + x3y3 – x1y3 + x2y3 – x3y3 + x2y2
– x3y2 – x2y1 + x1y1 – x2y2 + x1y2]
= 12 [x1y2 – x1y3 – x2y1 + x2y3 + x3y1 – x3y2]
= 12
x y
x y
x y
1 1
2 2
3 3
1
1
1
=′ ′
+′
f x f x
g x g x
f x f x
g x
1 2
1 2
1 2
1
( ) ( )
( ) ( )
( ) ( )
( ) ′′g x2( )
dW
dx
d
dx
f x f x
g x g x
( ) ( )
( ) ( )= 1 2
1 2
Supplementary Exercises 2 135
10. (b) The area of the triangle is
12 = 12[3(–1) + 2(3) + 4(3 + 1)] = 19
2
11. Let A be an n × n matrix for which the entries in each row add up tozeroand let x be then × 1 matrix each of whose entries is one. Then all of the entries in the n × 1 matrix Ax arezero since each of its entries is the sum of the entries of one of the rows of A. That is, thehomogeneous system of linear equations
has a nontrivial solution. Hence det(A) = 0. (See Theorem 2.3.6.)
12. To obtain B from A, we first interchange Row 1 and Row n. This changes the sign of det(A).If n = 2 or 3, we are done. If n > 3, then we interchange Rows 2 and n – 1, which againreverses the sign of the determinant. In general, we interchange Rows k + 1 and n – k untilall of the rows have been interchanged.
If n is even, then we will have made n/2 interchanges and the sign of det(A) will havechanged by a factor of (–1)n/2. On the other hand, if n is odd, then we will have made (n – 1)/2 interchanges and the sign of det(A) will have changed by a factor of (–1)(n–1)/2. Allof this can be summarized by the formula
Can you see why?
13. (a) If we interchange the ith and jth rows of A, then we claim that we must interchange theith and jth columns of A–1. To see this, let
A A=
−
Row 1
Row 2
Row n
Mand = Co1 ll.1, Col. 2, , Col. n
det ( ) ( ) det ( )( – )
B An n= − −1
1
2
Ax =
0
0
M
3 3 1
2 1 1
4 0 1
− −
136 Supplementary Exercises 2
where AA–1 = I. Thus, the sum of the products of corresponding entries from Row s inA and from Column r in A–1 must be 0 unless s = r, in which case it is 1. That is, ifRows i and j are interchanged in A, then Columns i and j must be interchanged in A–1
in order to insure that only 1’s will appear on the diagonal of the product AA–1.
(b) If we multiply the ith row of A by a nonzero scalar c, then we must divide the ith
column of A–1 by c. This will insure that the sum of the products of correspondingentries from the ith row of A and the ith column of A–1 will remain equal to 1.
(c) Suppose we add c times the ith row of A to the jth row of A. Call that matrix B. Nowsuppose that we add –c times the jth column of A–1 to the ith column of A–1. Call thatmatrix C. We claim that C = B – 1. To see that this is so, consider what happens when
Row j → Row j + c Row i [in A]
Column i → Column i – c Column j [in A–1]
The sum of the products of corresponding entries from the jth row of B and any kth
column of C will clearly be 0 unless k = i or k = j. If k = i, then the result will be c –c = 0. If k = j, then the result will be 1. The sum of the products of correspondingentries from any other row of B—say the rth row—and any column of C—say the kth
column—will be 1 if r = k and 0 otherwise. This follows because there have been nochanges unless k = i. In case k = i, the result is easily checked.
14. Let C be the matrix which is the same as A except for its ith row which is the sum of the ith
rows of B1 and B2, or twice the ith row of A. Then by Theorem 2.3.1,
det(B1) + det(B2) = det(C)
But det(C) = 2 det(A), so that
det(A) = 12 [det(B1) + det(B2)]
15. (a) We have
det( )λλ
λI A
a a a
a a a− =− − −
− − −−
11 12 13
21 22 23
aa a a31 32 33− −λ
Supplementary Exercises 2 137
If we calculate this determinant by any method, we find that
det(λI – A) = (λ – a11)(λ – a22)(λ – a33) – a23a32 (λ – a11)
–a13a31(λ – a22) – a12a21(λ – a33)
–a13a21a32 – a12a23a31
= λ3 + (–a11 – a22 – a33)λ2
+ (a11a22 + a11a33 + a22a33 – a12a21 – a13a31 – a23a32)λ
+ (a11a23a32 + a12a21a33 + a13a22a31
–a11a22a33 – a12a23a31 – a13a21a32)
15. (b) From Part (a) we see that b = –tr(A) and d = –det(A). (It is less obvious that c is thetrace of the matrix of minors of the entries of A; that is, the sum of the minors of thediagonal entries of A.)
16. Recall that sin(a + b) = sin a cos b + cos a sin b. If we multiply the 1st column of thedeterminant by cos δ and the 2nd column by sin δ and then add the results, we get the 3rd
column. Therefore the determinant is 0.
17. If we multiply Column 1 by 104, Column 2 by 103, Column 3 by 102, Column 4 by 10, andadd the results to Column 5, we obtain a new Column 5 whose entries are just the 5numbers listed in the problem. Since each is divisible by 19, so is the resulting determinant.
18. (b) Here we have
If we add λ times Row 2 to Row 1, we have
det( –
– –
–
– – –
λλ λ
λλ
I A =
0 1 1
1 1
1 5 3
2
+−− − −
= (λλ
λλ
– )1
0 1 1
1 1
1 5 3
A I A= −
− =0 1 1
1 0 1
1 5 3
so that λλ −− −
−− − −
1 1
1 1
1 5 3
λλ
138 Supplementary Exercises 2
(Factor λ – 1 from Row 1.) Now add –1 times Row 2 to Row 3 and interchange Rows1 and 2 to obtain the equation
To simplify, add Row 2 to Row 3 and then add (λ + 1)/4 times Row 3 to Row 2, getting
If we interchange Rows 2 and 3, the last determinant becomes lower triangular, sothat
det(λI – A) = –(λ – 1)(–1)(–1)(–4)
= (λ – 1)(λ2 – 2λ + 1) = (λ – 1)3
Thus, the characteristic equation is (λ – 1)3 = 0 and so λ = 1 is the only eigenvalue. Ifwe let λ = 1, then the eigenvectors are the nonzero solutions to the equation
The augmented matrix reduces to
If we put x3 = 2 t, then x2 = –t and x1 = t, so the eigenvectors are
x = [t –t 2t]T where t ≠ 0
1 1 1 0
0 11
20
0 0 0 0
− −
1 1 1
1 1 1
1 5 2
1
2
3
− −−− − −
x
x
x
=
0
0
0
2 1
4
2λ λ− +
det ( ) ( )λ λλ
λλ
λ
λ
I A− = − −−
+− −
= − −
−
1
1 1
0 1 1
0 4 3
1
1
( )
11
0 02 1
4
0 4 3
2λ λ
λ
− +
− −
det ( ) ( )λ λλ
λλ λ
I A− = − −−
+− − −
1
1 1
0 1 1
0 5 4
Supplementary Exercises 2 139
TECHNOLOGY EXERCISES 2.1
T3. Let y = ax3 + bx3 + cx + d be the polynomial of degree three to pass through the fourgiven points. Substitution of the x and y coordinates of these points into the equation of thepolynomial yields the system
7 = 27a + 9b + 3c + d
–1 = 8a + 4b + 2c + d
–1 = a + b + c + d
1 = 0a + 0b + 0c + d
Using Cramer’s Rule,
a =
−−
=
7 9 3 1
1 4 2 1
1 1 1 1
1 0 0 1
27 9 3 1
8 4 2 1
1 1 1 1
0 0 0 1
12
1221
27 7 3 1
8 1 2 1
1 1 1 1
0,= =
−−
b11 0 1
12
2
12
27 9 7 1
8 4 1 1
1 1 1 1
0 0
=−
= −
=
−−
c11 1
12
12
121
27 9 3 7
8 4 2 1
1 1
,=−
= − =
−
d
1 1
0 0 0 1
12
12
121
−
= =
141
Plot. y = x3 – 2x2 – x + 1
(1, −1) (2, −1)
(3, 7)
(0, 1)
142 Technology Exercises 2.1
EXERCISE SET 3.1
1.
2.
v1= (3, 6)
x
y
x
y
v3 = ( 4, 3)
(a) (c)
x
(3, 4, 5)
y
z
x
y
z
(3, 4, 5)
x
( 3, 4, 5)
y
z
x
z
y
(3, 0, 3)
(a) (c)
(e) (j)
143
3. (a) → = (3 – 4, 7 – 8) = (–1, –1)
(e)→→ = (–2 – 3, 5 + 7, –4 –2) = (–5, 12, –6)
5. (a) Let P = (x,y,z) be the initial point of the desired vector and assume that this vector
has the same length as v. Since has the same direction as v = (4, –2, –1), we have
the equation
= (3 – x, 0 – y, –5 – z) = (4, –2, –1)
If we equate components in the above equation, we obtain
x = –1, y = 2, and z = –4
Thus, we have found a vector which satisfies the given conditions. Any positive
multiple k will also work provided the terminal point remains fixed at Q. Thus, P
could be any point (3 – 4k, 2k, k – 5) where k > 0.
(b) Let P = (x, y, z) be the initial point of the desired vector and assume that this vector
has the same length as v. Since →→ is oppositely directed to v = (4, –2, –1), we have
the equation
→→ = (3 – x, 0 – y, –5 –z) = (–4, 2, 1)
If we equate components in the above equation, we obtain
x = 7, y = –2, and z = –6
Thus, we have found a vector →→ which satisfies the given conditions. Any positive
multiple k →→ will also work, provided the terminal point remains fixed at Q. Thus, P
could be any point (3 + 4k, –2k, –k – 5) where k > 0.
PQ PQ
PQ
PQ
PQ
PQ
PQ
PQ
P P1 2
P P1 2
v1= (3, 4, 5)
x
z
y
v9= (0, 0, 3)x
z
y
(i)(g)
144 Exercise Set 3.1
6. (a) v – w = (4 – 6, 0 – (–1), –8 – (–4)) = (–2, 1, –4)
(c) –v + u = (–4 – 3, 0 + 1, 8 + 2) = (–7, 1, 10)
(e) –3(v – 8w) = –3(4 – 48, 0 + 8, –8 + 32) = (132, –24, –72)
7. Let x = (x1, x2, x3). Then
2u – v + x = (–6, 2, 4) – (4, 0, –8) + (x1,x2, x3)
= (–10 + x1, 2 + x2, 12 + x3)
On the other hand,
7x + w = 7(x1, x2, x3) + (6, –1, –4)
= (7x1 + 6, 7x2 – 1, 7 x3 – 4)
If we equate the components of these two vectors, we obtain
7x1 + 6 = x1 – 10
7x2 – 1 = x2 + 2
7x3 – 4 = x3 + 12
Hence, x = (–8/3, 1/2, 8/3).
9. Suppose there are scalars c1, c2, and c3 which satisfy the given equation. If we equatecomponents on both sides, we obtain the following system of equations:
–2c1 – 3c2 + c3 = 0
9c1 + 2c2 + 7c3 = 5
6c1 + c2 + 5c3 = 4
The augmented matrix of this system of equations can be reduced to
The third row of the above matrix implies that 0c1 + 0c2 + 0c3 = –1. Clearly, there do notexist scalars c1, c2, and c3 which satisfy the above equation, and hence the system isinconsistent.
2 3 1 0
0 2 2 1
0 0 0 1
−− −
−
Exercise Set 3.1 145
10. If we equate components on both sides of the given equation, we obtain
c1 + 2c2 = 0
2c1 + c2 + c3 = 0
c2 + c3 = 0
The augmented matrix of this system of equations can be reduced to
Thus the only solution is c1 = c2 = c3 = 0.
11. We work in the plane determined by the three points O = (0, 0, 0), P = (2, 3, –2), and Q =
(7, –4, 1). Let X be a point on the line through P and Q and let t (where t is a positive,
real number) be the vector with initial point P and terminal point X. Note that the length
of t → is t times the length of . Referring to the figure below, we see that
and
Q
O
PX
tPQ
OP PQ OQ
+ =
OP tPQ OX
+ =
PQ
PQ
PQ
1 0 0 0
0 1 0 0
0 0 1 0
146 Exercise Set 3.1
Therefore,
(a) To obtain the midpoint of the line segment connecting P and Q, we set t =1/2. Thisgives
(b) Now set t = 3/4.This gives
12. The relationship between the xy-coordinate system and the x′y′-coordinate system is givenby
x′ = x –2, y′ = y –(–3) = y + 3
(a) x′ = 7 – 2 = 5and y′ = 5+ 3 = 8
(b) x = x′ + 2 = –1 and y = y′ –3 = 3
OX = − + − = −
1
42 3 2
3
47 4 1
23
4
9
4
1
4( , , ) ( , , ) , ,
OX OP OX= +
= − + −
1
2
1
21
22 3 2
1
27 4( , , ) ( , ,, )
, ,
1
9
2
1
2
1
2= − −
OX OP t OQ OP
t OP
= +
= +
( – )
( – )1 ttOQ
Exercise Set 3.1 147
(c)
(d) (1, 10)
(e) v1′ = v1 – 2, v2′ = v + 3
13. Q = (7, –3, –19)
14. Let (x0, y0, z0) denote the origin in the x′ y′ z′-coordinate system with respect to the xyz-coordinate system. Suppose that P1 and P2 are the initial and terminal points, respectively,for a vector v. Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of P1 and P2 in the xyz-coordinate system, and let (x′1, y′1, z′1) and (z′2, y′2, z′2) be the coordinates of P1 and P2 in thex′y′z′-coordinate system. Further, let (v1, v2, v3) and (v′1, v′2, v′3) denote the coordinates ofv with respect to the xyz- and x′y′z′-coordinate systems, respectively. Then
v1 = x2 – x1, v2 = y2 – y1, v3 = z2 – z1
and
v′1 = x′2 – x′1, v′2 = y′2 – y′1, v′3 = z′2 – z′1
However,
x′1 = x1 – x0, y′1 = y1 – y0, z′1 = z1 – z0
and
x′2 = x2 – x0, y′2 = y2 – y0, z′z′2 = z2 – z0
Q
y y'
x'
x
P85
6
7
5–3
148 Exercise Set 3.1
If we substitute the above expressions for x′1 and x′2 into the equation for v′1, we obtain
v′1 = (x2 – x0) – (x1 – x0 ) = x2 – x1 = v1
In a similar way, we can show that v′2 = v2 and v′3 = v3.
16. Let v = (v1, v2) and u = kv = (u1, u2). Let P and Q be the points (v1, v2) and (u1, u2),respectively. Since the triangles OPR and OQS are similar (see the diagram),
we have
Thus, u1 = kv1. Similarly, u2 = kv2.
17. The vector u has terminal point Q which is the midpoint of the line segment connecting P1and P2.
−OP1
OP2 − OP1
(OP2 − OP1)
O
Q P2P1
OP1OP2
12
u OS
OR
1
1υ= =
length of
length of
lengthh of
length of
OQ
OPk=
O R S
Q(u1, u2)
P(v1, v2)
kv
y
x
v
Exercise Set 3.1 149
18. We have x + y + z + w = 0 where
19. Geometrically, given 4 nonzero vectors attach the “tail” of one to the “head” of another andcontinue until all 4 have been strung together.The vector from the “tail” of the first vectorto the “head” of the last one will be their sum.
x
yz
w
x + y + z + w
x
yz
w
150 Exercise Set 3.1
EXERCISE SET 3.2
1. (a) v = (42 + (–3)2)1/2 = 5
(c) v = [(–5)2 + 02]1/2 = 5
(e) v = [(–7)2 + 22 + (–1)2]1/2 = 54
2. (a) d = [(3 – 5)2 + (4 – 7)2]1/2 = 13
(c) d = [7 – (–7)2 + (–5) – (–2)2 + 1 – (–1)2]1/2 = 209
3. (a) Since u + v = (3, –5, 7), then
u + v = [32 + (–5)2 + 72]1/2 = 83
(c) Since
– 2u = [(–4)2 + 42 + (–6)2]1/2 = 2 17
and
2u = 2 [22 + (–2)2 + 32]1/2 = 2 17
then
2u + 2u = 4 17
(e) Since w = [32 + 62 + (–4)2]1/2 = 61, then
, ,–1
ww=
3
61
6
61
4
61
151
4.
1 = v – w ≤ 5
5. (a) k = 1, l = 3
(b) no possible solution
6. l = 4
7. Since kv = (–k, 2k, 5k), then
kv = [k2 + 4k2 + 25k2]1/2 = |k| 30
If kv = 4, it follows that |k| 30 = 4 or k = ±4/ 30.
9. (b) From Part (a), we know that the norm of v/v is 1. But if v = (3, 4), then v = 5.Hence u = v/v = (3/5, 4/5) has norm 1 and has the same direction as v.
10. (b) By the result of Part (a), we have 4u = 4(u cos 30°, u sin 30°) = 4(3( 3/2),
3(1/2)) = (6 3, 6) and 5 v = 5(v cos(135°), v sin(135°)) = 5(2(–1/ 2), 2(1/
2)) = (–10/ 2, 10/ 2) = (–5 2, 5 2). Thus 4u – 5v = (6 3 + 5 2, 6 – 5 2).
11. Note that p – p0 = 1 if and only if p – p02 = 1. Thus
(x – x0)2 + (y – y0)
2 + (z – z0)2 = 1
The points (x, y, z) which satisfy these equations are just the points on the sphere of radius1 with center (x0, y0, z0); that is, they are all the points whose distance from (x0, y0, z0) is1.
22
33
0
0
w – v2 3w
w wv
v v
152 Exercise Set 3.2
12. First, suppose that u and v are neither similarly nor oppositely directed and that neither isthe zero vector.
If we place the initial point of v at the terminal point of u, then the vectors u, v, and u+ v form a triangle, as shown in (i) below.
Since the length of one side of a triangle—say u + v—cannot exceed the sum of thelengths of the other two sides, then u + v < u + v.
Now suppose that u and v have the same direction. From diagram (ii), we see that u+ v = u + v. If u and v have opposite directions, then (again, see diagram (ii)) u + v< u + v.
Finally, if either vector is zero, then u + v = u + v.
13. These proofs are for vectors in 3-space. To obtain proofs in 2-space, just delete the 3rdcomponent. Let u = (u1, u2, u3) and v = (v1, v2, v3). Then
(a) u + v = (u1 + v1, u2 + v2, u3 + v<I>3)
= (v1 + u1, v2 + u2, v3 + u3) = v + u
(c) u + 0 = (u1 + 0, u2 + 0, u3 + 0)
= (0+ u1, 0 + u2, 0 + u3)
= (u21, u
2, u
3) = 0 + u = u
(e) k(lu) = k(lu1, lu2, lu3) = (klu1, klu2, klu3) = (kl)u
14. Again, we work in 3-space. Let u = (u1, u2, u3).
(d) u + (–u) = (u1 + (–u1), u2 + (–u2), u3 + (–u3)) = (0, 0, 0) = 0
(g) (k + l)u = ((k + l)u1, (k + l)u2, (k + l)u3)
= (ku1 + lu1, ku2 + lu2, ku3 + lu3)
= (ku1, ku2, ku3) + (lu1, lu2, lu3)
= ku + lu
(h) 1u = (1u1, 1u2, 1u3) = (u1, u2, u3) = u
u
uu
u
u + vu + v
u + v
u + v
v
v
vv
(i) (ii)
Exercise Set 3.2 153
15. See Exercise 9. Equality occurs only when u and v have the same direction or when one isthe zero vector.
16. (a) For p = (a, b, c) to be equidistant from the origin and the xz-plane, we must have
a2 + b2 + c2 = |b|. Thus, a2 + b2 + c2 = b2 or a2 + c2 = 0, so that a = c = 0. That is, p
must lie on the y-axis.
(b) For p = (a, b, c)to be farther from the origin than the xz-plane, we must have a2 +
b2 + c2 > |b|. Thus a2 + b2 + c2 > b2 or a2 + c2 > 0, so that a and c are not both zero.
That is, p cannot lie on the y-axis.
17. (a) If x < 1, then the point x lies inside the circle or sphere of radius one with center atthe origin.
(b) Such points x must satisfy the inequality x – x0 > 1.
18. The two triangles pictured are similar since u and ku and v and kv are parallel and theirlengths are proportional; i.e., ku = ku and kv = kv. Therefore ku + kv = ku + v.That is, the corresponding sides of the two triangles are all proportional, where k is theconstant of proportionality. Thus, the vectors k(u + v) and ku + kv have the same length.Moreover, u + v and ku + kv have the same direction because corresponding sides of thetwo triangles are parallel. Hence, k(u + v) and ku + kv have the same direction. Therefore,k(u + v) = ku + kv.
154 Exercise Set 3.2
EXERCISE SET 3.3
1. (a) u • v = (2)(5) + (3)(–7) = –11
(c) u • v = (1)(3) + (–5)(3) + (4)(3) = 0
2. (a) We have u = [22 + 32]1/2 = 13 and v = [52 + (–7)2]1/2 = 74. From Problem
1(a), we know that u • v = –11. Hence,
(c) From Problem 1(c), we know that u • v = 0. Since neither u nor v is the zero vector,this implies that cos θ = 0.
3. (a) u • v = (6)(2)+ (1)(0)+ (4)(–3) =0. Thus the vectors are orthogonal.
(b) u • v = –1 < 0. Thus θ is obtuse.
4. (a) Since u • a = 0 and a ≠ 0, then w1 = (0, 0).
(c) Since u • a = (3)(1) + (1)(0) + (–7)(5) = –32 and a 2 = 12 + 02 + 52 = 26, we have
5. (a) From Problem 4(a), we have
w2 = u – w1 = u = (6, 2)
w132
261 0 5
16
130
80
13= − = − −
( , , ) , ,
cos–
.θ = ≈11
13 74110 8
155
5. (c) From Problem 4(c), we have
w2 = (3, 1, –7) – (–16/13, 0, –80/13) = (55/13, 1, –11/13)
6. (a) Since |u • a| = |–4 + 6| = |2| = 2 and a = 5, it follows from Formula (10) that projau
= 2/5.
(d) Since |u • a| = |3 –4 –42| = |–43 and a = 54, Formula (10) implies that projau =
43/ 54.
8. (c) From Part (a) it follows that the vectors (–4, –3) and (4, 3) are orthogonal to (–3, 4).Divide each component by the norm, 5, to obtain the desired unit vectors.
10. We look for vectors (x, y, z) such that (x, y, z) • (5, –2, 3) = 0. That is, such that 5x – 2y
+ 3z = 0. One such vector is (0, 3, 2). Another is (1, 1, –1). To find others, assign arbitraryvalues, not both zero, to any two of the variables x, y, and z and solve for the remainingvariable.
12. Since the three points are not collinear (Why?), they do form a triangle. Since
• = (1, 3, –2) • (4, –2, 1) = 0
the right angle is at B.
13. Let w = (x, y, z) be orthogonal to both u and v. Then u • w = 0 implies that x + z = 0 andv • w = 0 implies that y + z = 0. That is w = (x, x, –x). To transform into a unit vector, wedivide each component by w = 3x2. Thus either (1/ 3, 1/ 3, –1/ 3) or (–1/ 3,–1/ 3, 1/ 3) will work.
16. (a) We look for a value of k which will make p a nonzero multiple of q. If p = cq, then (2,k) = c(3, 5), so that k/5 = 2/3 or k = 10/3.
(c) We want k such that
p • q = p q cos π3
BC
AB
156 Exercise Set 3.3
or
If we square both sides and use the quadratic formula, we find that
The minus sign in the above equation is extraneous because it yields an angle of 2π–/3.
17. (b) Here
17. From the definition of the norm, we have
u + v2 = (u + v) • (u + v)
Using Theorem 3.3.2, we obtain
u + v2 = (u • u) + (u • v) + (v • u)+ (v • v)
or
(*) u + v2 = u2 + 2(u • v) + v2
Similarly,
(**) u – v2 = u2 – 2(u • v) + v2
If we add Equations (*)and (**), we obtain the desired result.
19. If we subtract Equation (**) from Equation (*) in the solution to Problem 16, we obtain
u + v2 – u – v2 = 4(u • v)
If we then divide both sides by 4, we obtain the desired result.
D =+ − −
+=
4 2 1 5 2
4 1
1
172 2
( ) ( )
( ) ( )
k =− ±60 34 3
33
6 5 4 3 51
22 2 2+ = + +
k k
Exercise Set 3.3 157
20. Let u1, u2, and u3 be three sides of thecube as shown. The diagonal of the faceof the cube determined by u1 and u2 is b= u1 + u2; the diagonal of the cube itselfis a = u1 + u2 + u3. The angle θ betweena and b is given by
Because u1, u2, and u3 are mutually orthogonal, we have ui
• uj= 0 whenever i ≠ j. Also u
i
• ui
= ui2 and u1 = u2 = u3. Thus
That is, θ =arccos(2/ 6).
21. (a) Let i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) denote the unit vectors along the x, y,and z axes, respectively.If v is the arbitrary vector (a, b, c), then we can write v = ai
+ bj + ck. Hence, the angle α between v and i is given by
since i = 1 and i • j = i • k = 0.
23. By the results of Exercise 21, we have that if vi
= (ai, b
i, c
i) for i = 1 and 2, then cos α
i=
ai/v
i, cos β
i= b
i/v
i, and cos γ
i= c
i/v
i. Now
cos α =⋅
=+ +
=vv ii
vv ii vv
a
a b c
a
2 2 2
cosθ =+
+ + +
=
uu uu
uu uu uu uu uu
12
22
12
22
32
12
22
2 uuuu
uu uu
22
12
12
3 2
2
6=
cos
( ) ( )
( ) (
θ =⋅
=+ + ⋅ +
+ + ⋅
a b
a b
u u u u u
u u u u1 2 3 1 2
1 2 3 11 2 3 1 2 1 2+ + + ⋅ +u u u u u u) ( ) ( )
158 Exercise Set 3.3
24. (a) area = 19 units
(b) β(–3, 12) area = 2 × 19 = 38 units
25. Note that
v • (k1w1 + k2w2) = k1(v • w1) + k2(v • w2) = 0
because, by hypothesis, v • w1 = v • w2 = 0. Therefore v is orthogonal to k1w1 + k2w2 forany scalars k1 and k2.
26. Note that w is a multiple of u plus a multiple of v and thus lies in the plane determined byu and v. Let α be the angle between u and w and let β be the angle between v and w. Then
Similarly, we can show that
Since both α and β lie between 0 and π, then cos α = cos β implies that α = β.
27. (a) The inner product x • y is defined only if both x and y are vectors, but here v • w isa scalar.
(b) We can add two vectors or two scalars, but not one of each.
cosβ = =+v w
v wu v
w⋅ ⋅ kl
cos α =⋅
=⋅ + ⋅
=⋅
uu
uu vv
uu vv uu uu
uu
ww
ww
k l
k
k vv
uu vv
+
=⋅ +
lk
k
lk
2
ww
ww
v v v v1 2 1 2and are orthogonal ⇔ =⇔ +
⋅ 0
1 2 1a a b bb c c
a a b b c c
2 1 2
1 2
1 2
1 2
1 2
1 2
1 2
0
0
+ =
⇔ + + =
⇔
v v v v v v
cosαα α β β γ γ1 2 1 2 1 2 0cos + + =cos cos cos cos
Exercise Set 3.3 159
27. (c) The norm of x is defined only for x a vector, but u • v is a scalar.
(d) Again, the dot product of a scalar and a vector is undefined.
28. Assume that neither a nor u is the zero vector.If projau = proj
ua, then
Thus, either u • a = 0 and the two vectors are orthogonal, or
In this case, a is clearly a scalar multiple of u. If we let a = ku, then
Hence, k = 1, so that a = u.
29. If, for instance, u = (1, 0, 0), v = (0, 1, 0) and w = (0, 0, 1), we have u • v = u • w = 0, butv ≠ w.
27. Suppose that r = c1u + c2v + c3w. Then
u • r = c1u • u + c2u • v + c3u • w = c1u2
since u • v = u • w = 0. Similarly, v • r = c2v–2 and w • r = c3w2. Since we know u • r,v • r, and w • r, this allows us to solve for c1, c2, and c3. We find that
31. This is just the Pythagorean Theorem.
u
u + vv
rruu rr
uuuu
vv rr
vvvv
ww rr
wwww= + +
⋅ ⋅ ⋅2 2 2
aa
aa
uu
uu
uu
uu2 2 2 2
= =k
k k
aa
aa
uu
uu2 2
=
uu aa
aaaa
aa uu
uuuu
uu aa
uuuu
⋅ ⋅ ⋅2 2 2
= =
160 Exercise Set 3.3
EXERCISE SET 3.4
1. (a)
(c) Since
we have
(e) Since
v – 2w = (0, 2, –3) – (4, 12, 14) = (–4, –10, –17)
we have
2. (a) By Theorem 3.4.1, u × v will be orthogonal to both u and v where
so (1, 2, –1), for instance, is orthogonal to both u and v.
3. (a) Since u × v = (–7, –1, 3), the area of the parallelogram is u × v = 59.
u v× = −− −
=, , ( , ,
4 2
1 5
6 2
3 5
6 4
3 118 36 )−18
u (v – 2w )× =−
− −−
−− − −
, ,2 1
10 17
3 1
4 17
3 2
4 −−
= − −
1044 55 22( , , )
(u v)× × = −− −
=, , (w
9 6
6 7
4 6
2 7
4 9
2 6277 40 42, , )−
u v× =−−
−−−
= −, , ( , ,
2 1
2 3
3 1
0 3
3 2
0 24 9 )6
v w× =−
−−
= −, , ( , ,
2 3
6 7
0 3
2 7
0 2
2 632 6 )−4
161
3. (c) Since u and v are proportional, they lie on the same line and hence the area of theparallelogram they determine is zero, which is, of course, u × v.
4. (a) We have u = →→ = (–1, –5, 2) and v = →→ = (2, 0, 3). Thus
so that
u × v = [(–15)2 + 72 + 102]1/2 = 374
Thus the area of the triangle is 374/2.
6. For Part (a), we have
and
Hence
u × v = –(v × u)
For Part (b), we have v + w = (7, 2, –3). Hence
On the other hand, from Part (a)we have
u × v = (2, 22, 6)
Also
u w× =−
−−
−−
= −, , ( ,
1 2
2 1
5 2
1 1
5 1
1 23 77 11, )
u (v w)× + =−
−−
−−
=, , (
1 2
2 3
5 2
7 3
5 1
7 2−−1 29 17, , )
v u× =−
−−
−−
= − −, , ( ,
0 2
1 2
6 2
5 2
6 0
5 12 222 6, )−
u v× =−
−−
−−
=, , ( , ,
1 2
0 2
5 2
6 2
5 1
6 02 22 )6
u v× =−
−− − −
= −, , ( ,
5 2
0 3
1 2
2 3
1 5
2 015 7,, )10
PR
PQ
162 Exercise Set 3.4
Thus
(u × v) + (u × w) = (2, 22, 6) + (–3, 7, 11)
= (–1, 29, 17)
and we have that
u × (v + w) = (u × v) + (u × w)
The proof of Part (c)is similar to the proof of Part (b).
For Part (d), we already have that u×v = (2, –22, 6). Hence k(u×v) = (–10, 110, –30).But ku = (–25, 5, –10) and kv = (–30, 0, 10). Thus
and
Thus
k(u × v) = (ku) × v = u × (kv)
For Part (e), we have
Similarly 0 × u = 0.
Finally, for Part (f), we have
u u× =−−
−−−
=, , ( , ,
1 2
1 2
5 2
5 2
5 1
5 10 0 )0 0=
u 0× =−
−−
=, , ( , ,
1 2
0 0
5 2
0 0
5 1
0 00 0 )0 0=
u ( v)× =−
−−
k , ,1 2
0 10
5 2
30 10
(– , – , – )
5 1
30 0
10 110 30
−−
=
( , ,ku) v× =−−
−− −
−−
5 10
0 2
25 10
6 2
25 5
6 0
= (– , – , – )10 110 30
Exercise Set 3.4 163
7. Choose any nonzero vector w which is not parallel to u. For instance, let w = (1, 0, 0) or(0, 1, 0). Then v = u × w will be orthogonal to u. Note that if u and w were parallel, thenv = u × w would be the zero vector.
Alternatively, let w = (x, y, z). Then w orthogonal to u implies 2x – 3y + 5z = 0. Nowassign nonzero values to any two of the variables x, y, and z and solve for the remainingvariable.
8. (a) We have
9. (e) Since (u × w) • v = v • (u × w) is a determinant whose rows are the components ofv, u, and w, respectively, we interchange Rows 1 and 2 to obtain the determinantwhich represents u • (v × w). Since the value of this determinant is 3, we have (u ×w) • v = –3.
10. (a) Call this volume V. Then, since
we have v = 16.
11. (a) Since the determinant
the vectors do not lie in the same plane.
12. For a vector to be parallel to the yz-plane, it must be perpendicular to the vector (1, 0, 0),
so we are looking for a vector which is perpendicular to both (1, 0, 0) and (3, –1, 2). Such
a vector is (1, 0, 0) × (3, –1, 2) = (0, –2, –1). Since we want a unit vector, we divide through
by the norm to obtain the vector (0, –2/ 5, –1/ 5). Obviously any vector parallel to this
one, such as (0, 2/ 5, 1/ 5), will also work.
− −−
−= ≠
1 2 1
3 0 2
5 4 0
16 0
2 6 2
0 4 2
2 2 4
24 2
2 42
6 2
4 216
−−−
=−−
+−
−= −
u v w⋅ × =−
−−
= − + −( ) ( ) (
1 2 4
3 4 2
1 2 5
20 4 2 15 −− + + = −2 4 6 4 10) ( )
164 Exercise Set 3.4
15. By Theorem 3.4.2, we have
(u + v) –(u – v) = u × (u – v) + v × (u – v)
= (u × u) + (u × (–v)) + (v × u) + (v × (–v))
= 0 – (u × v) – (u × v) – (v – v)
= –2(u × v)
17. (a) The area of the triangle with sides →→ and →→ is the same as the area of the trianglewith sides (–1, 2, 2) and (1, 1, –1) where we have “moved” A to the origin andtranslated B and C accordingly. This area is 12(–1, 2, 2) × (1, 1, –1) = 12(–4, 1, –3) =
26/2.
18. If the vector u and the given line make an angle θ, then the distance D between the pointand the line is given by
D = u sin θ (Why?)
= u × v/v (Why?)
19. (a) Let u = →→ = (–4,, 2) and v = →→ = (–3, 2, –4). Then the distance we want is
(–4, 0, 2) × (–3, 2, –4)/(–3, 2, –4) = (–4, –22, –8)/ 29. = 2 141/ 29
20. We have that u • v = u v cos θ and u × v = u v sin θ. Thus if u • v ≠ then u, v,and cos θ are all nonzero, and
21. (b) One vector n which is perpendicular to the plane containing v and w is given by
n = w × v = (1, 3, 3) × (1, 1, 2) = (3, 1, –2)
Therefore the angle φ between u and n is given by
φ =⋅
=
≈
− −cos cos
.
1 1 9
14
0 8726
uu nn
uu nn
rradians or( . )49 99
tansin
cosθ θ
θ= =
×=
×uu vv
uu vvuu vvuu vv
uu vv
uu vv
⋅⋅
AB
AP
AC
AB
Exercise Set 3.4 165
Hence the angle θ between u and the plane is given by
θ = π2 φ ≈ .6982 radians (or 40°19′′)
If we had interchanged the rôles of v and w in the formula for n so that n = v × w = (–3, –1, 2), then we would have obtained φ = cos–1(– 9
14) ≈ 2.269 radians or 130.0052°. In this case, θ = φ ≈ – π2.
In either case, note that θ may be computed using the formula
24. (a) By Theorems 3.4.2 and 3.2.1 and the definition of k0,
(u + kv) × v = (u × v) + (kv × v)
= (u × v) + k(v × v)
= (u × v) + k0
= u × v
(b) Let u = (u1, u2, u3), v = (v1, v2, v3), and z = (z1, z2, z3). Then we know from the textthat
Since x • y = y • x for any vectors x and y, we have
u (v z)⋅ × =u u u
v v v
z z z
1 2 3
1 2 3
1 2 3
θ sin .=⋅
−1 u nu n
166 Exercise Set 3.4
25. (a) By Theorem 3.4.1, we know that the vector v × w is perpendicular to both v and w.Hence v × w is perpendicular to every vector in the plane determined by v and w;moreover the only vectors perpendicular to v × w which share its initial point must bein this plane. But also by Theorem 3.4.1, u × (v × w) is perpendicular to v × w for anyvector u ≠ 0 and hence must lie in the plane determined by v and w.
(b) The argument is completely similar to Part (a), above.
26. Let u = (u1, u2, u3), v = (v1, v2, v3), and w = (w1, w2, w3). Following the hint, we find
v × i = (0, v3, –v2)
and
u × (v × i) = (–u2v2 – u3v3, u1v2, u1v3)
But
(u • i)v – (u • v)i = u1(v1, v2, v3) – (u1u1 + u2v2 + u3v3)(1, 0, 0)
= (u1v1, u1v2, u1v3) – (u1v1 + u2v2 + u3v3, 0, 0)
= (–u2v2 – u23v
3, u
1v
2, u
1v
3)
= u × (v × i)
Similarly,
u × (v × j) = (u • j)v – (u • v)j
and
u × (v × k) = (u • k)v – (u • v)k
(u z) v = v (u z) =× ⋅ ⋅ ×v v v1 2 33
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
u u u
u u u
v v v
–
z z z
z z z
= −
= ⋅u ((v z)×
Exercise Set 3.4 167
Now write w = (w1, w2, w3) = w1i + w2j + w3k. Then
u × (v × w) = u × (v × (w1i + w2j + w3k))
= w1[u × (v × i)] + w2[u × (v × j)] + w3[u × (v × k)]
= w1[(u • i)v – (u • v)i] + w2[(u • j)v –(u • v)j] + w3[(u • k)v – (u • v)k]
= (u • (w1i + w2j + w3k))v – (u • v)(w1i + w2j + w3k)
= (u • w)v – (u • v)w
29. If a, b, c, and d lie in the same plane, then (a × b) and (c × d) are both perpendicular tothis plane, and are therefore parallel. Hence, their cross-product is zero.
30. Recall that |a • (b × c)| represents the volume of the parallelepiped with sides a, b, and c,which is the area of its base times its height. The volume of the tetrahedron is just 13 (areaof its base) times (its height). The two heights are the same, but the area of the base of thetetrahedron is half of the area of the base of the parallelepiped. Hence, the volume of thetetrahedron is 13 (1
2|a • (b × c|).
31. (a) The required volume is
= 2/3
33. Let u = (u1, u2, u3), v = (v1, v2, v3), and w = (w1, w2, w3).
For Part (c), we have
u × w = (u2w3 – u3w2, u3w1 – u1w3, u1w2 – u2w1)
and
v × w = (v2w3 – v3w2, v3w1 – v1w3, u1w2 – v2w1)
1
61 3 2 2 0 3 2 3 1 2 3 3 1 3 0 2 1(– – , , – ) (( – , , – – ) ( – , ,+ ⋅ + × + –– ))
(– , , – ) ( , , )
3
1
64 4 3 6 10 4
( )
( )= ⋅
168 Exercise Set 3.4
Thus
(u × w) + (v × w)
= ([u2 + v2]w3 – [u3 + v3]w2, [u3 + v3]w1 – [u1 + u1]w3, [u1 + u1]w2 – [u2 + v2]w1)
But, by definition, this is just (u + v) × w.
For Part (d), we have
k(u × v) = (k[u2v3 – u3v2], k[u3u1 – u1v3], k[u1v2 – u2u1])
and
(ku) × v = (ku2v
3– ku
3v
2, ku
3u
1– ku
1v
3, ku
1v
2– ku
2v
1)
Thus, k(u × v) = (ku) × v. The identity k(u × v) = u × (kv) may be proved in an analogousway.
35. (a) Observe that u × v is perpendicular to both u and v, and hence to all vectors in theplane which they determine. Similarly, w = v × (u × v) is perpendicular to both v andto u × v. Hence, it must lie on the line through the origin perpendicular to v and in theplane determined by u and v.
(b) From the above, v • w = 0. Applying Part (d) of Theorem 3.7.1, we have
w = v × (u × v) = (v • v)u – (v • u)v
so that
u • w = (v • v)(u • u) – (v • u)(u • v)
= v2u2 – (u • v)2
36. It is not valid. For instance, let u = (1, 0, 0), v = (0, 1, 0), and w = (1, 1, 0). Then u × v =u × w = (0, 0, 1), which is a vector perpendicular to u, v, and w. However, v ≠ w.
37. The expression u • (v × w) is clearly well-defined.
Since the cross product is not associative, the expression u × v × w is not well-definedbecause the result is dependent upon the order in which we compute the cross products,i.e., upon the way in which we insert the parentheses. For example, (i × j) × j = k × j = –i
but i × (j × j) = i × 0 = 0.
Exercise Set 3.4 169
The expression u • v × w may be deemed to be acceptable because there is only onemeaningful way to insert parenthesis, namely, u • (v × w). The alternative, (u •v) –w, doesnot make sense because it is the cross product of a scalar with a vector.
38. If either u or v is the zero vector, then u × v = 0. If neither u nor v is the zero vector, thenu × v = u v sin θ where θ is the angle between u and v. Thus if u × v = 0, then sin θ= 0, so that u and v are parallel.
170 Exercise Set 3.4
EXERCISE SET 3.5
4. (a) We have
= (2, 1, 2) and = (3, –1, –2)
Thus × = (0, 10, –5) is perpendicular to the plane determined by and
and P, say, is a point in that plane. Hence, an equation for the plane is
0(x + 4)+ 10(y + 1) – 5(z + 1) = 0
or
2y – z + 1 = 0
5. (a) Normal vectors for the planes are (4, –1, 2) and (7, –3, 4). Since these vectors are notmultiples of one another, the planes are not parallel.
(b) The normal vectors are (1, –4, –3) and (3, –12, –9). Since one vector is three times theother, the planes are parallel.
6. (a) A normal vector for the plane is (1, 2, 3) and a direction vector for the line is (–4, –1,2). The inner product of these two vectors is –4 –2 + 6 = 0, and therefore they areperpendicular. This guarantees that the line and the plane are parallel.
7. (a) Normal vectors for the planes are (3, –1, 1) and (1, 0, 2). Since the inner product ofthese two vectors is not zero, the planes are not perpendicular.
8. (a) A normal vector for the plane is (2, 1, –1) and a direction vector for the line is (–4, –2,2). Since one of these vectors is a multiple of the other, the line and the plane areperpendicular.
PR PQ
PR
PQ
PR
PQ
171
10. (a) Call the points P and Q and call the line . Then the vector = (2, 4, –8) is parallelto and the point P = (5, –2, 4) lies on . Hence, one set of parametric equations for is: x = 5 + t, y = –2 + 2t, z = 4 –4t where t is any real number.
11. (a) As in Example 6, we solve the two equations simultaneously. If we eliminate y, wehave x + 7z + 12 = 0. Let, say, z = t, so that x = –12 – 7t, and substitute these valuesinto the equation for either plane to get y = –41 – 23t.
Alternatively, recall that a direction vector for the line is just the cross-product ofthe normal vectors for the two planes, i.e.,
(7, –2, 3) × (–3, 1, 2) = (–7, –23, 1)
Thus if we can find a point which lies on the line (that is, any point whose coordinatessatisfy the equations for both planes), we are done. If we set z = 0 and solve the twoequations simultaneously, we get x = –12 and y = –41, so that x = –12 – 7t, y = –41 –23t, z = 0 + t is one set of equations for the line (see above).
13. (a) Since the normal vectors (–1, 2, 4) and (2, –4, –8) are parallel, so are the planes.
(b) Since the normal vectors (3, 0, –1) and (–1, 0, 3) are not parallel, neither are theplanes.
14. (a) Since (–2, 1, 4) • (1, –2, 1) = 0, the planes are perpendicular.
(b) Since (3, 0, –2) • (1, 1, 1) = 1 ≠ 0, the planes are not perpendicular.
16. (a) Since 6(0) + 4t – 4t = 0 for all t, it follows that every point on the line also lies in theplane.
(b) The normal to the plane is n = (5, –3, 3); the line is parallel to v = (0, 1, 1). But n •v = 0, so n and v are perpendicular. Thus v, and therefore the line, are parallel to theplane. To conclude that the line lies below the plane, simply note that (0, 0, 1/3) is inthe plane and (0, 0, 0) is on the line.
(c) Here n = (6, 2, –2) and v is the same as before. Again, n • v = 0, so the line and theplane are parallel. Since (0, 0, 0) lies on the line and (0, 0, –3/2) lies in the plane,then the line is above the plane.
17. Since the plane is perpendicular to a line with direction (2, 3, –5), we can use that vectoras a normal to the plane. The point-normal form then yields the equation 2(x + 2) + 3(y –1)– 5(z –7) = 0, or 2x + 3y – 5z + 36 = 0.
PQ
172 Exercise Set 3.5
19. (a) Since the vector (0, 0, 1) is perpendicular to the xy-plane, we can use this as thenormal for the plane. The point-normal form then yields the equation z – z0 = 0. Thisequation could just as well have been derived by inspection, since it represents the setof all points with fixed z and x and y arbitrary.
20. The plane will have normal (7, 4, –2), so the point-normal form yields 7x + 4y – 2 z = 0.
21. A normal to the plane is n = (5, –2, 1) and the point (3, –6, 7) is in the desired plane.Hence, an equation for the plane is 5(x – 3) – 2(y + 6) + (z –7) = 0 or 5x – 2y + z – 34 =0.
22. If we substitute x = 9 – 5t, y = –1 – t, and z = 3 + t into the equation for the plane, we findthat t = 40/3. Substituting this value for t into the parametric equations for the line yieldsx = –173/3, y = –43/3, and z = 49/3.
24. The two planes intersect at points given by (–5 – 2t, –7 – 6t, t). Two such points are (–5, –7,0) and (–3, –1, –1). The plane ax + by + cz + d = 0 through these points and also through(2, 4, –1) will satisfy the equations
–5a – 7b + d = 0
–3a – b – c + d = 0
2a + 4b – c + d = 0
This system of equations has the solution a = (–1/2)t, b = (1/2)t, c = 2t, d = t where t isarbitrary. Thus, if we let t = –2, we obtain the equation
x – y – 4z – 2 = 0
for the desired plane.
Alternatively, note that the line of intersection of the two planes has direction given bythe vector v = (–2, –6, 1). It also contains the point (–5, –7, 0). The direction of the lineconnecting thispoint and (2, 4, –1) is given by w = (7, 11, –1). Thus the vector v – w = (5,–5, –20) is normal to the desired plane, so that a point-normal form for the plane is
(x – 2) – (y – 4) – 4(z + 1) = 0
or
x – y –4z – 2 = 0
Exercise Set 3.5 173
25. Call the points A, B, C, and D, respectively. Since the vectors = (–1, 2, 4) and =(–2, –1, –2) are not parallel, then the points A, B, and C do determine a plane (and not justa line). The normal to this plane is × = (0, –10, 5). Therefore an equation for theplane is
2y – z + 1 = 0
Since the coordinates of the point D satisfy this equation, all four points must lie in thesame plane.
Alternatively, it would suffice to show that (for instance) × and ×are parallel, so that the planes determined by A, B, and C and A, D, and C are parallel.Since they have points in common, they must coincide.
27. The normals to the two planes are (4, –2, 2) and (3, 3, –6) or, simplifying, n1 = (2, –1, 1)and n2 = (1, 1, –2). The normal n to a plane which is perpendicular to both of the givenplanes must be perpendicular to both n1 and n2. That is, n = n1 × n2 = (1, 5, 3). The planewith this normal which passes through the point (–2, 1, 5) has the equation
(x + 2) + 5(y – 1) + 3(z – 5) = 0
or
x + 5y + 3z – 18 = 0
28. The line of intersection of the given planes has the equations x = – 109 – t3 , y = 31
18 – t3, z = t and hence has direction (1, 1, –3). The plane with this normal vector which passesthrough the point (2, –1, 4) is the one we are looking for. Its equation is
(x – 2) + (y + 1) – 3(z – 4) = 0
Note that the normal vector for the desired plane can also be obtained by computing thecross product of the normal vectors of the two given planes.
30. The directions of the two lines are given by the vectors (–2, 1, –1) and (2, –1, 1). Sinceeach is a multiple of the other, they represent the same direction. The FIrst line passes(for example) through the point (3, 4, 1) and the second line passes through the points (5,1, 7) and (7, 0, 8), among others. Either of the methods of Example 2 will yield an equationfor the plane determined by these points.
31. If, for instance, we set t = 0 and t = –1 in the line equation, we obtain the points (0, 1, –3)and (–1, 0, –5). These, together with the given point and the methods of Example 2, willyield an equation for the desired plane.
DC
AD
BC
AB
BC
AB
BC
AB
174 Exercise Set 3.5
33. The plane we are looking for is just the set of all points P = (x, y, z) such that the distancesfrom P to the two fixed points are equal. If we equate the squares of these distances, wehave
(x + 1)2 + (y + 4)2 + (z + 2)2 = (x – 0)2 + (y + 2)2 + (z – 2)2
or
2x + 1 + 8y + 16 + 4z + 4 = 4y + 4 – 4z + 4
or
2x + 4y + 8z + 13 = 0
34. The vector v = (–1, 2, –5) is parallel to the line and the vector n = (–3, 1, 1) isperpendicular to the plane. But n • v = 0 and the result follows.
35. We change the parameter in the equations for the second line from t to s. The two lines willthen intersect if we can find values of s and t such that the x, y, and z coordinates for thetwo lines are equal; that is, if there are values for s and t such that
4t + 3 = 12s – 1
t + 4 = 6s + 7
1 = 3s + 5
This system of equations has the solution t = –5 and s = –4/3. If we then substitute t = –5into the equations for the first line or s = –4/3 into the equations for the second line, we findthat x = –17, y = –1, and z = 1 is the point of intersection.
36. The vector v1 = (4, 1, 0) is parallel to the first line and v2 = (4, 2, 1) is parallel to thesecond line. Hence n = v1 × v2 = (1, –4, 4) is perpendicular to both lines and is therefore anormal vector for the plane determined by the lines. If we substitute t = 0 into theparametric equations for the first line, we see that (3, 4, 1) must lie in the plane. Hence, anequation for the plane is
(x – 3) – 4(y – 4) + 4(z – 1) = 0
or
x – 4y + 4z + 9 = 0
Exercise Set 3.5 175
37. (a) If we set z = t and solve for x and y in terms of z, then we find that
38. Call the plane Ax + By + Cz + D/b = 0. Since the points (a, 0, 0), (0, b, 0), and (0, 0, c)
lie in this plane, we have
aA + D = 0
bB + D = 0
cC + D = 0
Thus A = –D/a, B = –D/b, and c = –D/c and an equation for the plane is
Alternatively, let P, Q, and R denote the points (a, 0, 0), (0, b, 0), and (0, 0, c),respectively. Then
= (–a, b, 0) and = (0, –b, c)
So
n = × = (bc, ac, ab)
is a normal vector for the plane. Since P, say, lies on the plane, we have
bc(x – a)+ acy + abz = 0
Dividing the above equation by abc yields the desired result.
39. (b) By Theorem 3.5.2, the distance is
D =−( ) + ( ) − ( ) −
+ + −( )=
2 1 3 2 4 1 1
2 3 4
1
292 2 2
QR
PQ
QR
PQ
x
a
y
b
z
c+ + = 1
x t y t z t= + = − − =11
23
7
23
41
23
1
23, ,
176 Exercise Set 3.5
40. (a) The point (0, 0, 1) lies in the first plane. The distance between this point and thesecond plane is
(b) The two planes coincide, so the distance between them is zero.
41. (a)
(b)
(c)
42. First observe that if we substitute the values for x, y, and z from the line equations into thesymmetric equations, we obtain the equations t = t = t, which hold for all values of t. Thusevery point on the line satisfies the symmetric equations.
It remains to show that every point (x, y, z) which satisfies the symmetric equationsalso lies on the line. That is, we must find a value of t for which the line equations hold.Clearly
44. (a) The symmetric equations for the line are
Thus, two of the planes are given by
or, equivalently,
x y x z− − = + − =2 17 0 4 27 0and
x y x z−=
+ −=
−−
7
4
5
2
7
4
5
1and
x y z−=
+=
−−
7
4
5
2
5
1
tx x
a
y y
b
z z
c=
−=
−=
−0 0 0
d
d
d
=
=
=
30
11
382
110, point is on the line
D =−
+ −( ) + ( )=
2 3
6 8 2
1
2 262 2 2
Exercise Set 3.5 177
45. (a) The normals to the two planes are (1, 0, 0) and (2, –1, 1). The angle between them isgiven by
Thus θ = cos–1 (2/ 6) ≈ 35°15′52′′.
47. If we substitute any value of the parameter—say t0—into r = r0 + tv and –t0 into r = r0 – tv,we clearly obtain the same point. Hence, the two lines coincide. They both pass through thepoint r0 and both are parallel to v.
47. Since the vector (a, b, c) is both the normal vector to the plane and the direction vector forthe line, the line must be perpendicular to the plane.
48. The equation r = (1 – t)r1 + tr2 can be rewritten as r = r1 + t(r2 – r1). This represents a linethrough the point P1 with direction r2 – r1. If t = 0, we have the point P1. If t = 1, we havethe point P2. If 0 < t < 1, we have a point on the line segment connecting P1 and P2. Hencethe given equation represents this line segment.
cos, , , ,
θ =( ) ⋅ −( )
+ +=
1 0 0 2 1 1
1 4 1 1
2
6
178 Exercise Set 3.5
EXERCISE SET 4.1
3. We must find numbers c1, c2, c3, and c4 such that
c1(–1, 3, 2, 0) + c2(2, 0, 4, –1) + c3(7, 1, 1, 4) + c4(6, 3, 1, 2) = (0, 5, 6, –3)
If we equate vector components, we obtain the following system of equations:
–c1 + 2c2 + 7c3+ 6c4 = 0
3c1 + c3 + 3c4 = 5
2c1 + 4c2 + c3 c4 = 6
–c2 + 4c3 2c4 = 3
The augmented matrix of this system is
The reduced row-echelon form of this matrix is
Thus c1 = 1, c2 = 1, c3 = –1, and c4 = 1.
1 0 0 0 1
0 1 0 0 1
0 0 1 0 1
0 0 0 1 1
−
−
− −
1 2 7 6 0
3 0 1 3 5
2 4 1 1 6
0 1 4 2 3
179
4. If we equate the second vector components in this equation, we have 0c1 + 0c2 + 0c3 = – 2.Hence, there do not exist scalars c1, c2, and c3 which satisfy the given equation.
5. (c) v = [32 + 42 + 02 + (–12)2]1/2 = 169 = 13
6. (a) u + v = (4, 4, 10, 1) = 42 + 42 + 102 + 12]1/2 = 133
(c) –2u + 2u = [(–8)2 + (–2)2 + (–4)2 + (–6)2]1/2 + 2 [42 + 12 + 22 + 32]1/2
= 1201/2 + 2[30]1/2 = 4 30
(e)
8. Since kv = [(–2k)2 + (3k)2 + 02 + (6k)2]1/2 = [49k2]1/2 = 7|k|, we have |kv| = 5 if and onlyif k = ±5/7.
9. (a) (2,5) • (–4,3) = (2)(–4) + (5)(3) = 7
(c) (3, 1, 4, –5) • (2, 2, –4, –3) = 6 + 2 –16 + 15 = 7
10. (a) Let v = (x, y) where v = 1. We are given that v • (3, –1) = 0. Thus, 3x – y = 0 or y
= 3x. But v = 1 implies that x2 + y2 = x2 + 9x2 = 1 or x = ±1/ 10. Thus, the only
possibilities are v = (1/ 10, 3/ 10) or v = (–1/ 10, –3/ 10).
You should graph these two vectors and the vector (3, –1) in an xy-coordinate system.
(b) Let v = (x, y, z) be a vector with norm 1 such that
x – 3y + 5z = 0
This equation represents a plane through (0,0,0) which is perpendicular to (1, –3, 5).There are infinitely many vectors v which lie in this plane and have norm 1 and initialpoint (0,0,0).
1 1
3 1 2 23 1 2 2
1
22 2 2 2 1 2ww =
+ + +
( ) =, , , ,/
, ,1
3 2
2
3
2
3
180 Exercise Set 4.1
11. (a) d(u,v) = [(1 – 2)2 + (–2 – 1)2]1/2 = 10
(c) d(u, v) = [(0 + 3)2 + (–2 – 2)2 + (–1 – 4)2 + (1 – 4)2]1/2 = 59
14. (e) Since u • v = 0 + 6 + 2 + 0 = 8, the vectors are not orthogonal.
15. (a) We look for values of k such that
u • v = 2 + 7 + 3k = 0
Clearly k = –3 is the only possiblity.
16. We must find two vectors x = (x1, x2, x3, x4) such that x • x = 1 and x • u = x • v = x • w
= 0. Thus x1, x2, x3, and x4 must satisfy the equations
x21 + x2
2 + x23 + x2
4 = 1
2x1 + x2 – 4x3 = 0
–x1 – x2 + 2x3+ 2x4 = 0
3x1+ 2x + 5x3+ 4x4 = 0
The solution to the three linear equations is x1 = –34t, x2 = 44t, x3 = –6t, and x4 = 11t. If wesubstitute these values into the quadratic equation, we get
[(–34)2 + (44)2 + (–6)2 + (11)2] t2 = 1
or
Therefore, the two vectors are
± 157(–34, 44, –6, 11)
t = ± = ±1
3249
1
57
Exercise Set 4.1 181
17. (a) We have |u • v| = |3(4) + 2(–1)| = 10, while
u v = [32 + 22]1/2[42 + (–1)2]1/2 = 221.
(d) Here |u • v| = 0 + 2 + 2 + 1 = 5, while
u v = [02 + (–2)2 + 22 + 12]1/2[(–1)2 + (–1)2 + 12 + 12]1/2 = 6.
18. (a) In this case
and
20. By Theorem 4.1.6, we have u • v = 14 u + v2 – 1
4 u – v2 = 14 (1)2 – 1
4 (52) = –6.
22. Note that u • a = 4 and a2 = 15. Hence, by Theorem 3.3.3, we have
and
uu uu uuuu aa
aaaa− = −
⋅= −( ) − −proja 2
2 1 4 14
15
4
15
8
15
4, , , , , ,
55
34
15
11
15
52
15
9
5
= −
, , ,
proja 2uu
uu aa
aaaa=
⋅= −( )4
151 1 2 3, , ,
AT u . v =
−
= −
9
1
2
612
u . vA =
⋅
−
= −
3
1
10
1812
u v. AT =
⋅
−
−
3
1
2 3
1 4
2
6 =
⋅
=
3
1
14
2668
Au . v =
⋅
−
=
5
13
2
668
182 Exercise Set 4.1
23. We must see if the system
3 + 4t = s
2 + 6t = 3 – 3s
3 + 4t = 5 – 4s
–1 – 2t = 4 – 2s
is consistent. Solving the first two equations yield t = – 4/9, s = 11/9. Substituting into 2rdequation yields 5/3 = 1/9. Thus the system is inconsistent, so the lines are skew.
24. Note that
v1 + … + vr2 = (v1 + …+ v
r) • (v1 + … + v
r)
If vi •vj = 0 whenever i ≠ j, then the result follows.
25. This is just the Cauchy-Schwarz Inequality applied to the vectors vT AT and uT AT withboth sides of the inequality squared. Why?
26. If we apply the Cauchy-Schwarz Inequality to the vectors u = (a, b) and v = (cos θ, sin θ),the result follows directly.
27. Let u = (u1, …, un), v = (v1, …, v
n), and w = (w1, …, w
n).
(a) u • (kv) = (u1, …, un) • (kv1, …, kv
n)
= u1 kv1 + … + un
kvn
= k(u1v1 + … + u
nv
n)
= k(u • v)
= ⋅ + ⋅
= + + + ⋅
= ≠
≠
∑ ∑
∑
v v v v
v v v v
ii
r
i i ji j
r i ji j
1
12 2
Exercise Set 4.1 183
27. (b) u • (v + w) = (u1, …, un) • (v1 + w1, …, v
n+ w
n)
= u1(v1 + w1) + … + un
(vn
+ wn)
= (u1v1 + … + unv
n) + (u1w1 + … + u
nw
n)
= u • v + u • w
30. Let u = (u1, …, un) and v = (v1, …, v
n).
(a) u • v = u1v1 + … + unv
n= v1u1 + … + v
nu
n= v • u
(c) (ku) • v = (ku1) + … + (kun)v
n= k(u
1v
1+ … + unvn) = k(u • v)
32. Let u = (u1, u2, …, un) and v = (v1, v2, …, v
n).
(a) Since d(u, v) is a square root of a sum of squares, it cannot be negative.
(b) Since d(u, v) = [(u1 – v1)2 + (u2 – v2)
2 + … + (un
– vn)2]1/2 = if and only if u
i– v
i= 0
for i = 1, 2, …, n, we have d(u, v) = 0 if and only if u = v.
(c) Here d(u, v) = [(u1 – v1)2 + … + (u
n– v
n)2]1/2 = [(v1 – u1)
2 + … + (vn
– un)2]1/2 = d
(v, u).
34. The result follows from the equations used to prove Theorem 4.1.6. It says that the sum ofthe squares of the lengths of the diagonals of the parallelogram with sides u and v is thesum of the squares of the lengths of the 4 sides.
35. (a) By theorem 4.1.7, we have
36. Since u – v2 = (u – v) • (u – v) = u2 + v>2 – 2(u • v) for all vectors u and v in Rn, we
need only show that for the formula to hold. Since this result is valid for cos
θ in R2 and R3, it would seem to be the logical way to define cos θ and hence θ in Rn for n
> 3. To show that this makes sense, we must check that so that cos θ is well
defined. But the above inequality is just the Cauchy-Schwarz inequality in Rn. Thus the
formula is valid in Rn.
uu vv
uu vv
⋅≤ 1
cosθ =⋅uu vv
uu vv
d( , ) .uu vv uu vv uu vv= + = + − =2 22
184 Exercise Set 4.1
37. (a) True. In general, we know that
u + v2 = u2 + v2 + 2(u • v)
So in this case u • v = 0 and the vectors are orthogonal.
(b) True. We are given that u • v = u • w = 0. But since u • (v + w) = u • v + u • w, itfollows that u is orthogonal to v + w.
(c) False. To obtain a counterexample, let u = (1, 0, 0), v = (1, 1, 0), and w = (–1, 1, 0).
Exercise Set 4.1 185
EXERCISE SET 4.2
1. (b) Since the transformation maps (x1, x2) to (w1, w2, w3), the domain is R2 and thecodomain is R3. The transformation is not linear because of the terms 2x1x2 and 3x1x2.
2. (a) Since we can write the transformation as
3. The standard matrix is A, where
so that
T( , , )− =−
−−
−
1 2 4
3 5 1
4 1 1
3 2 1
1
2
4
= −−
3
2
3
w = Ax =−
−−
3 5 1
4 1 1
3 2 1
1
2
3
x
x
x
the standard matrix is2 3 0 1
3 5 0 1
−−
w
w
x
x
x
x
1
2
1
2
3
4
2 3 0 1
3 5 0 1
=
−−
187
4. (a) The standard matrix is
It is interesting to note that T(1, 0) = (2, 1) and T(0, 1) = (–1, 1).
5. (a) The standard matrix is
Note that T(1, 0) = (0, –1, 1, 1) and T(0, 1) = (1, 0, 3, –1).
6. (b) We have
7. (b) Here
8. (c) Here
so the reflection of (–1, 2) is (2, –1).
T( , )− =
−
=
−
1 2
0 1
1 0
1
2
2
1
T( , )2 1
2 1 1
0 1 1
0 0 0
2
1
3
3− =
−
−
= −
0
2
0
T( )x =−
−
=
1 2 0
3 1 5
1
1
3
3
13
0 1
1 0
1 3
1 1
−
−
2 1
1 1
−
188 Exercise Set 4.2
9. (a) In this case,
so the reflection of (2, –5, 3) is (2, –5, –3).
10. (a)
12. (b) The image of (3, –4) is
(d) The image of (3, –4) is (4, 3), since
13. (b) The image of (–2, 1, 2) is (0, 1, 2 2), since
cos sin
sin cos
45 0 45
0 1 0
45 0 45
°( ) °( )
− °( ) °( )
−
=
−
2
1
2
1
20
1
20 1 0
1
20
1
2
−
=
2
1
2
0
1
2 2
cos sin
sin cos
90 90
90 90
3°( ) − °( )°( ) °( )
−−
=
−
−
=
4
0 1
1 0
3
4
4
3
or3 4 3
2
4 3 3
2
− +
, .
cos sin
sin cos
60 60
60 60
3
4
°( ) °( )− °( ) °( )
−
=−
−
=
1
2
3
2
3
2
1
2
3
4
3 4 3
2
4 3 3
2
−
−+
The desired projection is1 0
0 0
2
5
−
=
, , ).
2
092 0or
T ( , , )2 5 3
1 0 0
0 1 0
0 0 1
2
5
3
− =−
−
= −−
2
5
3
Exercise Set 4.2 189
15. (b) The image of (–2, 1, 2) is (0, 1, 2 2), since
16. (b)
17. (a) The standard matrix is
(c) The standard matrix for a counterclockwise rotation of 15° + 105° + 60° = 180° is
18. (a) The standard matrix is
1 0 0
0 0 0
0 0 1
1 0 0
0 1 0
0 0 1
−
=−−
1 0 0
0 0 0
0 0 1
cos sin
sin cos
180 180
180 180
°( ) − °( )°( ) °( )
=−
−
1 0
0 1
0 1
1 0
1 0
0 0
1
2
3
2
3
2
1
2
−
=−
0 0
1
2
3
2
The standard matrix is1 2 0
0 1 2
0 0
0 1
=
.
0 0
0 1 2
=
−
−
1
20
1
20 1 0
1
20
1
2
2
1
2
=
0
1
2 2
cos sin
sin cos
− °( ) − − °( )
− °( ) − °( )
45 0 45
0 1 0
45 0 45
−
2
1
2
190 Exercise Set 4.2
(b) The standard matrix is
19. (c) The standard matrix is
20. (a) Projection on either axis followed by projection on the other will yield the zero vector.Therefore the transformations commute. In matrix form, we have
(c) Geometrically, it is clear that the transformations do not commute. In matrix form, wehave
[ ] [ ]cos sin
sin cosT T1 2
1 0
0 0=
−
θ θθ θ
==−
cos sinθ θ0 0
[ ] [ ]T T1 21 0
0 0
0 0
0 1
0 0
0 0=
=
[ ] [ ]=
=
0 0
0 1
1 0
0 0 2 1T T
= −−
0 1 0
0 0 1
1 0 0
=−
−
−
1 0 0
0 1 0
0 0 1
0 0 1
0 1 0
1 0 0
1 0 0
0 0 1
0 1 0−
cos sin
sin cos
180 180 0
180 180 0
0 0 1
°( ) − °( )°( ) °( )
°( ) °( )
− °( )
cos sin
sin c
90 0 90
0 1 0
90 0 oos
cos sin
90
1 0 0
0 270 270
°( )
°( ) − °( )00 270 270sin cos°( ) °( )
2 0 0
0 2 0
0 0 2
1
20
1
20 1 0
1
20
1
2
−
=−
1 0 1
0 2 0
1 0 1
Exercise Set 4.2 191
and
21. (a) Geometrically, it doesn’t make any difference whether we rotate and then dilate orwhether we dilate and then rotate. In matrix terms, a dilation or contraction isrepresented by a scalar multiple of the identity matrix. Since such a matrix commuteswith any square matrix of the appropriate size, the transformations commute.
22. (a) We consider only T1. It is linear because it can be represented by the matrix
(b) Again we consider only T1. We have T1(x) = (x, 0, 0) and x – T1(x) = (x, y, z) – (x,0, 0) = (0, y, z), so that T1(x) • (x – T1x) = 0 and the vectors are orthogonal.
(c)
23. Set (a, b, c) equal to (1, 0, 0), (0, 1, 0), and (0, 0, 1) in turn.
24.
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
31
3
1
3
1
3
− +
+ −
− +1
3
1
3
−T(x)
x − T(x) x
x
y
T(x)
1 0 0
0 0 0
0 0 0
[ ] [ ]cos sin
sin cosT T2 1
1 0
0 0=
−
θ θθ θ =
cos
sin
θθ
0
0
192 Exercise Set 4.2
25. (a) Since T2(T1(x1, x2)) = (3(x1 + x2), 2(x1 + x2) + 4(x1 – x2)) = (3x1 + 3x2, 6x1 – 2x2),we have
We also have
27. Compute the trace of the matrix given in Formula (17) and use the fact that (a, b, c) is aunit vector.
28. (a) By inspection, c1 • c2 = c1 • c3 = c2 • c3 = 0 and c1 = c2 = c3 = 1 where c1, c2, andc3 are the column vectors of A. It is also easy to show that det(A) = 1.
(b) If we let , for instance, an axis of rotation is given by the vector
Thus 13, is an appropriate vector of length 1.
(c) From Exercise 27, we have . Substituting
into Formula (17), we find that θ = 90° + 360°n for n = 0, 1, 2, … .
cos , , , ,θ =−
= ° + ° =1 1
20 90 180 0 1 2so forn n
1
3
2
3
2
3, ,
=
−
+ −
1
98
94
9
1
94
98
9
=
=
2
94
94
9
2
33
1
32
32
3
u =
−
−
1
9
4
9
8
98
9
4
9
1
94
9
7
9
4
9
1
00
0
1
9
8
9
4
94
9
4
9
7
98
9
1
9
4
9
+
−
−
+ −
1
0
0
1 1
1
0
0
x =
1
0
0
[ ] [ ]T T2 13 0
2 4
1 1
1 1
3 3
6 2=
−
=
−
[ ]T T2 13 3
6 2 =
−
Exercise Set 4.2 193
29. (a) This is an orthogonal projection on the x-axis and a dilation by a factor of 2.
(b) This is a reflection about the x-axis and a dilation by a factor of 2.
30. (a) This transformation dilates the x coordinates by a factor of 2 and the y coordinates bya factor of 3.
(b) This is a rotation through angle of π/6.
31. Since cos(2θ) = cos2 θ – sin2 θ and sin(2θ) = 2 sin θ cos θ, this represents a rotationthrough an angle of 2θ.
32. We have
which represents a rotation through an angle –θ.
34. No, f(0) = b ≠ 0 unless b = 0; then f(x) = mx + b is not a linear function.
AT =
−
=
−( ) − −(cos sin
sin cos
cos sinθ θθ θ
θ θ ))−( ) −( )
sin cosθ θ
194 Exercise Set 4.2
EXERCISE SET 4.3
1. (a) Projections are not one-to-one since two distinct vectors can have the same imagevector.
(b) Since a reflection is its own inverse, it is a one-to-one mapping of R2 or R3 onto itself.
2. (a) The standard matrix is . Since det(A) = 0, A is not invertible and hence
the operator is not one-to-one.
(b) The standard matrix is . Since det(A) = 17 ≠ 0, A is invertible and hence
the operator is one-to-one.
3. If we reduce the system of equations to row-echelon form, we find that w1 = 2w2, so that anyvector in the range must be of the form (2w, w). Thus (3, 1), for example, is not in the range.
4. The range satisfies w1 – w2 + w3 = 0
Thus (1, 1, 1) Range.
5. (a) Since the determinant of the matrix
is 3, the transformation T is one-to-one with
Thus T w w w w w w− = − +
11 2 1 2 1 2
1
3
2
3
1
3
1
3( , ) , .
[ ]T− =
−
1
1
3
2
31
3
1
3
[ ]T =−
1 2
1 1
A =−
2 3
5 1
A =
8 4
2 1
195
5. (b) Since the determinant of the matrix
is zero, T is not one-to-one.
6. (a) Since the matrix
is invertible, T is one-to-one. Since
T–1(w1, w2, w3) = (w1 – 2w2 + 4w3, –w1 + 2w2 – 3w3, –w1 + 3w2 – 5w3)
(b) Since the matrix
is not invertible, T is not one-to-one.
8. (a) T is linear since
T((x1, y1) + (x2, y2))= (2(x21 + x2), y1 + y2)
= (2x1, y1) + (2x2, y2)
= T(x1, y1) + T(x2, y2)
and
[ ]T =−
−−
1 3 4
1 1 1
0 2 5
[ ]T− =
−− −− −
11 2 4
1 2 3
1 3 5
[ ]T =−
1 2 2
2 1 1
1 1 0
[ ]T =−
−
4 6
2 3
196 Exercise Set 4.3
T(k(x, y))= T(kx, ky) = (2kx, ky)
= k(2x, y) = kT(x, y)
(b) T is not linear since
T((x1, y1) + (x2, y2)) = (x1 + x2)2, y1 + y2
= (x21, y1) + (2x1x2, 0) + (x2
2, y2)
= T(x1, y1) + T(x2, y2) + (2x1x2, 0)
≠ T(x1, y1) + T(x2, y2) if x1x2 ≠ 0
and
T(k(x,y)) = (k2x2, ky) = k(kx2, y)
≠ kT(x, y) if k ≠ 0, 1 and x ≠ 0
9. (a) T is linear since
T((x1, y1) + (x2, y2) ) = (2(x1 + x2) + (y1 + y2), (x1 + x2) – (y1 + y2))
= (2x1 + y1, x1 – y1) + (2x2 + y2, x2 – y2)
= T(x1, y1) + T(x2, y2)
and
T(k(x, y)) = (2kx + ky, kx – ky)
= k(2x + y, x – y) = kT(x, y)
(b) Since
T((x1, y1) + (x2, y2))= (x1 + x2 + 1, y1 + y2)
= (x1 + 1, y1) + (x2, y2)
≠ T(x1, y1) + T(x2, y2)
and T(k(x, y)) = (kx + 1, ky) ≠ kT (x, y) unless k = 1, T is nonlinear.
Exercise Set 4.3 197
10. (a) T is linear since
T((x1, y1, z1) + (x2, y2, z2)) = (x1 + x2, x1 + x2 + y1 + y2 + z1 + z2)
= (x1, x1 + y1 + z1) + (x2, x2 + y2 + z2)
= T(x1, y1, z1) + T(x2, y2, z2)
and
T(k(x, y, z)) = (kx, kx + ky + kz)
= kT(x, y, z)
(b) Since for all vectors u and v,
T(u + v) = (1, 1) ≠ T(u) + T(v) = (2, 2)
and
T(ku) = (1, 1) ≠ kT(u) = (k, k) if k = 1
T is nonlinear.
13. (a) The projection sends e1 to itself and the reflection sends e1 to –e1, while the projection
sends e2 to the zero vector, which remains fixed under the reflection. Therefore
T(e1) = (–1, 0) and T(e2) = (0, 0), so that
(b) We have e1 = (1, 0) → (0, 1) → (0, –1) = 0e1 – e2 while e2 = (0, 1) → (1, 0) → (1, 0)
= e1 + 0e2. Hence
(c) Here e1 = (1, 0) → (3, 0) → (0, 3) → (0, 3) = 0e1 + 3e2 and e2 = (0, 1) → (0, 3) →
(3, 0) → (0, 0) = 0e1 + 0e2. Therefore
14. (a) Here e1 = (1, 0, 0) → (1, 0, 0) → (1/5, 0, 0) = 15e1, e2 = (0, 1, 0) → (0, –1, 0) → (0,
–1/5, 0) = – 15e2, and e3 = (0, 0, 1) → (0, 0, 1) → (0, 0, 1/5) = 15e3. Therefore
[ ]T = −
1 5 0 0
0 1 5 0
0 0 1 5
[ ]T =
0 0
3 0
[ ]T =−
0 1
1 0
[ ]T =−
1 0
0 0
198 Exercise Set 4.3
16. (c) The mapping is NOT 1 – 1.
For example,
yet
17. (a) By the result of Example 5,
or T(–1, 2) = (1/2, 1/2)
18. (a) This transformation maps vectors on the x-axis to themselves, vectors on the y-axisinto their negatives, and maps no other vectors into scalar multiples of themselves.Thus the eigenvalues are λ = ±1 and the eigenvectors are vectors (x, y) with either x = 0 or y = 0, but not both. To verify this, we observe that since e1 → e1 and
e2 → –e1, the standard matrix for the transformation is . Hence the char-acteristic equation is
or λ2 – 1 = 0 with solutions λ = ±1. If (x, y) is an eigenvector corresponding to λ = 1,then
or y = 0, so the vector must lie on the x-axis. If (x, y) is an eigenvector correspondingto λ = –1, then
or x = 0, so the vector must lie on the y-axis.
−
=
−
=
2 0
0 0
2
0
0
0
x
y
x
0 0
0 2
0
2 0
=
=
x
y y
0
det λ1 0
0 1
1 0
0 10
−
−
=
1 0
0 1−
T−
=
( ) ( )( )( )( ) (
1
2
1 2 1 2 1 2
1 2 1 2 1 2
2
))
−
=
2
1
2
1 2
1 2
A
1
1
1
0
0
0
0
04 1−
=
= ×
A
0
0
0
0
0
0
0
04 1
=
= ×
Exercise Set 4.3 199
18. (c) This transformation leaves vectors on the x-axis fixed and maps vectors on the y-axisto (0, 0). No other vectors are mapped into scalar multiples of themselves. Thus theonly eigenvalues are λ = 1 and λ = 0 with corresponding eigenvectors (x, 0) and (0, y),respectively, where xy ≠ 0. To verify this, observe that the characteristic equation is
or λ (λ – 1) = 0. Thus λ = 0 or λ = 1. If (x, y) is an eigenvector corresponding to λ =0, then
so that x = 0. If (x, y) is an eigenvector corresponding to λ = 1, then
so that y = 0. Thus the eigenvectors must indeed lie along the coordinate axes.
19. (a)
Eigenvalue λ1 = 4, eigenvector
Eigenvalue λ3 = 1, eigenvector
(b)
λ1 = 0,
λ2 = 1, ξ ξ21 22
1
0
0
0
0
1
=
=
,
, or in general
s
o
t
ξ1
0
1
0
=
A =
1 0 0
0 0 0
0 0 1
ξ ξ21 22
0
1
0
0
0
1
,=
=
ξ1
1
0
0
=
A =−
1 0 0
0 1 0
0 0 1
0 0
0 1
0 0
0
=
=
x
y y
−
=
−
=
1 0
0 0 0
0
0
x
y
x
det λ1 00 1
λ −1 00
−
=
1 0
0 0 λλ= 0
200 Exercise Set 4.3
(c) This transformation doubles the length of each vector while leaving its directionunchanged. Therefore λ = 2 is the only eigenvalue and every nonzero vector in R3 is acorresponding eigenvector. To verify this, observe that the characteristic equation is
or (λ – 2)3 = 0. Thus the only eigenvalue is λ = 2. If (x, y, z) is a correspondingeigenvector, then
Since the above equation holds for every vector (x, y, z), every nonzero vector is aneigenvector.
(d) Since the transformation leaves all vectors on the z-axis unchanged and alters (butdoes not reverse) the direction of all other vectors, its only eigenvalue is λ = 1 withcorresponding eigenvectors (0, 0, z) with z ≠ 0. To verify this, observe that thecharacteristic equation is
or
Since the quadratic (λ – 1/ 2)2 + 1/2 = 0 has no real roots, λ = 1 is the onlyeigenvalue. If (x, y, z) is a corresponding eigenvector, then
You should verify that the above equation is valid if and only if x = y = 0. Therefore thecorresponding eigenvectors are all of the form (0, 0, z) with z ≠ 0.
1 0
1 0
0 0 0
−
− −
1 2 1 2
1 2 1 2x
y
z
=
−( ) + ( )−( ) + ( )
1 1 2 1 2
1 2 1−1 2
x y
x y
0
0
0
0
=
( ) ( )λ λ λ− = − −( ) + ( )
1 1 1 2 1 22 2λ −
− λ −
1 2 1 2
1 2 1 2
= 0
λ −1 2
λ −1 2
λ −1
1 2 0
1 2 0
0 0
0− =
0 0 0
0 0 0
0 0 0
0
0
0
=
x
y
z
λ −λ −
λ −
2 0 0
0 2 0
0 0 2
0=
Exercise Set 4.3 201
20. (a) Yes. If T1 and T2 are both one-to-one (whether or not they are linear), then each mapsdistinct vectors to distinct vectors and therefore so does T2
2 T
1.
(b) If T2 is one-to-one and T1 is not, then T2 T1 cannot be one-to-one since there aredistinct vectors u and v such that T1(u) = T1(v) and therefore such that T2 T1 (u)= T2 T1(v). This is true whether or not the transformations are linear. However T1 T2 may indeed be one-to-one. To see this, consider, for instance, the transformationsgiven by
It is easy to check that T2 is one-to-one and T1 is not (since T1(e3) = 0). Moreover, T1 T2 is given by the matrix
which is invertible, so T1 T2 is one-to-one.
21. Since T(x, y) = (0, 0) has the standard matrix , it is linear. If T(x, y) = (1, 1) were
linear, then we would have
(1, 1) = T(0, 0) = T(0 + 0, 0 + 0) = T(0, 0) + T (0, 0) = (1, 1) + (1, 1) = (2, 2)
Since this is a contradiction, T cannot be linear.
22. (a) Since T is linear,
T (0) = T(0 + 0) = T(0) + T(0) = 2T(0)
so T(0) = 0.
(b) y1 = x21 + x2
2 is an R2 → R2 mapping, y2 = x1x2 where , but T is not linear.
23. From Figure 1, we see that T(e1) = (cos 2θ, sin 2θ) and from Figure 2, that T(e2) =
cos ,sin (sin3
22
3
22 2
π θ π θ+
+
= θθ θ, cos )− 2
T( ) ( )0 0=
0 0
0 0
1 0 0
0 1 0
1 0
0 1
0 0
1 0
0 1
=
[ ] [ ]T T1 21 0 0
0 1 0
1 0
0 1
0 0
=
=
and
202 Exercise Set 4.3
Figure 1 Figure 2
This, of course, should be checked for all possible diagrams, and in particular for the caseπ2 < θ < π. The resulting standard matrix is
25. (a) False. The transformation T(x1, x2) = x21 from R2 to R1 is not linear, but T(0) = 0.
(b) True. If not, T(u) = T(v) where u and v are distinct. Why?
(c) False. One must also demand that x ≠ 0.
(d) True. If c1 = c2 = 1, we obtain equation (a) of Theorem 4.3.2 and if c2 = 0, we obtainequation (b).
26. (a) Sometimes true
(b) Always false
(c) Sometimes true
27. (a) The range of T cannot be all of Rn, since otherwise T would be invertible and det(A)
≠ 0. For instance, the matrix sends the entire xy-plane to the x-axis.
(b) Since det(A) = 0, the equation T(x) = Ax = 0 will have a non-trivial solution andhence, T will map infinitely many vectors to 0.
1 0
0 0
cos sin
sin cos
2 2
2 2
θ θθ θ−
l
ly
y
x
x
(1, 0)
(0, 1)
(cos 2, sin 2)
Exercise Set 4.3 203
EXERCISE SET 4.4
1. (a) (x2 + 2x – 1) – 2(3x2 + 2) = –5x2 + 2x – 5
(b) 5/4x2 + 3x) + 6(x2 + 2x + 2) = 26x2 + 27x + 6
(c) (x4 + 2x3 + x2 – 2x + 1) – (2x3 – 2x) = x4 + x2 + 1
(d) π(4x3 – 3x2 + 7x + 1) = 4πx3 – 3πx2 + 7πx + π
2. (a) Yes,
(b) This is a linear transformation, corresponding to an R4 → R4 mapping (NOT R3 → R3
as in text).
3. (a) Note that the mapping fR3 → R given by f(a, b, c) = |a| has f(1, 0, 0) = 1, f(0, 1, 0) =
0, and f(0, 0, 1) = 0. So If f was a linear mapping, the matrix would be A = (1, 0, 0).
Thus, f(–1, 0, 0) would be found as . Yet, f(–1, 0, 0) = |–1| = 1 ≠ –1.
Thus f is not linear.
(b) Yes, and here A = (1, 0, 0) by reasoning in (a).
1 0 0
1
0
0
1, ,( )−
= −
A =−
−
0 0 1 0
0 1 0 0
1 0 0 0
0 0 0 1
A =
0 0 1
0 1 0
1 0 0
205
4. (a) Yes,
(b) Yes, N =
N is (n + 2) × (n + 1) in dimensions.
(c) No, note that when W(p(x)) = xp(λ) + 1, then w(0pn) = x(0p
n) + 1 = 1 ≠ 0p
n+1 so Wis not linear.
5. (a)
6.
(a0, a1, a2, a3) → (a1, a2, a3) (a0, a1, a2, a3, a4) → (a1, a2, a3, a4)
a b( )
( )0 1 0 0
0 0 2 0
0 0 0 3
0 1 0 0
0 0 2 0
0 0 0 33
A =
( )3 0 0 0
0 2 0 0
0 0 1 0
4 0 0 0 0
0 3 0 0b
00
0 0 2 0 0
0 0 0 1 0
5 0 0 0 0 0
0 4 0 0 0 0
0 0 3
( )c 00 0 0
0 0 0 2 0 0
0 0 0 0 1 0
1 0 0 0 0 0
1 1 0
0 1 1
−
−
0 1
0 1 0
−
00 0 0 1 1 0
0 0 0 0 1 1
0 0 0
−−
0 0 1−
A =
1 0 0
0 1 0
0 0 1
0 0 0
206 Exercise Set 4.4
(c)
(a0, a1, a2, a3, a4, a5) → (a1, 2a2, 3a3, 4a4, 5a5)
7. (a) T(ax + b) = (a + b)x + (a – b)
TP1 → P1
(b) T (ax + b) = ax2 + (a + b)x + (2a – b)
T P1 → P2
(c) T(ax3 + bx2 + cx + d) = (a + 2c – d)x + (2a + b + c + 3d)
TP3 → P1
7. (d) T(ax2 + bx + c) = bx
TP2 → P2
(e) T(ax2 + bx + c) = b
T2P2
→ P0
8. (a)
(b)
9. (a) 3et + 3e–t
(b) Yes, since cosh t = 12et + 12e–t, cosht corresponds to the vector (0, 0, 12, 12).
(c) A a b c=
−
0 1 0 0
0 0 0 0
0 0 1 0
0 0 0 1
( , , ,, ) ( , , , )d b c d→ −0
A a b c=−
→0 0 0
0 0 1
0 1 0
0( , , ) ( , cc b, )−
A a b c a c= −
→ −1 0 0
0 0 1
0 1 0
( , , ) ( , , bb)
0 1 0 0 0 0
0 0 2 0 0 0
0 0 0 3 0 0
0 0 0 0 4 0
0 0 0 0 0 5
Exercise Set 4.4 207
10. Assume Th 4.3.2 holds.
Then T(c1u + c2v) = T(c1u) + T(c2v) = c1T(u) + c2T(v).
Thus, the principle of superposition holds. Conversely, assume the principle of superpositionis true, i.e., T(c1u + c2v) = c1T(u) + c2T(v). Let c1 = c2 = 1. Then T(u + v) = T(u) + T(v).Next, let c2 = 0. Then T(c1u) = c1T(u). This is Th. 4.3.2.
11. If S(u) = T(u) + f, f ≠ 0, then S(0) = T(0) + f = f ≠ 0. Thus S is not linear.
12. The Vandermonde system is
Solving: a0 = 0, a1 = 0, a2 = 2.
Thus, p(x) = 2x2.
13. (a) The Vandermonde system is
Solving: a0 = 1, a1 = 2, A2 = 1
P(X) = X2 + 2X + 1 = (X + 1)2
(b) The system is:
Solving: b0 = 1, b1 = 0, b2 = 1.
Thus, p(x) = 1 • (x + 2)(x) + 0 • (x + 2) + 1 = (x + 2) • x + 1
= (x + 2) • x + 1
= x2 + 2x + 1
1 0 0
1 2 0
1 3 3
1
1
4
0
1
2
=b
b
b
1 2 4
1 0 0
1 1 1
0
1
2
−
a
a
a
==
1
1
4
1 1 1
1 0 0
1 1 1
0
1
2
−
a
a
a
=
2
2
0
208 Exercise Set 4.4
14. (a) The Vandermonde system is
Solving: a0 = 0, a1 = –1, a2 = 0, a2 = 1
Thus p(x) = x3 – x
(b) The system is
Solving: b0 = 0, b1 = 0, b2 = 0, b3 = 1
Thus p(x) = (x + 1)(x – 0)(x – 1)
(c) We have
(d) We have
(e) In this case a0 = a1 = a2 = a3 = 0, so p(x) = 0
Similarly b0 = b1 = b2 = b3 = 0
b
b
b
b
0
1
2
3
1 1 1 1
0 1 1 1
0
=
− −−
00 1 0
0 0 0 1
0
1
0
1
−
=
0
0
0
1
a
a
a
a
0
1
2
3
1 1 0 0
0 1 1 1
0 0 1 0
0 0 0
=−
1
0
0
0
1
0
1
=−
00
1
1 0 0 0 0
1 1 0 0 0
1 2 2 0 0
1 3 6 6 6
1 1 1 1 0
1 0 0 0 0
1 1 1 1 0
1 2 4 8 6
− −
Exercise Set 4.4 209
15. (a) The Vandermonde system is
Thus, p(x) = 2x3 – 2x + 2
(b) The system is
Thus, p(x) = 2(x + 2)(x + 1)(x – 1) – 4(x + 2)(x + 1) + 12(x + 2) – 10
(c) We have
(d) We have
16.
det
1
1
1
2
2
2
2 2 2 2 2 2
a a
b b
c c
bc ab a c a b ac b c= + + − − −
det1
1
a
bb a= −
b
b
b
b
0
1
2
3
1 2 4 8
0 1 3 7
0
=
− −−
00 1 2
0 0 0 1
2
2
0
2
−
−
=
−
−
10
12
4
2
a
a
a
a
0
1
2
3
1 2 2 2
0 1 3 1
0 0
=
−−
11 2
0 0 0 1
10
12
4
2
−
−
=−
2
2
0
2
1 0 0 0 10
1 1 0 0 2
1 3 6 0 2
1 4 12 12 14
−
= −== −=
Solving,
b
b
b
b
0
1
2
3
10
12
4
2
1 2 4 8 10
1 1 1 1 2
1 1 1 1 2
1 2 4
− − −− −
8 14
2
2
0
0
1
2
3
== −==
Solving, a
a
a
a 22
210 Exercise Set 4.4
= bc2 + ab2 + a2c – a2b – ac2 – b2c = (bc2 + ab2 – b2c – bac) + (–ac2 + A2C – a2b + abc)
= b(c2 + ab – bc – ac) – a(c2 – ac + ab – bc)
= (b – a) (c2 + ab – bc – ac)
= (b – a) [c(c – b) – a(c – b)]
= (b – a (c – b) (c – a).
The Vandermonde system for the curve through the two points (x0, y0), (x1, y1) with x0 ≠x, is
Since det = x1 – x0 ≠ 0, the above system has a unique solution for a0, a1.
Thus p(x) = a0 + a1x is a unique line.
The Vandermonde system for the curve through the data (x0, y0), (x1, y1), (x2, y2), with x0≠ x1, x0 ≠ x2, x1 ≠ x2 is
Since the determinant of = (x1 – x0)(x2 – x1) • (x2 – x0)
the above system has a unique solution for ≠0 a0, a1, a2. Thus p(x = a0 + a, x + a2x2 is
unique. If a2 ≠ 0, the curve is a parabola. If a2 = 0, the curve is a line.
17. (a)
(b)
a
a
a
x x x
x x
0
1
2
0 0 1
0 1
1
0 1
0 0 1
=−
− +
( )
b
b
b
0
1
2
a
a
x b
b
0
1
0 0
1
1
0 1
=
−
1
1
1
0 02
1 12
2 22
x x
x x
x x
1
1
1
0 02
1 12
2 22
0
1
2
x x
x x
x x
a
a
a
=
y
y
y
0
1
2
1
10
1
x
x
1
10
1
0
1
0
1
x
x
a
a
y
y
=
Exercise Set 4.4 211
(c)
where
= –(x1x2x3 + x0x2x3 + x0x1x3 + x0x1x2)
= x0x1 + x0x2 + x0x3 + x1x2 + x1x3 + x2x3
18. (a) No, it is not even a function. Ex. T(x) = 12x2 + c, c any constant.
(b) Yes. Let T(f/x)) = ba
a(x)dx
Then T(af(x) + bg(x)) = ba
af(x) + bg(x) dx
= a ba
f(x)dx + b ba
g(x) dx
= aT(f(x)) + bT(g/x).
19. (a) D2 = (2 0 0)
(b)
(c) No. For example, the matrix for 1st diff. from P3 → P2 is
D21 can not be formed.
20. The matrix is the mapping
T(ax2 + bx + c) = bx in P2 → P2
While the matrix [0 1 0] is the mapping
T(ax2 + bx + c) = b in P2 → P0
0 0 0
0 1 0
0 0 0
D2
3 0 0 0
0 2 0 0
0 0 1 0
=
D26 0 0 0
0 2 0 0=
2
1
a
a
a
a
a
x x x x x x0
1
2
3
4
0 0 1 0 11
=
− − 22 0 1 2 3
0 1 1 2 0 2 0 1 1
0
0 1
0 0 1
x x x x
x x x x x x x x
x
− +( ) + +
− +
xx x
x x
1 2 2
2 30 0 0 1
0 0 0 0 1
+( )− +( )
b
b
b
b
b
0
1
2
3
4
212 Exercise Set 4.4
21. (a) We first note P(xi) = 4
i. But from the expansion of the Vandermonde determinant,
where the xi
are unique, is nonzero. Hence the Vandermonde system has a uniquesolution that are the coefficients of the polynomial of nth degree through the n + 1data points. So, there is exactly one polynomial of the nth degree through n + 1 datapoints with the x
iunique. Thus, the Lagrange expression must be algebraically
equivalent to the Vandermonde form.
(b) Since ci
= yi, i = 0, 1, …, n, then the linear systems for the Vandermonde and iveritons
method remain the same.
(c) Newtons form alows for the easy addition of another point (xn+1, 4n+1) that does not
have to be in any order with respect to the other xi
values. This is done by adding anext term to p(i), pe.
p(x) = bn+1(x – x0)(x – x1)(x – x2) … (x – x
n)
+ bn(x – x0)(x – x1)(x – x2) … (x – x
n–1)
+ etc. + b1(x – x0) + b0,
where Pn(x) = b
n(x – x0)(x – x1)(x – x2) … (x – x
n–1) + … +b1(x – x0) + b0 is theinterpolant to the points (x0, y0) … (x
n, y
n). The coefficients for p
n+1(x) are found asin (4), giving an n + 1 degree polynomial. The extra point (x
n+1, y
n+1) may be the
desired interpolating value.
22.
det
23. We may assume in all cases that x = 1, since
Let (x1, x2) = (cos θ, sin θ) = since x = 1.
(a) we have T = + = + =max cos sin max cos3 1 3 22 2 2θ θ θ
x
T x
x
T x
x
T x
x
2
2
2
2
( )=
( )=
( )
1
1
1
1
0 02
0
1 12
1
1 12
1
x x x
x x x
x x x
x
n
n
n n nn
n
…
…
…− − −
xx x
x x
n nn
j k
j kj k
n
2 1…
= −
>=
( )
,
∏∏
Exercise Set 4.4 213
(b)
(c)
(d) T x x x x= +
+ −
max1
2
1
2
1
2
1
21 2
2
1 2
= + =
2
21
22 1max x x
T x x= + = + =max max sin4 9 4 5 321
22 2 θ
T x x= + =max 12
22 1
214 Exercise Set 4.4
EXERCISE SET 5.1
6. The pair (1, –2) is in the set but the pair (–1)(1, –2) = (–1, 2) is not because the firstcomponent is negative; hence Axiom fails. Axiom 5 also fails.
8. Axioms 1, 2, 3, 6, 9, and 10 are easily verified. Axiom 4 holds with 0 = (–1, –1) and Axiom5 holds with –(x, y) = (–x –2, –y –2). Axiom 7 fails because
k((x, y) + (x′, y′) = k(x + x′ + 1, y + y′ + 1)
= (kx + kx′ + k, ky + ky′ + k)
while
k(x, y) + k(x′, y′ = (kx, ky) + (kx′, ky)
= (kx + kx′ + 1, ky + ky′ + 1)
Hence, k(u + v) = ku + kv only if k = 1. Axiom 8 also fails, since if u = (x, y) , then
(k + )u = ((k + )x, (k + )y)
but
ku + u = (kx, ky) + (x, y)
= ((k + )x + 1, (k + )y + 1)
10. This is a vector space. Axioms 2, 3, 7, 8, 9, and 10 follow from properties of matrix additionand scalar multiplication. We verify the remaining axioms.
(1) If we add two matrices of this form, the result will again be a matrix of this form:
(*)a
b
c
d
a c
b d
0
0
0
0
0
0
+
=
++
215
(4) The 2 × 2 zero matrix is of the appropriate form and has the desired properties.
(5) If u is a matrix of the given form, then
is again of the desired form and u + (–u) = (–u) + u = 0.
(6) If u is any matrix of this form, then ku is
(**)
and ku has the desired form.
11. This is a vector space. We shall check only four of the axioms because the others followeasily from various properties of the real numbers.
(1) If f and g are real-valued functions defined everywhere, then so is f + g. We mustalso check that if f(1) = g(1) = 0, then (f + g)(1) = 0. But (f + g)(1) = f(1) + g(1)= 0 + 0 = 0.
(4) The zero vector is the function z which is zero everywhere on the real line. Inparticular, z(1) = 0.
(5) If f is a function in the set, then –f is also in the set since it is defined for all realnumbers and –f(1) = –0 = 0. Moreover, f + (–f) = (–f) + f = z.
(6) If f is in the set and k isany real number, then kf is a real valued function definedeverywhere. Moreover, kf(1) = k0 = 0.
12. This is a vector space and the proof is almost a direct repeat of that for Problem 10. In fact,we need only modify the two equations (*) and (**) in the following way:
Note that if u is any matrix of this form, then
− =− − +( )
− +( ) −
ua a b
a b b
a a b
a b b
c c d
c d d
a c a c++
+
++
=
+ +( ) + bb d
a c b d b d
ka a b
a b b
+( )+( ) + +( ) +
++
=
++
ka ka kb
ka kb kb
ka
b
ka
kb
0
0
0
0
=
− =−
−
u
a
b
0
0
216 Exercise Set 5.1
13. This is a vector space with 0 = (1, 0) and –x = (1, –x). The details are easily checked.
15. We must check all ten properties:
(1) If x and y are positive reals, so is x + y = xy.(2) x + y = xy = yx = y + x(3) x + (y + z) = x(yz) = (xy) = (x + y) + z(4) There is an object 0, the positive real number 1, which is such that
1 + x = 1 • x = x = x • 1 = x + 1
for all positive real numbers x.(5) For each positive real x, the positive real 1/x acts as a negative:
x + (1/x) = x(1/x) = 1 = 0 = 1 = (1/x)x = (1/x) + x
(6) If k is a real and x is a positive real, then kx = xk is again a positive real.(7) k(x + y) = (xy)k = xkyk = kx + ky
(8) (k + )x = xk+ = xkx = kx + x
(9) k(x) = (x)k = (x)k = xk = xk = (k)x
(10) 1x = x1 = x
16. This is not a vector space, since properties (7) and (8) fail. It is easy to show, for instance,that ku + kv = k2(u + v). Property (4) holds with 0 = (1, 1) , but Property (5) failsbecause, for instance, (0, 1) has no inverse.
17. (a) Only Axiom 8 fails to hold this space. Let k and m be scalars. Then
(k + m)(x, y, z) = ((k + m)2x, (k + m)2y, (k + m)2z) = (k2x, k2y, k2z) + (2kmx, 2kmy,2kmz) + m2x, m2y, m2z)
= k(x, y, z) + m(x, y, z) + (2kmx, 2kmy, 2kmz)
≠ k(x, y, z) + m(x, y, z),
and Axiom 8 fails to hold.
(b) Only Axioms 3 & 4 fail for this space.
Axiom 3: Using the obvious notation, we have
u + (v + w) = (u1, u2, u3) + v3 + w3, v2 + w2, v1 + w1)= (u3 + v1 + w1, u2 + v2 + w2, u1 + v3 + w3)
whereas
(u + v) + w = (u3 + v3, u2 + v2, u1 + v1) + (w1, w2, w3)= (u1 + v1 + w3, u2 + v2 + w2, u3 + v3 + w1)
Exercise Set 5.1 217
Thus, u + (v + w) ≠ (u + w) + w.
Axiom 4: There is no zero vector in this space. If we assume that there is, and let0 = (z1, z2, z3), then for any vector (a, b, c), we have (a, b, c) + (z1, z2,z3) = c + z3, b + z2, a + z1) = (a, b, c). Solving for the z′inferior/superioraligned
is, we have z3 = a – c, z2 = 0 adn z1 = c – a. Thus, there is no one
zero vector that will work for every vector (a, b, c) in R3.
(c) Let V be the set of all 2 × 2 invertible matrices and let A be a matrix in V. Since we areusing standard matrix addition adn scalar multiplication, the majority of axioms hold.However, the following axioms fail for this space V:
Axiom 1: Clearly if A is invertible, then so is –A. However, the matrix A + (–A) =0 is not invertible, and thus A + (–A) is not in V, meaning V is not closedunder addition.
Axiom 4: We’ve shown that the zero matrix is not in V, so this axiom fails.
Axiom 6: For any 2 × 2 invertible matrix A, det(kA) = k2 det(A), so for k ≠ 0, thematrix kA is also invertible. However, if k = 0, then kA is not invertible,so this axiom fails.
Thus, V is not a vector space.
18. Let V be the set of all matrices of the form . We will verify each of the 10 vector
space axioms using matrices A = , B = , and C = , and scalars
k and m.
(1) We can see from the given formula that the sum of two matrices in V is again amatrix in V. Thus, Axiom 1 holds.
(2) A + B = = B + A.
(3) We have
and Axiom 3 is satisfied.
A B Ca
a
b c
b c+ + =
+
++
=( ) 1
2
1 1
2 2
1
1
1
1
a b c
a b c
a b
a b
1 1 1
2 2 2
1 1
2 2
1
1
1
1
+ ++ +
=+
+
+
= + +( )
c
cA B C
1
2
1
1
a b
a b
b a
b a
1 1
2 2
1 1
2 2
1
1
1
1
++
=
++
c
c
1
2
1
1
b
b
1
2
1
1
a
a
1
2
1
1
a
b
1
1
218 Exercise Set 5.1
(4) The zero vector in this space is the matrix 0 = . Then
A + 0 = = A
And similarly, 0 + A = A. Thus, Axiom 4 holds.
(5) If A = , then –A = , and we have
A + (–A) = = 0
Similarly, we have (–A) + A = 0.
(6) We can see from the given formula that if A is in V, then kA is also a matrix in V.Thus, Axiom 6 holds.
(7) We have
and Axiom 7 is satisfied.
(8) We have
and Axiom 8 holds.
( )k m Ak m a
k m a
ka ma+ =
+( )+( )
=+1
2
1 11
1
1
1 kka ma
ka
ka
ma
ma
2 2
1
2
1
2
1
1
1
1
+
=
+
= +kA kB
k A Ba b
a b
k a b
k a b( )+ =
++
=
+( )+
1 1
2 2
1 1
2
1
1
1
1 22
1 1
2 2
11
1
1
1
( )
=+
+
=
ka kb
ka kb
ka
kka
kb
kb
kA kB
2
1
2
1
1
+
= +
a
a
a
a
1
2
1
2
1
1
1
1
0 1
1 0
+
−−
=
−−
a
a
1
2
1
1
a
a
1
2
1
1
a
a
a
a
1
2
1
2
1
1
0 1
1 0
0 1
1 0
+
=
++
0 1
1 0
Exercise Set 5.1 219
(9) We have
so Axiom 9 holds.
(10) Finally, we have
and Axiom 10 is also satisfied.
19. (a) Let V be the set of all ordered pairs (x, y) that satisfy the equation ax + by = c1 forfixed constants a, b and c. Since we are using the standard operations of addition andscalar multiplication, Axioms 2, 3, 5, 7, 9, 9, 10 will hold automatically. However, forAxiom 4 to hold, we need the zero vector (0, 0) to be in V. Thus a0 + b0 = c, whichforces c = 0. In this case, Axioms 1 and 6 are also satisfied. Thus, the set of all pointsin R2 lying on a line is a vector space exactly in the case when the line passes throughthe origin.
(b) Let V be the set of all ordered triples (x, y, z) that satisfy the equation ax + by + cz
= d, for fixed constants a, b, c and d. Since we are using the standard operations ofaddition and scalar multiplication, Axioms 2, 3, 5, 7, 8, 9, 10 will hold automatically.However, for Axiom 4 to hold, we need the zero vector (0, 0, 0) to be in V. Thus a0 +b0 + c0 = d, which forces d = 0. In this case, Axioms 1 and 6 are also satisfied. Thus,the set of all points in R3 lying on a plane is a vector space exactly in the case whenthe plane passes through the origin.
20. Let V be the set of all 2 × 2 invertible matrices. With the given operations, V is not a vectorspace. We will check all 10 axioms:
(1) The product of two invertible matrices is invertible, so Axiom 1 holds.
(2) In general, matrix multiplication is not commutative, so AB ≠ BA and Axiom 2 fails.
(3) Matrix multiplication is associative so A(BC) = (AB)C and Axiom 3 holds.
(4) The identity matrix functions as the zero vector here: AI = IA = A and Axiom 4holds.
(5) The inverse of A functions as –A in this case: A(A–1) = A–1A = I, so Axiom 5 holds.
a
a
a
a
a
a
1
2
1
2
1
2
1
1
1 1
1 1
1
1
=
=
k m A kma
ma
k ma
k ma( ( )) =
=
( )( )
1
2
1
2
1
1
1
1
=( )
( )
=km a
km akm A
1
2
1
1( )
220 Exercise Set 5.1
(6) If k = 0, then kA is not invertible, so kA is not in V and Axiom 6 fails.
(7) Axiom 7 fails: k(AB) ≠ (kA)(kB).
(8) Axiom 8 fails as well: (k + m)A ≠ (kA)(mA).
(9) Since we are using regular scalar multiplication, Axiom 9 holds.(10) Again, since we are using regular scalar multiplication, Axiom 10 holds.
Thus, V is not a vector space since Axioms 2, 6, 7, & 8 fail.
22. Properties (2), (3), and (7)–(10) all hold because a line passing through the origin in 3-space is a collection of triples of real numbers and the set of all such triples with the usualoperations is a vector space. To verify the remaining four properties, we need only checkthat
(1) If u and v lie on the line, so does u + v.
(4) The vector (0,0,0) lies on the line (which it does by hypothesis, since the linepasses through the origin).
(5) If u lies on the line, so does –u.
(6) If u lieson the line, so does any real multiple ku of u.
We check (1), leaving (5) and (6) to you.
The line passes through the origin and therefore has the parametric equations x = at,y = bt, z = ct where a, b, and c are fixed real numbers and t is the parameter. Thus, if u andv lie on the line, we have u = (at1, bt1, ct1) and v = (at2, bt2, ct2). Therefore u + v = (a(t1+ t2), b(t1 + t2), c(t1 + t2)), which is also on the line.
25. No. Planes which do not pass through the origin do not contain the zero vector.
26. No. The set of polynomials of exactly degree 1 does not contain the zero polynomial, andhence has no zero vector.
27. Since this space has only one element, it would have to be the zero vector. In fact, this isjust the zero vector space.
28. If a vector space V had just two distinct vectors, 0 and u, then we would have to definevector addition and scalar multiplication on V. Theorem 5.1.1 ensures that ku ≠ 0 unless k= 0. Thus we would have to define ku = u for all k ≠ 0. But then we would have
2 u = (1 + 1)u = 1u + 1u = u + u ≠ 0,
Exercise Set 5.1 221
so that u + u = u. However,
0u = (1 – 1)u = 1u + (–1)u = u + (–1)u = 0
which implies that (–1)u ≠ u. Hence (–1)u = 0, which is contrary to Theorem 5.1.1. ThusV cannot be a vector space.
30. We are given that ku = 0. Suppose that k ≠ 0. Then
1k
(ku) = 1kk u = (1)u = u (Axioms 9 and 10)
But
1k
(ku) = 1k0 = 0 (By hypothesis and Part (b))
Thus u = 0. That is, either k = 0 or u = 0.
33. Suppose that there are two zero vectors, 01 and 02. If we apply Axiom 4 to both of thesezero vectors, we have
01 = 01 + 02 = 02
Hence, the two zero vectorsare identical.
34. Suppose that u has two negatives, (–u)1 and (–u)2. Then
(–u)1 = (–u)1 + 0 = (–u)1 + (u + (–u)2) = ((–u)1 + u) + (–u)2 = 0 + (–u)2 = (–u)2
Axiom 5 guarantees that u must have at least one negative. We have proved that it has atmost one.
35. Following the hint, we have
(u + v) –(v + u) = (u + v) + (–(v + u)) by Theorem 5.1.1
= (u + v) + ((–1)v + (–1)u) by Property (7)
= (u + v) + (–v + (–u)) by Theorem 5.1.1
= ((u + v) + (–v)) + (–u) by Property (3)
= (u + (v + (–v)) + (–u) by Property (3)
= (u + 0) + (–u) by Property (5)
= u + (–u) by Property (4)
= 0 by Property (5)
222 Exercise Set 5.1
Thus (u + v) + (–1)(v + u) = 0, from which it follows that
(u + v) + (–(v + u)) = 0 by Theorem 5.1.1
so that
[(u + v) + (–(v + u))] + (v + u) = 0 + (v + u) adding to both sides of the equation
and thus
(u + v) + (–(v + u) + (v + u)) = 0 + (v + u) by Property (3)
or
(u + v) + 0 = 0 + (v + u) by Property (5)
so that finally u + v = v + u by Property (4).
Exercise Set 5.1 223
EXERCISE SET 5.2
1. (a) The set is closed under vector addition because
(a, 0, 0) + (b, 0, 0) = (a + b, 0, 0)
It is closed under scalar multiplication because
k(a, 0, 0) = (ka, 0, 0)
Therefore it is a subspace of R3.
(b) This set is not closed under either addition or scalar multiplication. For example, (a,1, 1) + (b, 1, 1) = (a + b, 2, 2) and (a + b, 2, 2) does not belong to the set. Thus it isnot a subspace.
(c) This set is closed under vector addition because
(a1, b1, 0) + (a2, b2, 0) = (a1 + a2, b1 + b2, 0).
It is also closed under scalar multiplication because
k(a, b, 0) = (ka, kb, 0).
Therefore, it is a subspace of R3.
2. (a) This set is closed under addition since the sum of two integers is again an integer.However, it is not closed under scalar multiplication since the product ku where k isreal and a is an integer need not be an integer. Thus, the set is not a subspace.
(c) If det(A) = det(B) = 0, it does not necessarily follow that det(A + B) = 0. For instance,
let A = and B = . Thus the set is not a subspace.0 0
0 1
1 0
0 0
225
(e) This set is closed under vector addition because
It is also closed under scalar multiplication because
Therefore, it is a subspace of M22.
3. (a) This is the set of all polynominals with degree ≤ 3 and with a constant term which isequal to zero. Certainly, the sum of any two such polynomials is a polynomial withdegree ≤ 3 and with a constant term which is equal to zero. The same is true of aconstant multiple of such a polynomial. Hence, this set is a subspace of P3.
(c) The sum of two polynomials, each with degree ≤ 3 and each with integral coefficients,is again a polynomial with degree ≤ 3 and with integral coefficients. Hence, the subsetis closed under vector addition. However, a constant multiple of such a polynomialwill not necessarily have integral coefficients since the constant need not be an integer.Thus, the subset is not closed under scalar multiplication and is therefore not asubspace.
4. (a) The function f(x) = –1 for all x belongs to the set, but the function (–1)f(x) = 1 for allx does not. Hence, the set is not closed under scalar multiplication and is therefore nota subspace.
(c) Suppose that f and g are in the set. Then
(f + g)(0) = f(0) + g(0) = 2 + 2 = 4
and
–2f(0) = (–2)(2) = –4
This set is therefore not closed under either operation.
(e) Let f(x) = a + b sin x and g(x) = c + d sin x be two functions in this set. Then
(f + g)(x) = (a + c) + (b + d) sin x
a a
a a
ka ka
k a k a
ka ka
− −
=
−( ) −( )
=
− kka ka( ) −( )
.
a a
a a
b b
b b
a b a b
a b a− −
+
− −
=
+ +− − − − bb
a b a b
a b a b
=
+ +− +( ) − +( )
226 Exercise Set 5.2
and
k(f(x)) = ka + kb sin x
Thus, both closure properties are satisfied and the set is a subspace.
5. (b) If A and B are in the set, then aij
= –aji
and bij
= –bji
for all i and j. Thus aij
+ bij
=–(a
ji+ b
ji) so that A + B is also in the set. Also a
ij= –a
jiimplies that ka
ij= –(ka
ji), so
that kA is in the set for all real k. Thus the set is a subspace.
(c) For A and B to be in the set it is necessary and sufficient for both to be invertible, butthe sum of 2 invertible matrices need not be invertible. (For instance, let B = –A.)Thus A + B need not be in the set, so the set is not a subspace.
6. (b) The matrix reduces to
so the solution space is the line x = 2t, y = t, z = 0.
(d) The matrix reduces to the identity matrix, so the solution space is the origin.
(f) The matrix reduces to
so the solution space is the plane x –3y + z = 0.
7. (a) We look for constants a and b such that au + bv = (2, 2, 2), or
a(0, –2, 2) + b(1, 3, –1) = (2, 2, 2)
Equating corresponding vector components gives the following system of equations:
b = 2
–2a+ 3b = 2
2a – b = 2
1 3 1
0 0 0
0 0 0
−
1 2 0
0 0 1
0 0 0
−
Exercise Set 5.2 227
From the first equation, we see that b = 2. Substituting this value into the remainingequations yields a = 2. Thus (2, 2, 2) is a linear combination of u and v.
(c) We look for constants a and b such that au + bv = (0, 4, 5), or
a(0, –2, 2) + b(1, 3, –1) = (0, 4, 5)
Equating corresponding components gives the following system of equations:
b = 0
–2a + 3b = 4
2a – b = 5
From the first equation, we see that b = 0. If we substitute this value into theremaining equations, we find that a = –2 and a = 5/2. Thus, the system of equations isinconsistent and therefore (0, 4, 5) is not a linear combination of u and v.
8. (a) We look for constants a, b, and c such that au + bv + cw = (–9, –7, –15); that is, such that
a(2, 1, 4) + b(1, –1, 3) + c(3, 2, 5) = (–9, –7, –15)
If we equate corresponding components, we obtain the system
2a + b + 3c = –9
a – b + 2c = –7
4a + 3b + 5c = –15
The augmented matrix for this system is
The reduced row-echelon form of this matrix is
Thus a = –2, b = 1, and c = –2 and (–9, –7, –15) is therefore a linear combination ofu, v, and w.
1 0 0 2
0 1 0 1
0 0 1 2
−
−
2 1 3 9
1 1 2 7
4 3 5 15
−− −
−
228 Exercise Set 5.2
8. (c) This time we look for constants a, b, and c such that
au + bv + cw = (0, 0, 0)
If we choose a = b = c = 0, then it is obvious that au + bv + cw = (0, 0, 0). We nowproceed to show that a = b = c = 0 is the only choice. To this end, we equatecomponents to obtain a system of equations whose augmented matrix is
From Part (a), we know that this matrix can be reduced to
Thus, a = b = c = 0 is the only solution.
9. (a) We look for constants a, b, and c such that
ap1
+ bp2
+ cp3
= –9 –7x –15x2
If we substitute the expressions for p1, p2, and p3 into the above equation and equatecorresponding coefficients, we find that we have exactly the same system of equationsthat we had in Problem 8(a), above. Thus, we know that a = –2, b = 1, and c = –2 andthus –2p1 + 1p2 – 2p3 = –9 – 7x – 15x2.
(c) Just as Problem 9(a) was Problem 8(a) in disguise, Problem 9(c) is Problem 8(c) indifferent dress. The constants are the same, so that 0 = 0p1 + 0p2 + 0p3.
10. (a) We ask if there are constants a, b, and c such that
4 0
2 2
1 1
2 3
0 2
1 4− −
+
−
+
=b c
6 8
1 8
−− −
1 0 0 0
0 1 0 0
0 0 1 0
2 1 3 0
1 1 2 0
4 3 5 0
−
Exercise Set 5.2 229
If we multiply, add, and equate corresponding matrix entries, we obtain the followingsystem of equations:
4a + b = 6
–b + 2c = –8
–2a + 2b + c = –1
–2a + 3b + 4c = –8
This system has the solution a = 1, b = 2, c = –3; thus, the matrix is a linearcombination of the three given matrices.
(b) Clearly the zero matrix is a linear combination of any set of matrices since we canalways choose the scalars to be zero.
11. (a) Given any vector (x, y, z) in R3, we must determine whether or not there areconstants a, b, and c such that
(x, y, z) = av1 + bv2 + cv3
= a(2, 2, 2) + b(0, 0, 3) + c(0, 1, 1)
= (2a, 2a + c, 2a + 3b + c)
or
x = 2a
y = 2a + c
z = 2a + 3b + c
This is a system of equations for a, b, and c. Since the determinant of the system isnonzero, the system of equations must have a solution for any values of x, y, and z,whatsoever. Therefore, v1, v2, and v3 do indeed span R3.
Note that we can also show that the system of equations has a solution by solvingfor a, b, and c explicitly.
11. (c) We follow the same procedure that we used in Part (a). This time we obtain thesystem of equations
3a + 2b + 5c + d = x
a – 3b – 2c + 4d = y
4a + 5b + 9c – d = z
230 Exercise Set 5.2
The augmented matrix of this system is
which reduces to
Thus the system is inconsistent unless the last entry in the last row of the abovematrix is zero. Since this is not the case for all values of x, y, and z, the given vectorsdo not span R3.
12. (a) Since cos(2x) = (1) cos2 x + (–1) sin2 x for all x, it follows that cos(2x) lies in thespace spanned by cos2 x and sin2 x.
(b) Suppose that 3 + x2 is in the space spanned by cos2 x and sin2 x; that is, 3 + x2 = acos2 x + b sin2 x for some constants a and b. This equation must hold for all x. If weset x = 0, we find that a = 3. However, if we set x = π, we find a = 3 + π2. Thus wehave a contradiction, so 3 + x2 is not in the space spanned by cos2 x and sin2 x.
13. Given an arbitrary polynomial a0 + a1x + a2x2 in P2, we ask whether there are numbers a,
b, c and d such that
a0 + a1x + a2x2 = ap1 + bp2 + cp3 + dp4
If we equate coefficients, we obtain the system of equations:
a0 = a + 3b + 5c – 2d
a1 = –a + b – c – 2d
a2 = 2a + 4c + 2d
1 3 2 4
0 1 1 13
11
0 0 0 0
− −
−−
−
y
x y
z 44
17
3
11
y x y−
−
3 2 5 1
1 3 2 4
4 5 9 1
x
y
z
− −−
Exercise Set 5.2 231
A row-echelon form of the augmented matrix of this system is
Thus the system is inconsistent whenever –a0 + 3a1 + 2a2 ≠ 0 (for example, when a0 = 0,a1 = 0, and a2 = 1). Hence the given polynomials do not span P2.
14. (a) As before, we look for constants a, b, and c such that
(2, 3, –7, 3) = av1 + bv2 + cv3
If we equate components, we obtain the following system of equations:
2a + 3b – c = 2
a – b = 3
5b + 2c = –7
3a + 2b + c = 3
The augmented matrix of this system is
This reduces to
Thus a = 2, b = –1, and c = –1, and the existence of a solution guarantees that thegiven vector is in spanv1, v2, v3.
1 0 0 2
0 1 0 1
0 0 1 1
0 0 0 0
−−
2 3 1 2
1 1 0 3
0 5 2 7
3 2 1 3
−−
−
1 3 5 2
0 1 1 14
0 0 0 0 3
0
0 1
0
−
−+
− +
a
a a
a aa a1 22+
232 Exercise Set 5.2
14. (c) Proceeding as in Part (a), we obtain the matrix
This reduces to a matrix whose last row is [0 0 0 1]. Thus the system is inconsistentand hence the given vector is not in spanv1, v2, v3.
15. The plane has the vector u × v = (0, 7, –7) as a normal and passes through the point(0,0,0). Thus its equation is y – z = 0.
Alternatively, we look for conditions on a vector (x, y, z) which will insure that it liesin spanu, v. That is, we look for numbers a and b such that
(x, y, z) = au + bv
= a(–1, 1, 1) + b(3, 4, 4)
If we expand and equate components, we obtain a system whose augmented matrix is
This reduces to the matrix
Thus the system is consistent if and only if = 0 or y = z.
17. The set of solution vectors of such a system does not contain the zero vector. Hence itcannot be a subspace of Rn.
7
− +y z
1 3
0 17
0 07
− −+
− +
x
x y
y z
−
1 3
1 4
1 4
x
y
z
2 3 1 1
1 1 0 1
0 5 2 1
3 2 1 1
−−
Exercise Set 5.2 233
18. Suppose that spanv1, v2, …, vr = spanw1, w2, …, w
k. Since each vector v
iin S belongs to
spanv1, v2, …, vr, it must, by the definition of spanw1, w2, …, w
k, be a linear combination
of the vectors in S′. The converse must also hold.
Now suppose that each vector in S is a linear combination of those in S′ and conversely.Then we can express each vector v
ias a linear combination of the vectors w1, w2, …, w
k,
so spanv1, v2, …, vr ⊆ spanw1, w2, …, w
k. But conversely we have spanw1, w2, …, w
k ⊆
spanv1, v2, …, vr, so the two sets are equal.
19. Note that if we solve the system v1 = aw1 + bw2, we find that v1 = w1 + w2. Similarly, v2 =2w1 + w2, v3 = –w1 + 0w2, w1 = 0v1 + 0v2 – v3, and w2 = v1 + 0v2 + v3.
21. (a) We simply note that the sum of two continuous functions is a continuous function andthat a constant times a continuous function is a continuous function.
(b) We recall that the sum of two differentiable functions is a differentiable function andthat a constant times a differentiable function is a differentiable function.
23. (a) False. The system has the form Ax = b where b has at least one nonzero entry.Suppose that x1 and x2 are two solutions of this system; that is, Ax1 = b and Ax2 = b.Then
A(x1 + x2) = Ax1 + Ax2 = b + b ≠ b
Thus the solution set is not closed under addition and so cannot form a subspace ofR
n. Alternatively, we could show that it is not closed under scalar multiplication.
(b) True. Let u and v be vectors in W. Then we are given that ku + v is in W for all scalarsk. If k = 1, this shows that W is closed under addition. If k = –1 and u = v, then thezero vector of V must be in W. Thus, we can let v = 0 to show that W is closed underscalar multiplication.
(d) True. Let W1 and W2 be subspaces of V. Then if u and v are in W1 ∩ W2, we know thatu + v must be in both W1 and W2, as must ku for every scalar k. This follows from theclosure of both W1 and W2 under addition and scalar multiplication.
(e) False. Spanv = span2v, but v ≠ 2 v in general.
24. (a) Two vectors in R3 will span a plane if and only if one is not a constant multiple of theother. They will span a line if and only if one is a constant multiple of the other.
(b) Spanu = spanv if and only if u is a nonzero multiple of v. Why?
(c) The solution set will be a subspace of Rn if and only if b = 0. See Exercise 23(a).
234 Exercise Set 5.2
25. No. For instance, (1, 1) is in W1 and (1, –1) is in W2, but (1, 1) + (1, –1) = (2, 0) is inneither W1 nor W2.
26. (a) The matrices and span M22.
27. They cannot all lie in the same plane.
0 0
0 1
1 0
0 0
0 1
0 0
0 0
1 0
, ,
Exercise Set 5.2 235
EXERCISE SET 5.3
2. (a) Clearly neither of these vectors is a multiple of the other. Thus they are linearlyindependent.
(b) Following the technique used in Example 4, we consider the system of linear equations
–3k1 + 5k2 + k3 = 0
–k22
+ k3
= 0
4k1 + 2k2 + 3k3 = 0
Since the determinant of the coefficient matrix is nonzero, the system has only thetrivial solution. Therefore, the three vectors are linearly independent.
(d) By Theorem 5.3.3, any four vectors in R3 are linearly dependent.
3. (a) Following the technique used in Example 4, we obtain the system of equations
3k1 + k2+ 2k3 + k4 = 0
8k1 + 5k2 – k3 + 4k4 = 0
7k1 + 3k2+ 2k3 = 0
–3k1 – k2+ 6k3 + 3k4 = 0
Since the determinant of the coefficient matrix is nonzero, the system has only thetrivial solution. Hence, the four vectors are linearly independent.
3. (b) Again following the technique of Example 4, we obtain the system of equations
3k2 + k3 = 0
3k2 + k3 = 0
2k1 = 0
2k1 – k3 = 0
237
The third equation, above, implies that k1 = 0. This implies that k3 and hence k2 mustalso equal zero. Thus the three vectors are linearly independent.
4. (a) We ask whether there exist constants a, b, and c such that
a(2 – x + 4x2) + b(3 + 6x + 2x2) + c(2 + 10x – 4x2) = 0
If we equate the coefficients of x0, x, and x2 in the above polynomial to zero, we obtainthe following system of equations:
2a + 3b + 2c = 0
–a + 6b+ 10c = 0
4a + 2b – 4c = 0
Since the coefficient matrix of this system is invertible, the trivial solution is the onlysolution. Hence, the polynomials are linearly independent.
(d) If we set up this problem in the same way we set up Part (a), above, we obtain threeequations in four unknowns. Since this is equivalent to having four vectors in R3, thevectors are linearly dependent by Theorem 5.3.3.
5. (a) The vectors lie in the same plane through the origin if and only if they are linearlydependent. Since the determinant of the matrix
is not zero, the matrix is invertible and the vectors are linearly independent. Thusthey do not lie in the same plane.
6. (a) Since v2 = –2v1, the vectors v1 and v2 lie on the same line. But since v3 is not amultiple of v1 or v2, the three vectors do not lie on the same line through the origin.
(c) Since v1 = 2v2 = –2v3, these vectors all lie on the same line through the origin.
7. (a) Note that 7v1 – 2v2 + 3v3 = 0.
2 6 2
2 1 0
0 4 4
−−
238 Exercise Set 5.3
8. (a) By inspection, we see that
(1, 2, 3, 4) + (0, 1, 0, –1) = (1, 3, 3, 3)
so we have the linear combination (1, 2, 3, 4) + (0, 1, 0, –1) – (1, 3, 3, 3) = 0 and thisis a linearly dependent set.
(b) Using the notation v1 = (1, 2, 3, 4), v2 = (0, 1, 0, –1), and v3 = (1, 3, 3, 3), the equationfrom Part (a) becomes v3 = v1 + v2. Solving this for v1 and then v2, we have the threedependence relations
v1 = –v2 + v3
v2 = –v1 + v3
v3 = v1 + v2.
9. If there are constants a, b, and c such that
a(λ, –1/2, –1/2) + b(–1/2, λ, –1/2) + c(–1/2, –1/2, λ) = (0, 0, 0)
then
The determinant of the coefficient matrix is
This equals zero if and only if λ = 1 or λ = –1/2. Thus the vectors are linearly dependent forthese two values of λ and linearly independent for all other values.
11. Suppose that S has a linearly dependent subset T. Denote its vectors by w1,…, wm
. Thenthere exist constants k
i, not all zero, such that
k1w1 + … + km
wm
= 0
But if we let u1, …, un–m
denote the vectors which are in S but not in T, then
k1w1 + … + km
wm
+ 0u1 + … + 0un–m
= 0
λ λ λ λ32
3
4
1
41
1
2( )− − = − +
λλ
λ
− −− −− −
1 2 1 2
1 2 1 2
1 2 1 2
a
b
c =
0
0
0
Exercise Set 5.3 239
Thus we have a linear combination of the vectors v1, …, vn
which equals 0. Since not all ofthe constants are zero, it follows that S is not a linearly independent set of vectors, contraryto the hypothesis. That is, if S is a linearly independent set, then so is every non-emptysubset T.
13. This is similar to Problem 10. Since v1, v2, …, vr is a linearly dependent set of vectors,
there exist constants c1, c2, …, cr
not all zero such that
c1v1 + c2v2 + … + cr
vr
= 0
But then
c1v1 + c2v2 + … + crv
r+ 0v
r+1 + … + 0vn
= 0
The above equation impliesthat the vectors v1, …, vn
are linearly dependent.
15. Suppose that v1, v2, v3 is linearly dependent. Then there exist constants a, b, and c not allzero such that
(*) av1 + bv2 + cv3 = 0
Case 1: c = 0. Then (*) becomes
av1 + bv2 = 0
where not both a and b are zero. But then v1, v2 is linearly dependent, contrary tohypothesis.
Case 2: c ≠ 0. Then solving (*) for v3 yields
v3 = – ac
v1 – bc
v2
This equation implies that v3 is in spanv1, v2, contrary to hypothesis. Thus, v1, v2, v3is linearly independent.
16. Note that (u – v) + (v – w) + (w – u) = 0.
18. Any nonzero vector forms a linearly independent set. The only scalar multiple of a nonzerovector which can equal the zero vector is the zero scalar times the vector.
240 Exercise Set 5.3
20. (a) Since sin2 x + cos2 x = 1, we observe that
2(3 sin2 x) + 3(2 cos2 x) + (–1)(6) = 0
Hence the vectors are linearly dependent.
(c) Suppose that there are constants a, b, and c such that
a(1) + b sin x + c sin 2x = 0
Setting x = 0 yields a = 0. Setting x = π/2 yields b = 0, and thus, since sin 2x 0, wemust also have c = 0. Therefore the vectors are linearly independent.
(e) Suppose that there are constants a, b, and c such that
a(3 – x)2 + b(x2 – 6x) + c(5) = 0
or
(9a + 5c) + (–6a – 6b)x + (a + b)x2 = 0
Clearly a = –b = –(5/9)c. Thus a = 5, b = –5, c = –9 is one solution and the vectors arelinearly dependent.
This conclusion may also be reached by noting that the determinant of the systemof equations
9a + 5c = 0
–6a – 6b = 0
a + b = 0
is zero.
21. (a) The Wronskian is
Thus the vectors are linearly independent.
1
0 1
0 0
0
x e
e
e
e
x
x
x
x= ≡
Exercise Set 5.3 241
21. (b) The Wronskian is
Thus the vectors are linearly independent.
23. Use Theorem 5.3.1, Part (a).
24. (a) False. There are 6 such matrices, namely
Since M22 has dimension 4, at least two of these matrices must be linear combinationsof the other four. In fact, it is easy to show that A1, A2, A3, and A4 are linearlyindependent and that A5 = –A2 + A3 + A4 and A6 = –A1 + A3 + A4.
(b) False. One of the vectors might be the zero vector. Otherwise it would be true.
(d) False. A finite set of vectors can be linearly dependent without containing the zerovector.
26. We could think of any 3 linearly independent vectors in R3 as spanning R3. That is, theycould determine directions for 3 (not necessarily orthogonal) coordinate axes. Then anyfourth vector would represent a point in this coordinate system and hence be a linearcombination of the other 3 vectors.
A A A4 5 60 1
1 0
0 1
0 1
0 0
1 1=
=
=
A A A1 2 31 1
0 0
1 0
1 0
1 0
0 1=
=
=
sin cos sin
cos sin sin cos
sin cos
x x x x
x x x x x
x x
− +− − 2ccos sin
sin cos sin
cos sin sin cos
x x x
x x x x
x x x x
−= − + xx
x
x x x x
0 0 2
2 2 02 2
cos
cos sin cos ) cos= − − = − ≡(
242 Exercise Set 5.3
EXERCISE SET 5.4
2. (a) This set is a basis. It has two vectors and neither is a multiple of the other.
(c) This set is not a basis since one vector is a multiple of the other.
3. (a) This set has the correct number of vectors and they are linearly independent because
= 6 ≠ 0
Hence, the set is a basis.
(c) The vectors in this set are linearly dependent because
= 0
Hence, the set is not a basis.
4. (a) The vectors in this set are linearly dependent because
= 0
Thus, the set is not a basis. (Compare with Problem 3(c), above.)
1 1 1
3 1 7
2 4 0
− −
2 4 0
3 1 7
1 1 1
− −
1 2 3
0 2 3
0 0 3
243
4. (c) This set has the correct number of vectors and
= 1 ≠ 0
Hence, the vectors are linearly independent and therefore are a basis.
5. The set has the correct number of vectors. To show that they are linearly independent, weconsider the equation
If we add matrices and equate corresponding entries, we obtain the following system ofequations:
3a + d = 0
6a – b – 8c = 0
3a – b – 12c – d = 0
–6a – 4c + 2d = 0
Since the determinant of the coefficient matrix is nonzero, the system of equations hasonly the trivial solution; hence, the vectors are linearly independent.
6. (a) Recall that cos 2x = cos2 x – sin2 x; that is,
1v1 + (–1)v2 + (–1)v3 = 0
Hence, S is not a linearly independent set of vectors.
(b) We can use the above identity to write any one of the vectors vi
as a linearcombination of the other two. Since no one of these vectors is a multiple of any other,they are pairwise linearly independent. Thus any two of these vectors form a basis forV.
7. (a) Clearly w = 3u1 – 7u2, so the coordinate vector relative to u1, u2 is (3, –7)
a b c3 6
3 6
0 1
1 0
0 8
12 4−
+
−−
+
−− −
+
−
=
d
1 0
1 2
0 0
0 0
1 0 0
1 1 0
1 1 1
244 Exercise Set 5.4
(b) If w = au1 + bu2, then equating coordinates yields the system of equations
2a + 3b = 1
–4a + 8b = 1
This system has the solution a = 5/28, b = 3/14. Thus the desired coordinate vector is(5/28, 3/14).
8. (a) Since w = 12(u1 + u2), the coordinate vector of w relative to S = u1, u2 is (w)S
= (12, 12).
(b) Since w = 12(–u1 + u2), the coordinate vector of w relative to S = u1, uθ2 is (w)S
=
(– 12, 12).
(c) Since w = u2 = 0u1 + u2, the coordinate vector of w relative to S = u1, u2 is (w)S
=(0, 1).
9. (a) If v = av1 + bv2 + cv3, then
a + 2b + 3c = 2
2b + 3c =–1
3c = 3
From the third equation, c = 1. Plugging this value into the second equation yields b= –2, and finally, the first equation yields a = 3. Thus the desired coordinate vector is(3, –2, 1).
10. (b) If p = ap1 + bp2 + cp3, then equating coefficients yields the system of equations
a + b = 2
a + c = –1
b + c = 1
This system has the solution a = 0, b = 2, and c = –1. Thus the desired coordinatevector is (0, 2, –1).
12. The augmented matrix of the system reduces to
Hence, x1 = s, x2 = 0, and x3 = s. Thus the solution space is spanned by (1, 0, 1) and hasdimension 1.
1 0 1 0
0 1 0 0
0 0 0 0
−
Exercise Set 5.4 245
15. If we reduce the augmented matrix to row-echelon form, we obtain
Thus x1 = 3r – s, x2 = r, and x3 = s, and the solution vector is
Since (3, 1, 0) and (–1, 0, 1) are linearly independent, they form a basis for the solutionspace and the dimension of the solution space is 2.
16. Since the determinant of the system is not zero, the only solution is x1 = x2 = x3 = 0.
Hence there is no basis for the solution space and its dimension is zero.
19. (a) Any two linearly independent vectors in the plane form a basis. For instance, (1, –1,–1) and (0, 5, 2) are a basis because they satisfy the plane equation and neither is amultiple of the other.
(c) Any nonzero vector which lies on the line forms a basis. For instance, (2, –1, 4) willwork, as will any nonzero multiple of this vector.
(d) The vectors (1, 1, 0) and (0, 1, 1) form a basis because they are linearly independentand
a(1, 1, 0) + c(0, 1, 1) = (a, a + c, c)
20. This space is spanned by the vectors 0, x, x2 and x3. Only the last three vectors form alinearly independent triple. Thus the space has dimension 3.
21. (a) We consider the three linear systems
− + =− =− =
k k
k k
k k
1 2
1 2
1 2
1 0 0
2 2 0 1 0
3 2 0 0 1
x
x
x
r s
r
s
1
2
3
3 3
1
0
=−
=
+−
r s
1
0
1
1 3 1 0
0 0 0 0
0 0 0 0
−
246 Exercise Set 5.4
which give rise to the matrix
A row-echelon form of the matrix is
from which we conclude that e3 is in the span of v1, v2, but e1 and e2 are not. Thusv1, v2, e1 and v1, v2, e2 are both bases for R3.
22. We consider the linear system of equations k1v1 + k2v2 + k3e1 + k4e2 + k5e3 + k6e4 = 0which give rise to the coefficient matrix
A row-echelon form of the matrix is
If we eliminate any two of the final four columns of this matrix, we obtain a 4 × 4 matrix.If the determinant of this matrix is zero, then the system of equations has a nontrivialsolution, so the corresponding vectors are linearly dependent. Otherwise, they are linearlyindependent. If we include column 3 (which corresponds to e1) in the determinant, itsvalue is zero. Otherwise, it is not. Thus we may add any two of the vectors e2, e3, and e4 tothe set v1, v2 to obtain a basis for R4.
1 3 1 0 0 0
0 1 1 1 4 0 0
0 0 0 1 2 0
0 0 0 0 1 2 3
−− −
1 3 1 0 0 0
4 8 0 1 0 0
2 4 0 0 1 0
3 6 0 0 0 1
−−
−−
1 1 1 0 0
0 1 3 0 1
0 0 1 1 2 0
− −
−−−
1 1 1 0 0
2 2 0 1 0
3 2 0 0 1
Exercise Set 5.4 247
23. Since u1, u2, u3 has the correct number of elements, we need only show that they arelinearly independent. Let
au1 + bu2 + cu3 = 0
Thus
av1 + b(v1 + v2) + c(v1 + v2 + v3) = 0
or
(a + b + c)v1 + (b + c)v2 + cv3 = 0
Since v1, v2, v3 is a linearly independent set, the above equation implies that a + b + c =b + c = c = 0. Thus, a = b = c = 0 and u1, u2, u3 is also linearly independent.
24. (a) Note that the polynomials 1, x, x2, …, xn form a> set of n + 1 linearly independentfunctions in F(–∞, ∞).
(b) From Part (a), dim F(–∞, ∞) > n for any positive integer n. Thus F(–∞, ∞) is infinite-dimensional.
25. First notice that if v and w are vectors in V and a and b are scalars, then (av + bw)S
=a(v)
S+ b(w)
S. This follows from the definition of coordinate vectors. Clearly, this result
applies to any finite sum of vectors. Also notice that if (v)S
= (0)S, then v = 0. Why?
Now suppose that k1v1 + … + krv
r= 0. Then
(k1v1 + … + krv
r)
S= k1(v1)S
+ … + kr(v
r)
S
= (0)S
Conversely, if k1(v1)S+ … + kr(vr)S = (0)S, then
(k1v1 + … + krvr)S
= (0)S, or k1v1 + … + k
rv
r= 0
Thus the vectors v1, …, vr
are linearly independent in V if and only if the coordinate vectors(v1)S
, …, (vr)
Sare linearly independent in Rn.
248 Exercise Set 5.4
26. If every vector v in V can be written as a linear combination v = a1v1 + … + arv
rof v1, …,
vr, then, as in Exercise 24, we have (v)S
= a1(v1)S+ … + a
r(v
r)
S. Hence, the vectors (v1)S
,…, (v
r)
Sspan a subspace of Rn. But since V is an n-dimensional space with, say, the basis
S = u1, …, un, then if u = b1u1 + … + b
nu
n, we have (u)
S= (b1, …, b
n); that is, every
vector in Rn represents a vector in V. Hence (v1)S, …, (v
r)
S spans Rn.
Conversely, if (v1)S, …, (v
r)
S spans Rn, then for every vector (b1, …, b
n) in Rn, there
is an r-tuple (a1, …, ar) such that
(b1, …, bn) = a1(v1)S
+ … + ar (vr)S
= (a1v1 + … + arvr)S
Thus a1v1 + … + arv
r= b1u1 + … + b
nu
n, so that every vector in V can be represented as
a linear combination of v1, …, vr.
27. (a) Let v1, v2, and v3 denote the vectors. SinceS = 1, x, x2 is the standard basis for P2, wehave (v1)S
= (–1, 1, –2), (v2)S= (3, 3, 6), and (v3)S
= (9, 0, 0). Since (–1, 1, –2), (3,3, 6), (9, 0, 0) is a linearly independent set of three vectors in R3, then it spans R3.Thus, by Exercises 24 and 25, v1, v2, v3 is linearly independent and spans P2. Henceit is a basis for P2.
28. (a) It is clear from the picture that the x′-y′ coordinates of (1, 1) are (0, 2).
(d) Let (a, b) and (a′, b′) denote the coordinates of a point with respect to the x-y and x′-y′ coordinate systems, respectively. If (a, b) is positioned as in Figure (1), then wehave
a′ = a – b and b′ = 2b
y
l
l
y'
x and x'45∞
2
Exercise Set 5.4 249
(d)
Figure 1 Figure 2
These formulas hold no matter where (a,b) lies in relation to the coordinate axes.Figure (2) shows another configuration, and you should draw similar pictures for all ofthe remaining cases.
30. See Theorem 5.4.7.
31. There is. Consider, for instance, the set of matrices
Each of these matrices is clearly invertible. To show that they are linearly independent,consider the equation
aA + bB + cC + dD =
This implies that
0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0
a
b
c
d
=
0
0
0
0
0 0
0 0
A B
C D
=
=
=
=
0 1
1 1
1 0
1 1
1 11 1
0 1and
11 0
y'
y
(a, b)
a
b
x and x'
(a', b')
45∞45∞
2b
a' b
y'
y
(a, b)
x and x'
(a', b')
45∞aa
b
2b
45∞
250 Exercise Set 5.4
The above 4 × 4 matrix is invertible, and hence a = b = c = d = 0 is the only solution. Andsince the set, A, B, C, and D consists of 4 linearly independent vectors, it forms a basis forM22.
32. (b) The most general 2 × 2 symmetric matrix has the form . Hence, if n = 2,the matrices
form a basis and the dimension is therefore 3.
If n = 3, the matrices
form a basis and the dimension is therefore 6.
In general, there are n2 elements in an n × n matrix with on or
above the main diagonal. Therefore the dimension of the subspace of n × n symmetric
matrices is n(n + 1)/2.
(c) Since there are n2 elements in each n × n matrix, with n elements on the maindiagonal, there are (n2 – n)/2 elements above the diagonal. Thus, any triangular matrixcan have at most (n2 – n)/2 + n = (n2 + n)/2 nonzero elements. The set of n × n
matrices consisting of all zeros except for a 1 in each of these spots will form a basisfor the space. Consequently, the space will have dimension n(n + 1)/2.
33. (a) The set has 10 elements in a 9 dimensional space.
n n n n2
2
1
2
+=
+( )
1 0 0
0 0 0
0 0 0
, ,
0 0 0
0 1 0
0 0 0
,
0 0 0
0 0 0
0 0 1
0 1 0
1 0 0
0 0 0
00 0 1
0 0 0
1 0 0
0 0 0
0 0 1
0 1 0
, and
1 0
0 0and
, ,
0 0
0 1
0 1
1 0
a c
c b
Exercise Set 5.4 251
35. (b) The equation x1 + x2 + … + xn
= 0 can be written as x1 = –x2 –x3 – … –xn
where x2,x3, …, x
ncan all be assigned arbitrary values. Thus, its solution space should have
dimension n – 1. To see this, we can write
The n – 1 vectors in the above equation are linearly independent, so the vectors doform a basis for the solution space.
36. (a) If p1(1) = p2(1) = 0, then (p1 + p2)(1) = p1(1) + p2(1) = 0. Also, kp1(1) = 0. Hence Wis closed under both vector addition and scalar multiplication.
(b) We are looking at the set W of polynomials ax2 + bx + c in P2 where a + b + c = 0 orc = –(a + b). Thus it would appear that the dimension of W is 2.
(c) The polynomial p(x) = ax2 + bx + c will be in the subspace W if and only if p(1) = a+ b + c = 0, or c = –a – b. Thus,
where a and b are arbitrary. The vectors and are linearly independent and
hence the polynomials x2 – 1 and x – 1 form a basis for W.
0
1
1−
1
0
1−
a
b
c
a
b
a b
a
=− −
=−
1
0
1
+−
b
0
1
1
−
+
−
1
1
0
0
0
1
0
1
0
0
3
x
+ +
−
xn
1
0
0
0
1
x
x
x
x
x x x
n
n1
2
3
2 3
=
− − − −x
x
2
3
xn
252 Exercise Set 5.4
EXERCISE SET 5.5
3. (b) Since the equation Ax = b has no solution, b is not in the column space of A.
(c) Since A = b, we have b = c1 – 3c2 + c3.
(d) Since A = b, we have b = c1 + (t – 1)c2 + tc3 for all real numbers t.
5. (a) The general solution is x1 = 1 + 3t, x2 = t. Its vector form is
Thus the vector form of the general solution to Ax = 0 is
(c) The general solution is x1 = – 1 + 2r – s – 2t, x2 = r, x3 = s, x4 = t. Its vector form is
−
+
+
−1
0
0
0
2
1
0
0
r s
11
0
1
0
2
0
0
1
+
−
t
t3
1
1
0
3
1
+
t
1
1t
t
−
1
3
1
−
253
Thus the vector form of the general solution to Ax = 0 is
6. (a) Since the reduced row-echelon form of A is
the solution to the equation Ax = 0 is x1 = 16t, x2 = 19t, x3 = t. Thus
is a basis for the nullspace.
(c) Since the reduced row-echelon form of A is
one solution to the equation Ax = 0 is x1 = –s + 2t, x2 = –s – 4t, x3 = s, x4 = 7t. Thusthe set of vectors
is a basis for the nullspace.
−−
−
1
1
1
0
2
4
0
7
and
1 0 1 2 7
0 1 1 4 7
0 0 0 0
−
16
19
1
1 0 16
0 1 19
0 0 0
−−
r s t
2
1
0
0
1
0
1
1
+
−
+
−22
0
0
1
254 Exercise Set 5.5
(e) The reduced row-echelon form of A is
One solution to the equation Ax = 0 is x1 = –2s – 16t, x2 = 2t, x3 = 5t, x4 = s, and x5= 12t. Hence the set of vectors
is a basis for the nullspace of A.
8. (a) From a row-echelon form of A, we have that the vectors (1, –1, 3) and (0, 1, –19) area basis for the row space of A.
(c) From a row-echelon form of A, we have that the vectors (1, 4, 5, 2) and (0, 1, 1, 4/7)are a basis for the row space of A.
9. (a) One row-echelon form of AT is
Thus a basis for the column space of A is
1
5
7
0
1
1
and
1 5 7
0 1 1
0 0 0
−
−
2
0
0
1
0
16
2
5
0
12
and
1 0 0 2 4 3
0 1 0 0 1 6
0 0 1 0 5 12
0 0 0 0 0
0 0 0 0 0
−−
Exercise Set 5.5 255
9. (c) One row-echelon form of AT is
Thus a basis for the column space of A is
10. (a) Since, by 8(a), the row space of A has dimension 2, any two linearly independent rowvectors of A will form a basis for the row space. Because no row of A is a multiple ofanother, any two rows will do. In particular, the first two rows form a basis and Row 3= Row2 + 2(Row 1).
(c) Refer to 8(c) and use the solution to 10(a), above. In particular, the first two rowsform a basis and Row 3 = Row 1 – Row 2.
(e) Let r1, r2, …, r5 denote the rows of A. If we observe, for instance, that r1 = –r3 + r4 andthat r2 = 2r1 + r5, then we see that r3, r4, r5 spans the row space. Since the dimensionof this space is 3 (see the solution to Exercise 6(e)), the set forms a basis.
For those who don’t wish to rely on insight, set ar1 + br2 + cr3 + dr4 + er5 = 0 andsolve the resulting homogeneous system of equations by finding the reduced row-echelon form for AT. This yields
a = –s + 2t
b = –t
c = –s
d = s
e = t
so that (–s + 2t)r1 – tr2 – sr3 + sr4 + tr5 = 0, or
s(–r1 – r3 + r4) + t(2r1 – r2 + r5) = 0
1
2
1
0
1
1−
−
and
1 2 1
0 1 1
0 0 0
0 0 0
−−
256 Exercise Set 5.5
Since this equation must hold for all values of s and t, we have
r1 = –r3 + r4 and r2 = 2r1 + r5
which is the result obtained above.
11. (a) The space spanned by these vectors is the row space of the matrix
One row-echelon form of the above matrix is
and the reduced row-echelon form is
Thus (1, 1, –4, –3), (0, 1, –5, –2), (0, 0, 1, –1/2) is one basis. Another basis is (1, 0,0, –1/2), (0, 1, 0, –9/2), (0, 0, 1, –1/2).
12. (a) If we solve the vector equation
(*) av1 + bv2 + cv3 + dv4 = 0
we obtain the homogeneous system
a – 3b – c – 5d = 0
3b + 3c + 3d = 0
a + 7b + 9c + 5d = 0
a + b + 3c – d = 0
1 0 0 1 2
0 1 0 9 2
0 0 1 1 2
−−−
1 1 4 3
0 1 5 2
0 0 1 1 2
− −− −
−
1 1 4 3
2 0 2 2
2 1 3 2
− −−
−
Exercise Set 5.5 257
The reduced row-echelon form of the augmented matrix is
Thus v1, v2 forms a basis for the space. The solution is a = –2s + 2t, b = –s – t, c = s,d = t. This yields
(–2s + 2t)v1 + (–s – t)v2 + sv3 + tv4 = 0
or
s(–2v1 – v2 + v3) + t(2v1 – v2 + v4) = 0
Since s and t are arbitrary, set s = 1, t = 0 and then s = 0, t = 1 to obtain thedependency equations
–2v1 – v2 + v3 = 0
2v1 – v2 + v4 = 0
Thus
v3 = 2v1 + v2
and
v4 = –2v1 + v2
12. (c) If we solve the vector equation
(*) av1 + bv2 + cv3 + dv4 + ev5 = 0
we obtain the homogeneous system
a – 2b + 4c – 7e = 0
–a + 3b – 5c + 4d+ 18e = 0
5a + b + 9c + 2d + 2e = 0
2a + 4c – 3d – 8e = 0
1 0 2 2 0
0 1 1 1 0
0 0 0 0 0
0 0 0 0 0
−
258 Exercise Set 5.5
The reduced row-echelon form of the augmented matrix is
This tells us that v1, v2, v4 is the desired basis. The solution is a = –2s + t, b = s – 3t,c = s, d = –2t, and e = t. This yields
(–2s + t)v1 + (s – 3t)v2 + sv3 – 2tv4 + tv5 = 0
or
s(–2v1 + v2 + v3) + t(v1 – 3v2 – 2v4 + v5) = 0
Since s and t are arbitrary, set s = 1, t = 0 and then s = 0, t = 1 to obtain thedependency equations
–2v1 + v2 + v3 = 0
v1 – 3v2 – 2v4 + v5 = 0
Thus
v3 = 2v1 – v2
and
v5 = –v1 + 3v2 + 2v4
13. Let A be an n × n invertible matrix. Since AT is also invertible, it is row equivalent to In. It
is clear that the column vectors of In
are linearly independent. Hence, by virtue of Theorem5.5.5, the column vectors of AT, which are just the row vectors of A, are also linearlyindependent. Therefore the rows of A form a set of n linearly independent vectors in Rn,and consequently form a basis for Rn.
15. (a) We are looking for a matrix so that the only solution to the equation Ax = 0 is x = 0.
Any invertible matrix will satisfy this condition. For example, the nullspace of the
matrix A = is the single point (0, 0, 0).1 0 0
0 1 0
0 0 1
1 0 2 0 1 0
0 1 1 0 3 0
0 0 0 1 2 0
0 0 0 0 0 0
−−
Exercise Set 5.5 259
(b) In this case, we are looking for a matrix so that the solution of Ax = 0 is
one-dimensional. Thus, the reduced row-echelon form of A has one column without
a leading one. As an example, the nullspace of the matrix A = is
span , a plane in R3.
(c) In this case, we are looking for a matrix so that the solution space of Ax = 0 is
two-dimensional. Thus, the reduced row-echelon form of A has two columns without
leading ones. As an example, the nullspace of the matrix A = is
span , a plane in R3.
16. (a) True. Since premultiplication by an elementary matrix is equivalent to an elementaryrow operation, the result follows from Theorem 5.5.3.
15. (c) False. For instance, let A = and EA = . Then the column space of A is
span and the column space of EA is span . These are not the same
spaces.
(d) True by Theorem 5.5.1
(e) False. The row space of an invertible n × n matrix is the same as the row space of In,
which is Rn. The nullspace of an invertible matrix is just the zero vector.
1
0
1
2
1 1
0 0
1 1
2 2
−
1
1
0
1
0
1
,
1 1 1
0 0 0
0 0 0
−
1
1
1
1 0 1
0 1 1
0 0 0
−−
260 Exercise Set 5.5
17. (a) The matrices will all have the form where s andt are any real numbers.
(b) Since A and B are invertible, their nullspaces are the origin. The nullspace of C is theline 3x + y = 0. The nullspace of D is the entire xy-plane.
18. Let A = [1 1 1]. Then Ax = [1] has the particular solution [1 0 0]T and Ax = [0] has thegeneral solution [s t – s – t]T. Thus the general solution can be written as
19. Theorem: If A and B are n × n matrices and A is invertible, then the row space of AB is therow space of B.
Proof: If A is invertible, then there exist elementary matrices E1, E2, …, Ek
such that
A = E1E2 … Ekin
or
AB = E1E2 … EkB
Thus, Theorem 5.5.4 guarantees that AB and B will have the same row spaces.
x
x
x
s
1
2
3
1
0
0
1
0
1
=
+
+−
t
0
1
1
3 5
3 5
3 5
0 0
0 0
3 5
s s
t ts t
−−
=
−
+
−
Exercise Set 5.5 261
EXERCISE SET 5.6
2. (a) The reduced row-echelon form for A is
Thus rank (A) = 2. The solution to Ax = 0 is x = 16t, y = 19t, z = t, so that the nullityis one. There are three columns, so we have 2 + 1 = 3.
(c) The reduced row-echelon form for A is
Thus rank (A) = 2. The null space will have dimension two since the solution to Ax =0 has two parameters. There are four columns, so we have 2 + 2 = 4.
4. Recall that rank(A) is the dimension of both the row and column spaces of A. Use theDimension Theorem to find the dimensions of the nullspace of A and of AT, recalling that ifA is m × n, then AT is n × m, or just refer to the chart in the text.
7. Use Theorems 5.6.5 and 5.6.7.
(a) The system is consistent because the two ranks are equal. Since n = r = 3, n – r = 0and therefore the number of parameters is 0.
(b) The system is inconsistent because the two ranks are not equal.
1 0 1 2 7
0 1 1 4 7
0 0 0 0
−
1 0 16
0 1 19
0 0 0
−−
263
7. (d) The system is consistent because the two ranks are equal. Here n = 9 and r = 2, sothat n – r = 7 parameters will appear in the solution.
(f) Since the ranks are equal, the system is consistent. However A must be the zeromatrix, so the system gives no information at all about its solution. This is reflected inthe fact that n – r = 4 – 0 = 4, so that there will be 4 parameters in the solution for the4 variables.
9. The system is of the form Ax = b where rank(A) = 2. Therefore it will be consistent if andonly if rank([A|b]) = 2. Since [A|b] reduces to
the system will be consistent if and only if b3 = 4b2 – 3b1, b4 = –b2 + 2b1, and b5 = 8b2 – 7b1,where b1 and b2 can assume any values.
10. Suppose that A has rank 2. Then two of its column vectors are linearly independent. Thus,by Theorem 5.6.9, at least one of the 2 × 2 submatrices has nonzero determinant.
Conversely, if at least one of the determinants of the 2 × 2 submatrices is nonzero, then,by Theorem 5.6.9, at least two of the column vectors must be linearly independent. Thusthe rank of A must be at least 2. But since the dimension of the row space of A is at most2, A has rank at most 2. Thus, the rank of A is exactly 2.
11. If the nullspace of A is a line through the origin, then it has the form x = at, y = bt, z = ct
where t is the only parameter. Thus nullity(A) = 3 – rank(A) = 1. That is, the row andcolumn spaces of A have dimension 2, so neither space can be a line. Why?
1 3
0 1
0 0 4 3
0 0 2
0 0 8
1
2 1
3 2 1
4 2 1
5 2
−−
− ++ −
−
b
b b
b b b
b b b
b b ++
7 1b
264 Exercise Set 5.6
12. (a) If we attempt to reduce A to row-echelon form, we find that
A→ if t ≠ 0
→ if t ≠ 1
→ if t ≠ – 2
Thus rank(A) = 3 if t ≠ 0, 1, –2. If t = 0, rank(A) = 3 by direct computation. If t = 1,rank(A) = 1 by inspection, and if t = –2, rank(A) = 2 by the above reduction.
13. Call the matrix A. If r = 2 and s = 1, then clearly rank(A) = 2. Otherwise, either r – 2 or s– 1 = 0 and rank(A) = 3. Rank(A) can never be 1.
14. Call the matrix A and note that rank(A) is either 1 or 2. Why? By Exercise 10, rank(A) =1 if and only if
Thus we must have x2 – y = xy – z = y2 – xz = 0. If we let x = t, the result follows.
16. (a) The column space of the matrix
is the xy-plane.
1 0 1
0 1 0
0 0 0
x y
x
x z
y
y z
x y10
10 0, ,= = =and
1 1
0 1 1
0 0 2
1 1
0 1 1
0 0 1
t
t
t
−+
→ −
1 1
0 1 1
0 1 1
t
t
−− − +
( )
1 1
0 1 1
1 1
1 1
0 1 1
0 1 1
t
t t
t
t
t t
t t
− −
→ − −
− − 22
Exercise Set 5.6 265
16. (b) The nullspace is the line x = t, y = 0, z = –t.
(c) The row space is the plane x – z = 0.
(d) In general, if the column space of a 3 × 3 matrix is a plane through the origin then thenullspace will be a line through the origin and the row space will be a plane throughthe origin. This follows from the fact that if the column space is a plane, the matrixmust have rank 2 and therefore nullity 1. The one dimensional subspaces of R3 arelines through the origin and the two dimensional subspaces are planes through theorigin.
Similarly, if the column space represents a line through the origin, then so does therow space. In this case, the nullspace will represent a plane through the origin.
17. (a) False. Let A =
(c) True. If A were an m × n matrix where, say, m > n, then it would have m rows, eachof which would be a vector in R
n. Thus, by Theorem 5.4.2, they would form a linearly
dependent set.
18. (a) Since the row rank equals the column rank and since A has 3 rows and more than 3columns, its maximum rank is 3. Hence, the number of leading 1’s in its reduced row-echelon form is less than or equal to 3.
(b) Since nullity (A) ≤ 5, there could be as many as 5 parameters in the general solutionof Ax = 0. The maximum of 5 would occur if A were the zero matrix.
(c) Since A has 3 columns and more than 3 rows, its maximum rank is 3. Hence themaximum number of leading 1’s in the reduced row-echelon form of A is 3.
(d) Since nullity (A) ≤ 3, there could be as many as 3 parameters in the general solutionto Ax = 0.
1 0 0
0 1 0
266 Exercise Set 5.6
SUPPLEMENTARY EXERCISES 5
1. (b) The augmented matrix of this system reduces to
Therefore, the solution space is a plane with equation 2x – 3y + z = 0
(c) The solution is x = 2t, y = t, z = 0, which is a line.
2. Let A be the coefficient matrix. Since det(A) = –(1 – s)2(2 + s), the solution space is theorigin unless s = 1 or s = –2. If s = 1, the solution space is the plane x1 + x2 + x3 = 0. If s= –2, the solution space is the line x1 = t, x2 = t, x3 = t.
Alternative Solution: Let A be the coefficient matrix. We can use the DimensionTheorem and the result of Exercise 12(a) of Section 5.6 to solve this problem. Recall thatrank(A) = 1 if s = 1, rank(A) = 2 if s = –2, and rank(A) = 3 for all other values of s. Hencenullity(A) = 2 if s = 1, nullity(A) = 1 if s = –2, and nullity(A) = 0 for all other values of s.Thus, if s = 1, then the solution space is a two-dimensional subspace of R3, i.e., a planethrough the origin. If s = –2, then the solution space is a one-dimensional subspace of R3,i.e., a line through the origin. If s ≠ 1 and s ≠ –2, then the solution space is the zero-dimensional subspace of R3, i.e., 0.
4. (a) The identities
sin(x + θ) = cos θ sin x + sin θ cos x
cos(x + θ) = cos θ cos x – sin θ sin x
hold for all values of x and θ. Hence
2 3 1 0
0 0 0 0
0 0 0 0
−
267
(*)f1 = (cos θ)f + (sin θ)g
g1 = (– sin θ)f + (cos θ)g
That is, f1 and g1 are linear combinations of f and g and therefore belong to W.
4. (b) If we solve the system (*) for f and g, we obtain
f = (cos θ)f1 + (– sin θ)g1
g = (sin θ)f1 + (cos θ)g1
Hence, any linear combination of f and g is also a linear combination of f1 and g1 andthus f1 and g1 span W. Since the dimension of W is 2 (it is spanned by 2 linearlyindependent vectors), Theorem 5.4.6(b) implies that f1 and g1 form a basis for W.
5. (a) We look for constants a, b, and c such that v = av1 + bv2 + cv3, or
a + 3b + 2c = 1
–a + c = 1
This system has the solution
a = t – 1 b = 23 – t c = t
where t is arbitrary. If we set t = 0 and t = 1, we obtain v = (–1)v1 + (2/3)v2 and v =(–1/3)v2 + v3, respectively. There are infinitely many other possibilities.
(b) Since v1, v2, and v3 all belong to R2 and dim(R2) = 2, it follows from Theorem 5.4.2 thatthese three vectors do not form a basis for R2. Hence, Theorem 5.4.1 does not apply.
6. Suppose that there are constants c1, …, cn, not all zero, such that c1Av1 + … + c
nA v
n= 0.
Then
A(c1v1 + … + cnv
n) = 0
Since the vectors v1, …, vn
are linearly independent, the n × 1 matrix c1v1 + … + cnv
n
cannot equal 0. Thus the equation Ax = 0 has a non-trivial solution, and so A is notinvertible. Therefore, by Theorem 1.5.3, A is invertible if and only if Av1, Av2, …, Av
nare
linearly independent.
268 Supplementary Exercises 5
7. Consider the polynomials x and x + 1 in P1. Verify that these polynomials form a basis forP1.
8. (c) Since the odd numbered rows are all repeats of Row 1 and the even numbered rowsare all repeats of Row 2, while Rows 1 and 2 are linearly independent, an n × n
checker board matrix has rank 2 whenever n ≥ 2. Since the nullity is n minus therank, we have nullity = n – 2.
10. (a) If p belongs to the set, then it contains only even powers of x. Since this set is closedunder polynomial addition and scalar multiplication (Why?), it is a subspace of P
n.
One basis is the set 1, x2, x4, …, x2m where 2m = n if n is even and 2m = n – 1 if nis odd.
(b) If p belongs to this set, then its constant term must be zero. Since this set is closedunder polynomial addition and scalar multiplication (Why?), it is a subspace of P
n.
One basis is the set x, x2, …, xn.
12. (a) A 3 × 3 symmetric matrix has the form A = , so a basis is formed by the
six matrices
(b) A 3 × 3 skew-symmetric matrix has the form A = , so a basis is formed
by the three matrices
13. (a) Since = – 1≠ 0, the rank is 2.
(b) Since all three 2 × 2 subdeterminants are zero, the rank is 1.
1 0
2 1−
0 1 0
1 0 0
1 0 0
0 0 1
0 0 0
1 0 0
−
−
, ,,
0 0 0
0 0 1
1 1 0−
0
0
0
a b
a c
b c
−− −
1 0 0
0 0 0
0 0 0
0 1 0
1 0 0
0 0 0
, , ,
0 0 1
0 0 0
1 0 0
0 0 0
0 1 0
0 0 0
,, ,
0 0 0
0 0 1
0 1 0
0 0 0
0 0 0
0 0 1
a b c
b d e
c e f
Supplementary Exercises 5 269
(c) Since the determinant of the matrix is zero, its rank is less than 3. Since = –1≠ 0, the rank is 2.
(d) Since the determinant of the 3 × 3 submatrix obtained by deleting the last column is30 ≠ 0, the rank of the matrix is 3.
14. Call the matrix A. Since the determinant of every 5 × 5, 4 × 4, and 3 × 3 submatrix is zero,rank(A) ≤ 2. Since
det = –a5ja
i6
for i = 1, 2, …, 4 and j = 1, …, 5, then rank(A) = 2 if any of these determinants is nonzero.Otherwise, if anyof the numbers a
ij≠ 0, then rank(A) = 1 and if a
ij= 0 for all i and j, then
rank(A) = 0.
15. (b) Let S
= v1, …, vn and let u = u1v1 + … + u
nv
n. Thus (u)
S= (u1, …, u
n). We have
ku = ku1v1 + … + kunv
n
so that (ku)S
= (ku1, …, kun) = k(u
1, …, un). Therefore (ku)S = k(u)S.
0 6
5 56
a
a a
i
j
1 0
2 1−
270 Supplementary Exercises 5
EXERCISE SET 6.1
1. (c) Since v + w = (3, 11), we have
⟨u, v + w⟩ = 3(3) + (–2)(11) = –13
On the other hand,
⟨u, v⟩ = 3(4) + (–2)(5) = 2
and
⟨u, w⟩ = 3(–1) + (–2)(6) = –15
(d) Since ku = (–12, 8) and kv = (–16, –20) , we have
⟨ku, v⟩ = (–12)(4) + (8)(5) = –8
and
⟨u, kv⟩ = 3(–16) + (–2)(–20) = –8
Since ⟨u,v⟩ = 2, k⟨u, v⟩ = –8.
2. (c) Since v + w = (3, 11), we have
⟨u, v + w⟩ = 4(3)(3) + 5(–2)(11) = –74
On the other hand,
⟨u, v⟩ = 4(3)(4) + 5(–2)(5) = –2
and
⟨u, w⟩ = 4(3)(–1) + 5(–2)(6) = –72
271
3. (a) ⟨u, v⟩ = 3(–1) – 2(3) + 4(1) + 8(1) = 3
4. (a) ⟨p, q⟩ = (–2)(4) + 1(0) + 3(–7) = –29
5. (a) By Formula (4),
(b) We have ⟨u, v⟩ = 9(–3)(1) + 4(2)(7) = 29.
6. (a) Since
then
= [ 5v1 – v2 –v1 + 10v2 ]
= 5u1v1 – u1v2 – u2v1 + 10u2v2
7. (a) By Formula (4), we have ⟨u, v⟩ = vT AT Au where
A =
3 0
0 5
u
u
1
2
v v1 2
1
2
5 1
1 10,
−−
u
u
A AT =
−−
5 1
1 10
uu,,vv =
=
v vu
u1 2
1
2
3 0
0 2
3 0
0 2
vv vu
u
v vu
u
1 21
2
1 21
9 0
0 4
9 4
= 22
1 1 2 29 4
= +u v u v
272 Exercise Set 6.1
8. (a) (1) ⟨u, v⟩ = 3u1v1 + 5u2v2 = 3v1u1 + 5v2u2 = ⟨v, u⟩
(2) If w = (w1, w2), then
⟨u + v, w⟩ = 3(u1 + v1)w1 + 5(u2 + v2)w2
= (3u1w1 + 5u2w2) + (3v1w1 + 5v2w2)
= ⟨u, w⟩ + ⟨v, w⟩
(3) ⟨ku, v⟩ = 3(ku1)v1 + 5(ku2)v2
= k(3u1v1 + 5u2v2)
= k⟨u, v⟩
(4) ⟨v, v⟩ = 3v12 + 5v2
2 ≥ 0
Moreover, ⟨v, v⟩ = 0 if and only if v1 = v2 = 0, or v = 0.
9. (b) Axioms 1 and 4 are easily checked. However, if w = (w1, w2, w3), then
⟨u + v, w⟩ = (u1 + v1)2w1
2 + (u2 + v2)2w2
2 + (u3 + v3)2w3
2
= ⟨u, w⟩ + ⟨v, w⟩ + 2u1v1w12 + 2u2v2w2
2 + 2u3v3w32
If, for instance, u = v = w = (1, 0, 0), then Axiom 2 fails.
To check Axiom 3, we note that ⟨ku, v⟩ = k2⟨u, v⟩. Thus ⟨ku, v⟩ ≠ k⟨u, v⟩ unless k= 0 or k = 1, so Axiom fails.
(c) (1) Axiom 1 follows from the commutativity of multiplication in R.
(2) If w = (w1, w2, w3), then
⟨u + v, w⟩ = 2(u1 + v1)w1 + (u2 + v2)w2 + 4(u3 + v3)w3
= 2u1w1 + u2w2 + 4u3w3 + 2v1w1 + v2w2 + 4v3w3
= ⟨u, w⟩ + ⟨v, w⟩
(3) ⟨ku, v = 2(ku1)v1 + (ku2)v2 + 4(ku3)v3 = k⟨u, v⟩
(4) ⟨v, v⟩ = 2v12 + v2
2 + 4v32 ≥ 0
= 0 if and only if v1 = v2 = v3 = 0, or v = 0
Thus this is an inner product for R3.
Exercise Set 6.1 273
10. (c) Since , we have
Thus w = 125 = 5 5.
11. We have u – v = (–3, –3).
(b) d(u, v) = (–3, –3) = [3(9) + 2(9)]1/2 = 45 = 3 5
(c) From Problem 10(c), we have
Thus
d(u, v) = 117 = 3 13
12. (a) p = [(–2)2 + (3)2 + (2)2]1/2 = 17
13. (a) A = [(–2)2 + (5)2 + (3)2 + (6)2]1/2 = 74
14. Since p – q = 1 – x – 4x2, we have
d(p, q) = ⟨p – q, p – q⟩1/2 = [12 + (–1)2 + (–4)2]1/2 = 3 2
d( , )u v[ ] = − − −
−
−−
23 3
2 1
1 13
3
3
= 117
w2 1 3
2 1
1 13
1
3= −
−−
−
= [[ ]−−
=5 40
1
3125
A AT =
−−
2 1
1 13
274 Exercise Set 6.1
15. (a) Since , we have
d(A, B) = ⟨A – B, A – B⟩1/2 = [62 + (–1)2 + 82 + (–2)2]1/2 = 105
16. (a) We have ⟨u + v, v + w⟩ = ⟨u, v⟩ + ⟨u, w⟩ + ⟨v, v⟩ + ⟨v, w⟩ = 2 + 5 + (2)2 –3 = 8.
(d) We have u + v2 = ⟨u + v, u + v⟩ = ⟨u, u⟩ + 2⟨u, v⟩ + ⟨v, v⟩ = (1)2 + 2(2) + (2)2 = 9.Hence u + v = 3.
17. (a) For instance,
(b) We have
d(p, q)= p – q
= 1 – x
= −
= − +
−
−
∫
∫
( )
( )
/
1
1 2
21
1 1 2
21
1
x dx
x x dx
= − +
=
−
1 2
23
1
1 1 2
3
22
3
/
/
x xx
=1 2
2
36
/
x x dxx
=
=
=−
−∫ 2
1
1 1 2 3
1
1 1 2
3
2
3
1 2
A B− =−−
6 1
8 2
Exercise Set 6.1 275
18. (a) Since we are looking for points (x, y) with (x, y) = 1, we have
or 1 . This is the ellipse shown in the figure.
20. Observe that
(*) u + v2 = ⟨u + v, u + v⟩ = ⟨u, u⟩ + 2⟨u, v⟩ + ⟨v, v⟩
and
(**) u – v2 = ⟨u –v, u – v⟩ = ⟨u, u⟩ – 2⟨u, v⟩ + ⟨v, v⟩
Thus if we add (*) and (**), we obtain the equation
u + v2 + u – v2 = 2u2 + 2v2
21. If, in the solution to Exercise 20, we subtract (**) from (*) and divide by 4, we obtain thedesired result.
22. If, for instance, , then ⟨V, V⟩ = –2 ⟨ 0. Thus Axiom 4 fails.V =−
0 1
1 0
y
4
-2
-4
2
x
1
4
1
1612 2
x y+ = .
1
4
1
1612 2
1 2
x y+
=
276 Exercise Set 6.1
23. Axioms 1 and 3 are easily verified. So is Axiom 2, as shown: Let r = r(x) be a polynomialin P2. Then
⟨p + q, r⟩ = [(p + q)(0)]r(0) + [(p + q)(1/2)]r(1/2) + [(p + q)(1)]r(1)
= p(0)r(0) + p(1/2)r(1/2) + p(1)r(1) + q(0)r(0) + q(1/2)r(1/2) + q(1)r(1)
= ⟨p, r⟩ + ⟨q, r⟩
It remains to verify Axiom 4:
⟨p, p⟩ = [p(0)]2 + [p(1/2)]2 + [p(1)]2 ≥ 0
and
⟨p, p⟩ = 0 if and only if p(0) = p(1/2) = p(1) = 0
But a quadratic polynomial can have at most two zeros unless it is identically zero. Thus⟨p, p⟩ = 0 if and only if p is identically zero, or p = 0.
24. Using the hint and properties of the transpose, we have
⟨u, Av⟩ = (Av)Tu = (vTAT)u = vT(ATu) = ⟨ATu, v⟩
26. (1) Axiom 1 follows immediately from the commutativity of multiplication in the realnumbers.
(2) If z = (z1, …, zn), then
⟨u + v, z⟩ = w1 (u1 + v1)z1 + … + wn(u
n+ v
n)z
n
= (w1u1z1 + … + wnu
nz
n) + (w1v1z1 + … + w
nv
nz
n)
= ⟨u, z⟩ + ⟨v, z⟩
(3) If k is any constant, then
⟨ku, v⟩ = w1(ku1)v1 + … + wn(ku
n)v
n
= k(w1u1v1 + … + wnu
nv
n)
= k⟨u, v⟩
Exercise Set 6.1 277
(4) Since ⟨v, v⟩ = w1v12 + … + w
nv
n2 where w1, …, w
nare positive reals, then ⟨v, v⟩ ≥
0 and ⟨v, v⟩ = 0 if and only if v1 = … = vn
= 0 or v = 0. If one of the constants, wi,
were negative or zero, then Axiom 4 would fail to hold.
27. (b) ⟨p, q⟩ = (x – 5x3) (2 + 8x2) dx = (2x – 2x3 – 40x5) dx
= x2 – x4/2 – 20x6/3 = 0
28. (a) ⟨f, g⟩ = cos 2πx sin 2πx dx = (1/4π) sin2 2πx = 0
29. We have ⟨U, V⟩ = u1v1 + u2v2 + u3v3 + u4v4 and
which does, indeed, equal ⟨U, V⟩.
30. We use the formula ⟨u, v⟩ = vT AT Au as suggested in the hint.
(1) We must show that ⟨u, v⟩ = ⟨v, u⟩. Since ⟨v, u⟩ is a real number which can bethought of as a 1 × 1 matrix, it will suffice to show that ⟨u, v⟩ = ⟨v, u⟩T. But
⟨v, u⟩T = (uT AT Av)T = vT AT Au = ⟨u, v⟩
tr tr 1( )U Vu u
u u
v v
v v
T =
3
2 4
1 2
3 4
=+ ++ +
tru v u v u v u v
u v u v u v u
1 1 3 3 1 2 3 4
1 1 4 3 2 2 4vv
u v u v u v u v
4
1 1 3 3 2 2 4 4
= + + +
0
1
∫0
1
∫
0
1
∫
-1
1
∫-1
1
∫
278 Exercise Set 6.1
(2) We have
⟨u + v, w⟩ = wT AT A(u + v)
= wT AT Au + wT AT Av
= ⟨u, w⟩ + ⟨v, w⟩
(3) Here ⟨ku, v⟩ = vT AT A(ku) = k(vT AT Au) = k⟨u, v⟩.
(4) Since ⟨v, v⟩ = vT AT Av= Av • Av ≥ 0, we need only show that ⟨v, v⟩ = 0 if and onlyif v = 0. But ⟨v, v⟩ = 0 if and only if Av = 0, or v = A–10 = 0, since A is invertible.
Therefore ⟨v, v⟩ ≥ 0 for every vector v. It is clear that if v = 0, then ⟨v, v⟩ = 0. It remainsto show that if v ≠ 0, then ⟨v, v⟩ ≠ 0.
To see this, we notice that the only way that ⟨v, v⟩ can equal zero is for v • ri
to equalzero for every row r
iof A. But this implies that a linear combination of the columns of A is
the n × 1 zero matrix. That is, if we let ci
be the ith column of A, then
v1c1 + v2c2 + … + vnc
n= 0
Therefore the column vectors of A are linearly dependent, and hence A is not invertible.This is why the matrix A has to be invertible to generate an inner product. If A is invertible,then ⟨v, v⟩ can only = 0 when v = 0.
31. Calling the matrix A, we have
⟨u, v⟩ = vT AT Au = vT A2u = w1u1
v1 + … + w
nu
nv
n
Exercise Set 6.1 279
33. To prove Part (a) of Theorem 6.1.1 first observe that ⟨0, v⟩ = ⟨v, 0⟩ by the symmetry axiom.Moreover,
⟨0, v⟩ = ⟨00, v⟩ by Theorem 5.1.1
= 0⟨0, v⟩ by the homogeneity axiom
= 0
Alternatively,
⟨0, v⟩ + ⟨0, v⟩ = ⟨0 + 0, v⟩ by additivity
= ⟨0, v⟩ by definition of the zero vector
But ⟨0, v⟩ = 2⟨0, v⟩ only if ⟨0, v⟩ = 0.
To prove Part (d), observe that, by Theorem 5.1.1, –v (the inverse of v) and (–1)v arethe same vector. Thus,
⟨u – v, w⟩ = ⟨u + (–v), w⟩
= ⟨u, w⟩ + ⟨–v, w⟩ by additivity
= ⟨u, w⟩ – ⟨v, w by homogeneity
34. The set of points ⟨(x,y), (x,y)⟩ = x2/52 + y2/42 = 1 is the unit circle with respect to theweighted Euclidean inner product ⟨u, v⟩ = (1/25)u1v1 + (1/16)u2v2.
280 Exercise Set 6.1
EXERCISE SET 6.2
1. (e) Since u • v = 0 + 6 + 2 + 0 = 8, the vectors are not orthogonal.
2. Let’s assume that these scalars k and l exist. Then for these three vectors to be mutuallyorthogonal we have
⟨u, v⟩ = ⟨(2, k, 6), (l, 5, 3)⟩ = 2l + 5k + 18 = 0
⟨u, w⟩ = ⟨(2, k, 6), (1, 2, 3)⟩ = 2 + 2k + 18 = 0
⟨v, w⟩ = ⟨(l, 5, 3), (1, 2, 3)⟩ = l + 10 + 9 = 0.
Solving the second and third equations we have k = –10 and l = –19. However, if wesubstitute these values into the first equation, we have ⟨u, v⟩ = 2(–19) + 5(–10) + 18 = –70≠ 0. Thus, there are no scalars k and l so that u, v, and w are mutually orthogonal.
3. We have ku + v = (k + 6, k + 7, –k – 15), so
ku + v = ⟨(ku + v), (ku + v)⟩1/2
= [(k + 6)2 + (k + 7)2 + (–k – 15)2]1/2
= (3k2 + 56k + 310)1/2
Since ku + v = 13 exactly when ku + v2 = 169, we need to solve the quadratic equation3k2 + 56k + 310 = 169 to find k. Thus, values of k that give ku + v = 13 are k = –3 or k =–47/3.
4. We have ⟨u, w1⟩ = ⟨u, w3⟩ = 0 but ⟨u, w2⟩ = –2 ≠ 0. Hence, u is not orthogonal to W.
281
5. (a)
(c)
(e)
6. (a)
(Compare with 3(c), above.)
7. ⟨p, q⟩ = (1)(0) + (–1)(2) + (2)(1) = 0
8. (a)
9. (b) = (2)(1) + (1)(1) + (–1)(0) + (3)(–1) = 0
Thus the matrices are orthogonal.
9. (d) = 4 + 1 – 5 + 6 = 6 = 0
Thus the matrices are not orthogonal.
10. (a) We look for values of k such that
⟨u, v⟩ = 2 + 7 + 3k = 0
Clearly k = –3 is the only possible value.
2 1
1 3
2 1
5 2−
,
2 1
1 3
1 1
0 1−
−
,
cos,
θ = =+ + +
=A B
A B
6 12 1 0
50 14
19
10 7
cos,
θ =− + + + −
− + + + −
1 5 2 2 4 9
1 5 2 2 4 9
2 2
2
x x x x
x x x x22
2 20 18
30 1010=
− + −=
1 0 1 0 3 3 3 3
1 0 1 0 3 3 3 3
, , , , , , ,
, , , , , ,
( ) − − − −( )( ) − − − −
==− −
=−3 3
2 36
1
2
cos, , , , ,
, , , ,θ =
−( ) −( )−( ) −
=−1 5 2 2 4 9
1 5 2 2 4 9
22 20 18
30 1010
+ −=
cos( , ),( , )
( , ) ,θ =
−−
=−
=1 3 2 4
1 3 2 4
2 12
10 20
−−1
2
282 Exercise Set 6.2
11. We must find two vectors x = (x1, x2, x3, x4) such that ⟨x, x⟩ = 1 and ⟨x, u⟩ = ⟨x, v⟩ = ⟨x,w⟩ = 0. Thus x1, x2, x3, and x4 must satisfy the equations
x12 + x2
2 + x32 + x4
2 = 1
2x1 + x2 – 4x3
= 0
–x1
– x2+ 2x
3+ 2x
4= 0
3x1+ 2x2+ 5x3 + 4x4 = 0
The solution to the three linear equations is x1 = –34t, x2 = 4t, x3 = –6t, and x4 = 11t. If wesubstitute these values into the quadratic equation, we get
[(–34)2 + (44)2 + (–6)2 + (11)2] t2 = 1
or
Therefore, the two vectors are
12. (a) Here ⟨u, v⟩2 = (3(4) + 2(–1))2 = 100, while ⟨u, u⟩ ⟨v, v⟩ = (32 + 22) (42 + (–1)2) = 221.
13. (a) Here ⟨u, v⟩2 = (3(–2)(1) + 2(1)(0))2 = 36, while, on the other hand, ⟨u, u⟩⟨v, v⟩ =(3(–2)2 + 2(1)2) 3(1)2 + 2(0)2 = 42.
14. Since W is the line y = 2x, then the space W of all vectors which are perpendicular to
every vector in W is just the line y = – 12x. Note that .
15. (a) Here W is the line which isnormal to the plane and which passes through the origin.By inspection, a normal vector to the plane is (1, –2, –3). Hence this line hasparametric equations x = t, y = –2t, z = –3t.
( , ) ,x x x x21
20⋅ −
=
± − −1
5734 44 6 11( , , , )
t = ±1
57
Exercise Set 6.2 283
16. (a) A row-echelon form of A is
Therefore the vectors v1 = (1, 2, –1, 2) and v2 = (0, 1, –3, 2) form a basis for the rowspace of A. From the above matrix, we see that the nullspace of A consists of allvectors of the form (–5s + 2t, 3s – 2t, s, t) where s and t are parameters. If we set s =1, t = 0 and s = 0, t = 1, we obtain a basis w1 = (–5, 3, 1, 0), w2 = (2, –2, 0, 1) for thenullspace of A. It is easy to check that v1
• w1 = v1• w2 = v2
• w1 = v2• w2 = 0, which
gives the desired result.
17. (a) The subspace of R3 spanned by the given vectors is the row space of the matrix
which reduces to
The space we are looking for is the nullspace of this matrix. From the reduced form,we see that the nullspace consists of all vectors of the form (16, 19, 1)t, so that thevector (16, 19, 1) is a basis for this space.
Alternatively the vectors w1 = (1, –1, 3) and w2 = (0, 1, –19) form a basis for therow space of the matrix. They also span a plane, and the orthogonal complement ofthis plane is the line spanned by the normal vector w1 × w2 = (16, 19, 1).
18. (b) Since v2 = 2v1, we are looking for a basis for the orthogonal complement of the spacespanned by v1. (This space is the line containing v1 and its orthogonal complement isthe plane 2x – z = 0.) The solution space for the equation 2x1 + 0x2 – x3 = 0 is the setof vectors (Alternatively, the subspace W of R3 spanned by the given vectors v1 and v2is the row space of the matrix A, where
Since (2, 0, –1) is a basis for the row space, then (2, 0, –1) is a basis for W. To find abasis for W, note that (1, 0, 2) and (0, 1, 0) are a basis for the null-space of A andhence for W. It is obvious that W is the plane 2x – z = 0.
A =−−
→
−
2 0 1
4 0 2
2 0 1
0 0 0
1 1 3
0 1 19
0 0 0
−−
1 1 3
5 4 4
7 6 2
−− −−
1 2 1 2
0 1 3 2
0 0 0 0
−−
284 Exercise Set 6.2
(c) The orthogonal complement of the subspace of R4 spanned by the given vectors is thenullspace of the matrix
which reduces to
Thus the nullspace consists of all vectors of the form (–s + 2t, –s – 4t, s, 7t) and thevectors (–1, –1, 1, 0) and (2, –, 0, 7) form a basis for this space.
19. If u and v are orthogonal vectors with norm 1, then
u – v = ⟨u – v, u – v⟩1/2
= [⟨u, u – 2⟨u, v⟩ + ⟨v, v⟩]1/2
= [1 – 2(0) + 1]1/2
= 2
20. If ⟨w, u1⟩ = ⟨w, u2⟩ = 0, then
⟨w, k1u1 + k2u2⟩ = ⟨w, k1u1⟩ + ⟨w, k2u2⟩
= k1 ⟨w, u1⟩ + k2 ⟨w, u2⟩
= 0
Thus w is orthogonal to k1u1 + k2u2.
Now consider R3 with the Euclidean inner product. If w is perpendicular to both u1 andu2 and they determine a plane, then w is perpendicular to every vector in that plane. If u1and u2 determine a line, then w must be perpendicular to every vector in that line. If u1 andu2 determine neither a plane nor a line, then u1 = u2 = 0 and the result is not of interest.
21. By definition, u is in span u1, u2, …, ur if and only if there exist constants c1, c2, …, c
rsuch
that
u = c1u1 + c2u2 + … + cru
r
But if ⟨w, u1⟩ = ⟨w, u2⟩ = … = ⟨w, ur⟩ = 0, then ⟨w, u⟩ = 0.
1 4 5 2
0 1 1 4 7
0 0 0 0
/
1 4 5 2
2 1 3 0
1 3 2 2−
Exercise Set 6.2 285
22. If a vector z is orthogonal to all of the basis vectors, then, by the result of Exercise 19, itis orthogonal to every linear combination of the basis vectors, and hence to every vector inV. Thus, in particular, ⟨z, z⟩ = 0, so, by Axiom 4, z = 0.
23. We have that W = spanw1, w2, …, wk
Suppose that w is in W. Then, by definition, ⟨w, wi⟩ = 0 for each basis vector w
iof W.
Conversely, if a vector w of V is orthogonal to each basis vector of W, then, by Problem20, it is orthogonal to every vector in W.
24. Note that
v1 + … + vr2 = ⟨v1 + … + v
r, v1 + … + v
r⟩f=
=
If vi, v
j= 0 whenever i ≠ j, then the result follows.
25. (c) By Property (3) in the definition of inner product, we have
ku2 = ⟨ku, ku⟩ = k2⟨u, u⟩ = k2u2
Therefore ku = |k| u.
26. (d) We must show that for all u, v, and w
u – v ≤ u – w + w – v
If we apply the triangle inequality to the vectors u – w and w – v, then we have
((u – w) + (w – v) ≤ u – w + w – v
and the result follows.
27. This is just the Cauchy-Schwarz Inequality using the inner product on Rn generated by A(see Formula (4) of Section 6.1).
= + + +≠∑v v v v1
2 2L r i j
i j
,
v v v vi i
i
r
i j
i j
r
, ,= ≠∑ ∑+
1
286 Exercise Set 6.2
28. If we let R2 have the Euclidean inner product and apply the Cauchy-Schwarz Inequality tothe vectors u = (a, b) and v = (cos θ, sin θ), the result follows directly.
30. Suppose that
(*) ⟨u, v⟩2 = ⟨u, u⟩ ⟨v, v⟩
Following the proof of Theorem 6.2.1, we see that (*) can hold if and only if either u = 0or the quadratic equation ⟨tu + v, tu + v⟩ = 0 has just one real root. If u = 0, then u and vare linearly dependent. If the equation has one real root, t0, then ⟨t0u + v, t0u + v⟩ = 0.Thus, by Axiom 4, t0u + v = 0, and again u and v are linearly dependent.
Conversely, suppose that u and v are linearly dependent. Then either u = 0 or v can beuniquely expressed as a multiple, –t20, of u; that is, t0u + v = 0. Thus if we work backwardthrough the above argument, we can verify that ⟨u, v⟩2 = ⟨u, u⟩ ⟨v, v⟩.
31. We wish to show that ABC is a right angle, or that and are orthogonal. Observe
that = u – (–v) and = v – u where u and v are radii of the circle, as shown in the
figure. Thus u = v. Hence
⟨ , ⟩ = ⟨u + v, v – u⟩
= ⟨u, v⟩ + ⟨v, v⟩ + ⟨u, –u⟩ + ⟨v, –u⟩
= ⟨v, u⟩ + ⟨v, v⟩ – ⟨u, u⟩ – ⟨v, u⟩
= v2 – u2
= 0
32. Suppose that ⟨u, v⟩ = au1v1 + bu2v2 is a weighted Euclidean inner product on R2. For u andv to be orthogonal unit vectorswith respect to this inner product, we must have
⟨u, u⟩ = a + 3b = 1
⟨v, v⟩ = a + 3b = 1
⟨u, v⟩ = –a + 3b = 0
which yields a = 1/2 and b = 1/6.
33. (a) As noted in Example 9 of Section 6.1, 0
1f(x)g(x)dx is an inner product on C[0, 1].
Thus the Cauchy-Schwarz inequality must hold, and that is exactly what we’re askedto prove.
BC
AB
BC
AB
BC
AB
Exercise Set 6.2 287
(b) In the inner product notation, we must show that
⟨f + g, f + g⟩1/2 ≤ ⟨f, f⟩1/2 + ⟨g, g⟩1/2
or, squaring both sides, that
⟨f + g, f + g⟩ ≤ ⟨f, f⟩ + 2⟨f, f⟩1/2 ⟨g, g⟩1/2 + ⟨g, g⟩
For any inner product, we know that
⟨f + g, f + g⟩ = ⟨f, f⟩ + 2⟨f, g⟩ + ⟨g, g⟩
By the Cauchy-Schwarz inequality
⟨f, g⟩2 ≤ ⟨f, f⟩ ⟨g, g⟩
or
⟨f, g⟩ ≤ ⟨f, f⟩1/2 ⟨g, g⟩1/2
If we substitute the above inequality into the equation for ⟨f + g, f + g⟩, we obtain
⟨f + g, f + g⟩ ≤ ⟨f, f⟩ + 2⟨f, f⟩1/2 ⟨g, g⟩1/2 + ⟨g, g⟩
as required.
34. Recall that cos (α + β) = cos α cos β – sin α sin β. Thus
cos(α – β) = cos α cos(–β) – sin α sin(–β)
= cos α cos β + sin α sin β
Adding these two equations gives the identity
cos(α + β) + cos(α – β) = 2 cos α cos β
288 Exercise Set 6.2
Thus if k =, then k + and k – are nonzero and
= 0
35. (a) W is the line y = –x.
(b) W is the xz-plane.
(c) W is the x-axis.
36. (a) If the solution space is a line in R3, then, by Theorem 6.2.6, the row space is theorthogonal complement to that line, or the plane through the origin orthogonal to theline.
(c) If the homogeneous system AT x = 0 has a unique solution, then that solution must bethe zero vector. Therefore the nullspace of AT is the vector space consisting of just thezero vector. Its orthogonal complement, which is all of R3, is the column space andhence also, because they have the same dimension, the row space of A. This alsofollows from the fact that AT and hence also A must be invertible.
37. (b) False. Let n = 3, let V be the xy-plane, and let W be the x-axis. Then V is the z-axisand W is the yz-plane. In fact V is a subspace of W
(c) True. The two spaces are orthogonal complements and the only vector orthogonal toitself is the zero vector.
(d) False. For instance, if A is invertible, then both its row space and its column space areall of Rn.
=
= + +
∫ cos( )cos( )
cos( ) cos(
kx lx dx
k l x
0
1
2
π
kk l x dx
k l x
k l
k l
)
sin sin
−
=+( )
++
−
∫0
1
2
π
(( )−
x
k l0
π
Exercise Set 6.2 289
38. (a) Let W be the subspace of all diagonal matrices in M22. Thus, a vector D = in W
is fully described by its diagonal elements a and b. A vector M22 is in W
W if and only if
⟨D, V⟩ = tr(DTV) = ax + bw = 0
for all possible values of a and b. This can only happen if x = w = 0. Thus, W is thesubspace of all matrices in M22 with only zeros on the diagonal.
(b) Let W be the subspace of all symmetric matrices in M22. Thus, the elements of W are
matrices of the form S = . A matrix V = in M22 is in W if and only if
⟨S, V⟩ = tr(STV) = tr(SV)
= ax + cz + cy + bw = 0
for all values of a, b, and c. This implies that x = w = 0 and y = –z. Thus, W is thesubspace of all skew-symmetric matrices in M22.
x y
z w
a c
c b
Vx y
z w=
a
b
0
0
290 Exercise Set 6.2
EXERCISE SET 6.3
2. (a) The vectors are orthogonal, but since (2, 0) ≠ 1, they are not orthonormal.
(c) The vectors are not orthogonal and therefore not orthonormal.
4. (a) The set is not orthonormal because the last two vectors are not orthogonal.
(c) The set is not orthonormal because the last two vectors are not orthogonal.
5. See Exercise 3, Parts (b) and (c).
6. (a) Denote the vectors by v1, v2, v3, and v4. Then
⟨v1, vi⟩ = 0 for i = 2, 3, 4
⟨v2, v3⟩ = 49 – 29 – 29 = 0
⟨v2, v4⟩ = 29 + 29 – 49 = 0
⟨v3, v4⟩ = 29 – 49 + 29 = 0
Thus, the vectors are orthogonal. Moreover, v1 = 1 and v2 = v3 = v4 =
49 + 19 + 49 = 1. The vectors are therefore orthonormal.
291
7. (b) Call the vectors u1, u2 and u3. Then ⟨u1, u2⟩ = 2 – 2 = 0 and ⟨u1, u3⟩ = ⟨u2, u3⟩ = 0. The
set is therefore orthogonal. Moreover, u1 = 2, u2 = 8 = 2 2, and u3 = 25
= 5. Thus is an orthonormal set.
9. It is easy to verify that v1• v2 = v1
• v3 = v2• v3 = 0 and that v3 = 1. Moreover, v1
2 =(–3/5)2 + (4/5)2 = 1 and v2 = (4/5)2 + (3/5)2 = 1. Thus v1, v2, v3 is an orthonormal set inR3. It will be an orthonormal basis provided that the three vectors are linearly independent,which is guaranteed by Theorem 6.3.3.
(b) By Theorem 6.3.1, we have
10. It is easy to check that v1• v2 = v1
• v3 = v1• v4 = v2
• v3 = v2• v3
4= v
3• v
4= 0, so that
we have 4 orthogonal vectors in R4. Thus, by Theorem 6.3.3, they form an orthogonal basis.
(b) By Formula (1), we have
11. (a) We have (w)S
= (⟨w, u1⟩, ⟨w, u2⟩) .=−
= −( )4
2
10
22 2 5 2, ,
2 3 2 5 2 22 3 2 10 2 2
1 1 4 1
2 2 6, , ,− −( ) =
+ + ++ + +
+− −
v122 15 2 2 2
4 4 9 4
2 6 2 0 2
1 4 0 1
2 0
+ −+ + +
+− + +
+ + ++
+
v
v
2
3++ −
+ + +
= + − +
0 2
1 0 0 1
15 2
7
5 2
21
2 201 2 3 4
v
v v3
v v
4
( , , )3 7 49
5
28
50
12
5
21
51− = − − +
+ − +v 00 4
37 5 9 5 4
2 3
1 2 3
+
= −( ) + −( ) +
v v
v v v
1
2
1
2
1
21 2 3u u u, ,
292 Exercise Set 6.3
12. (a) We have u = w1 + w2 = and v = –w1 + 4w2 = .
(b) By Theorem 6.3.2, these numbers can be computed directly using the Euclidean inner
product without reference to the basis vectors. We find that u = 2, d(u, v) = 13,
and ⟨u, v⟩ = 3. Similar computations with the versions of u and v obtained in Part (b)
fortunately yield the same results.
14. (a) u = (1 + 4 + 1 + 9)1/2 = 15
v – w = (2, 1, –2, 4) = (4 + 1 + 4 + 16)1/2 = 5
v + w = (–2, –7, 4, 6) = (4 + 49 + 16 + 36)1/2 = 105
⟨v, w⟩ = 0 + 12 + 3 + 5 = 20
16. (a) Let
and v1, v2 is the desired orthonormal basis.
17. (a) Let
Since ⟨u2, v1⟩ = 0, we have
vu
u22
2
1
2
1
20, ,= = −
vu
u11
1
1
3
1
3
1
3, ,= =
vu
u
Since u v
11
1
2 1
1
10
3
10
2
10
6
10
= = −
= −
,
, == −
− = +
4
10
2 24
10
12 2 1 1
,
, ( , )
we have
u u v v110
3
10
12
5
4
5, ,−
=
This vector hhas norm12
5Thus
v
, .4
5
4 10
5
5
4 10
12
52
=
= ,, ,4
5
3
10
1
10
=
13
6
16
5,
7
5
1
5, −
Exercise Set 6.3 293
Since ⟨u3, v1⟩ = and ⟨u3, v2⟩ = , we have
u3 – ⟨u3, v1⟩v1 – ⟨u3, v2⟩v2
This vector has norm Thus
and v1, v2, v3 is the desired orthonormal basis.
19. Since the third vector is the sum of the first two, we ignore it. Let u1 = (0, 1, 2) and u2 =(–1, 0, 1). Then
Since ⟨u2, v1⟩ = , then
where . Hence
Thus v1, v2 is an orthonormal basis.
v25
30
2
30
1
30= −
−
, ,
, ,− −
=12
5
1
5
30
5
u u v v2 2 1 1 12
5
1
5, , ,− = − −
2
5
vu
u11
1
01
5
2
5, ,= =
v31
6
1
6
2
6, ,= −
1
6
1
6
1
3
1
6, , .−
=
= −
1
6
1
6
1
3, ,
= −
− −( , , ) , , ,1 2 14
3
1
3
1
3
1
3
1
2
1
2
1
220,
1
2
4
3
294 Exercise Set 6.3
21. Note that u1 and u2 are orthonormal. Thus we apply Theorem 6.3.5 to obtain
and
22. Since u1 and u2 are not orthogonal, we must apply the Gram-Schmidt process before we caninvoke Theorem 6.3.5. Thus we let
Since ⟨u2, v1⟩ = , we have
where . Thus
Hence v1, v2 is the desired orthonormal set. Next, we compute ⟨w, v1⟩ = 2 3 and
⟨w, v2⟩.
Theorem 6.3.5 then gives
w w v v w v v1 1 1 2 213
14
31
14
40
14= + =
, , , ,
= −7 42
6
v25
42
1
42
4
42= − −
, ,
5
3
1
3
4
3
42
3, ,− −
=
u u v v2 2 1 15
3
1
3
4
3, , ,− = − −
1
3
vu
u11
1
1
3
1
3
1
3, ,= =
w w w2 1
9
50
12
5
= −
=
, ,
w w u u w u u1 1 1 2 2
4
50
3
52 0 1
= +
= − −
+
, ,
, , , , 00
4
52
3
5
( )
= −
, ,
Exercise Set 6.3 295
and
24. (a) If we apply the Gram-Schmidt process to the column vectors
Thus
and
(c) If we apply the Gram-Schmidt process to the column vectors
Thus
and
A = −
1 3 8 3 26
2 3 11 3 26
2 3 7 3 26
3 1 3
0 26 3
R, ,
,=
=
u q u q
u q1 1 2 1
2 20
3 1 3
0 26 3
u u we obta1 2
1
2
2
1
1
= −
=
,
1
iin q q1 2
1 3
2 3
2 3
8 3 26
11 3 26
7 3 26
= −
=
,
A =−
1 5 2 5
2 5 1 5
5 5
0 5
R, ,
,=
=
u q u q
0 u q1 1 2 1
2 2
5 5
0 5
u u we obtain q1 2 11
2
1
3
1 5
2 5=
=
−
=
,
=−
, q22 5
1 5
w w w2 11
14
1
14
2
14= − = −
, ,
296 Exercise Set 6.3
25. By Theorem 6.3.1, we know that
w = a1v1 + a2v2 + a3v3
where ai
= ⟨w, vi⟩. Thus
w2 = ⟨w, w⟩
But ⟨vi, v
j⟩ = 0 if i ≠ j and ⟨v
i, v
i⟩ = 1 because the set v1, v2, v3 is orthonormal. Hence
w2 = a12 + a2
2 + a32
= ⟨w, v1⟩2 + ⟨w, v2⟩
2 + ⟨w, v3⟩2
26. Generalize the solution of Exercise 25.
27. Suppose the contrary; that is, suppose that
(*) u3 – ⟨u3, v1⟩v1 – ⟨u3, v2⟩v2 = 0
Then (*) implies that u3 is a linear combination of v1 and v2. But v1 is a multiple of u1while v2 is a linear combination of 1 and 2. Hence, (*) implies that u3 is a linear combinationof 1 and 2 and therefore that u1, u2, u3 is linearly dependent, contrary to the hypothesisthat u1, …, u
n is linearly independent. Thus, the assumption that (*) holds leads to a
contradiction.
28. The diagonal entries of R are ⟨ui, q
i⟩ for i = 1, …, n, where q
i= v
i/v
i is a vector resulting
from an application of the Gram-Schmidt process. That process first generates anorthogonal set of vectors v1, …, v
nwhere each vector v
iis u
iminus a linear combination of
the vectors v1, …, vi–1. Therefore ⟨u
i, v
i⟩ = ⟨v
i, v
i⟩ for i = 1, …, n. But since ⟨u
i, q
i⟩ = ⟨u
i,
vi⟩/v
i = ⟨v
i, v
i⟩/v
i = v
i and each of the vectors v
iis nonzero, we have that each diagonal
entry of R is nonzero.
= += ≠∑ ∑, ,a a ai i i
i
i j i j
i
2
1
3
1
v v v v
Exercise Set 6.3 297
29. We have u1 = 1, u2 = x, and u3 = x2. Since
we let
Then
and thus v2 = u2/u2 where
Hence
In order to compute v3, we note that
and
Thus
and
u u v v u v3 3 1 1 3 22 1
3− − = −, , x
u v3 22
1
11
20, = =
−∫ x dx
u v3 12
1
11
2
2
3, = =
−∫ x dx
v23
2= x
u22
1
1 2
3= =
−∫ x dx
u v2 1 1
11
20, = =
−∫ x dx
v11
2=
u u u112
1 1
11 2= = =
−∫, dx
298 Exercise Set 6.3
Hence,
30. (b) If we call the three polynomials v1, v2, and v3, and let u = 2 – 7x2, then, by virtue ofTheorem 6.3.1,
u =⟨u, v1⟩v1 + ⟨u, v2⟩v2 + ⟨u, v3⟩v3
Here
It is easy to check that
31. This is similar to Exercise 29 except that the lower limit of integration is changed from –1to 0. If we again set u1 = 1, u2 = x, and u3 = x2, then u1 = 1 and thus
v1 = 1
Then ⟨u2, v1⟩ = x dx = 12 and thus
v
or
v
2 =−−
= −
= −( )
x
xx
x
1 2
1 212 1 2
3 2 12
( )
0
1
∫
u = − −2
3
28
15
5
21 3v v .
u v
u v
,
,
12
1
1
22
1
22 7
2
3
3
22 7 0
= −( ) = −
= −( ) =
−
−
∫ x dx
x dx11
1
32 2
1
11
2
5
22 7 3 1
28
15
5
2
∫
∫= −( ) −( ) = −−
u v, x x dx
v or v3 3= −
= −( )45
8
1
3
5
2 23 12 2
x x
x x dx2
22
2
1
11
3
1
3
8
45− = −
=−∫
Exercise Set 6.3 299
Finally,
and
Thus
or
v3 = 5 (6x2 – 6x + 1)
32. (a) Let S = w1, w2, …, wn be the orthonormal basis. Then (u)
S= (u1, u2, …, u
n) means
that u = u1w1 + u2w2 + … + unw
n. Thus
Taking positive square roots yields Theorem 6.3.2(b).
Note that this result also follows from Theorem 6.3.1 and Exercise 26.
(b) Since d(u, v) = u – v, we can apply Theorem 6.3.2(b) to u – v to obtain Theorem6.3.2(b).
(c) Let S = w1, w2, …, wn be the orthonormal basis. Then
u w and v= w== =∑ ∑u vi i
i
n
i i
i
n
1 1
u w w w w
w w w
21 1 1 1= + + + +
= +
u u u u
u u u u
n n n n
i i i i i j i
L L,
, ,,
(
w
because S is orthono
j
i ji
n
i
i
n
u
≠=
=
∑∑
∑=
1
2
1
rrmal)
v3
2
2
2
13
12
2 1
13
12
2 1
6 51
6=
− − −( )
− − −( )= − +
x x
x x
x x
u v3 23 2
0
13 2
3
6, ( )= − =∫ x x dx
u v3 12
0
1 1
3, = =∫ x dx
300 Exercise Set 6.3
Hence
33. Let W be a finite dimensional subspace of the inner product space V and let v1, v2, …, vr
be an orthonormal basis for W. Then if u is any vector in V, we know from Theorem 6.3.4that u = w1 + w2 where w1 is in W and w2 is in W Moreover, this decomposition of u isunique. Theorem 6.3.5 gives us a candidate for w1. To prove the theorem, we must showthat if w1 = ⟨u, v1⟩v1 + … + ⟨u, v
r⟩v
rand, therefore, that w2 = u – w1 then
(i) w1 is in W
and
(ii) w2 is orthogonal to W.
That is, we must show that this candidate “works.” Then, since w1 is unique, it will be proj
Wu.
Part (i) follows immediately because w1 is, by definition, a linear combination of thevectors v1, v2, …, v
r.
⟨w2, vi⟩ = ⟨u – w1, vi
⟩
= ⟨u, vi⟩ – ⟨w1, vi
⟩
= ⟨u, vi⟩ – ⟨u, v
i⟩⟨v
i, v
i⟩
= ⟨u, vi⟩ – ⟨u, v
i⟩
= 0
Thus, w2 is orthogonal to each of the vectors v1, v2, …, vr
and hence w2 is in W.
If the vectors vi
form an orthogonal set, not necessarily orthonormal, then we mustnormalize them to obtain Part (b) of the theorem.
u v w w w
w w w
, ,
, ,
= + +
= +=∑
u u v
u v u v
n n n n
i i i i
i
n
i i i
1 1
1
ww
because is othonormal
i
i j
i i
i
n
u v S
≠
=
∑
∑=1
( )
Exercise Set 6.3 301
34. (a) Call the plane and note that n = (a, b, c) is the normal vector to . Choose anarbitrary vector u in and note that u • n = 0. Hence u × n (and n × u) must lie in and be perpendicular to both u and n. Therefore u and u × n (or n × u) form anorthonogonal basis for . Normalize each of these vectors to obtain the requiredorthonormal basis.
(b) In this case, n = (1, 2, –1). The vector u = (1, 1, 3) lies in the plane given by x + 2y
– z = 0. Note that n × u = (7, –4, –1) lies in the plane and that u • (n × u) = 0. So anorthonormal basis for consists of the vectors
and
35. The vectors x = (1/ 3, 0) and y = (0, 1/ 2) are orthonormal with respect to the given
inner product. However, although they are orthogonal with respect to the Euclidean inner
product, they are not orthonormal.
The vectors x = (2/ 30, 3/ 30) and y = (1/ 5, –1/ 5) are orthonormal with respect
to the given inner product. However, they are neither orthogonal or of unit length with
respect to the Euclidean inner product.
37. (a) True. Suppose that v1, v2, …, vn
is an orthonormal set of vectors. If they were linearlydependent, then there would be a linear combination
c1v1 + c2v2 + … + cnv
n= 0
where at least one of the numbers ci
≠ 0. But
ci
= ⟨vi, c1v1 + c2v2 + … + c
nv
n⟩ = ⟨v
i, 0⟩ = 0
for i = 1, …, n. Thus, the orthonormal set of vectors cannot be linearly dependent.
n u
n u
××
=− −
, ,7
66
4
66
1
66
u
u, ,=
1
11
1
11
3
11
302 Exercise Set 6.3
(b) False. The zero vector space has basis 0, which cannot be converted to a unit vector.
(c) True, since projW
u is in W and projW u is in W.
(d) True. If A is a (necessarily square) matrix with a nonzero determinant, then A haslinearly independent column vectors. Thus, by Theorem 6.3.7, A has a QR
decomposition.
38. If we project a vector uk
onto a space W that contains it, then the projection does notchange u
k. That is, proj
Wu
k= u
k. Thus, if we start with a linearly dependent set u1, u2, …,
um
so that uk
is a linear combination of the vectors that come before it, then when wecalculate v
k= u
k– proj
Wk – 1u
kwe have v
k= 0. Thus, the set constructed by the Gram-
Schmidt process will contain the zero vector, and will be a linearly dependent set.
Exercise Set 6.3 303
EXERCISE SET 6.4
1. (a) If we call the system Ax = b, then the associated normal system is AT Ax = ATb, or
which simplifies to
2. (a) We have
so that det(AT A) = 0. Hence A does not have linearly independent column vectors.
3. (a) The associated normal system is AT Ax = ATb, or
1 1 1
1 1 2
1 1
1 1
1 2
1
2
− −
−
−
x
x
=
− −
−
1 1 1
1 1 2
7
0
7
A AT =
−
−
1 2 0
3 1 1
2 3 1
1 3 2
2 1 3
0 1 1
=−
−
5 1 4
1 11 10
4 10 14
21 25
25 35
20
201
2
=
x
x
1 2 4
1 3 5
1 1
2 3
4 5
1
2−
−
x
x =
−
−
1 2 4
1 3 5
2
1
5
305
or
This system has solution x1 = 5, x2 = 1/2, which is the least squares solution ofAx = b.
The orthogonal projection of b on the column space of A is Ax, or
3. (c) The associated normal system is
or
This system has solution x1 = 12, x2 = –3, x3 = 9, which is the least squares solution ofAx = b.
The orthogonal projection of b on the column space of A is Ax, or
which can be written as (3, 3, 9, 0).
1 0 1
2 1 2
1 1 0
1 1 1
12
3
9
−−
−
−
=
3
3
9
0
7 4 6
4 3 3
6 3 6
11
2
3
−−
− −
=x
x
x
88
12
9−
=− − −
1 2 1 1
0 1 1 1
1 2 0 1
6
0
9
3
1 2 1 1
0 1 1 1
1 2 0 1
1 0 1
2 1 2
1 1 0
1 1 1− − −
−−
−
x
x
x
1
2
3
1 1
1 1
1 2
5
1 2
11 2
9 2
4
−−
= −
−
3 2
2 6
14
71
2
−−
=
−
x
x
306 Exercise Set 6.4
4. (a) First we find a least squares solution of Ax = u where A = [u1T|v2
T]. The associatednormal system is
or
This system has solution x1 = –1, x2 = 5/3, which is the least squares solution. Thedesired orthogonal projection is Ax, or
or (2/3, 7/3, 5/3).
5. (a) First we find a least squares solution of Ax = u where A = [v1T|v2
T|v3T]. The associated
normal system is
or
7 4 6
4 3 3
6 3 6
1
2
3
−−
− −
=x
x
x
30
21
21−
=− − −
2 1 1 1
1 0 1 1
2 1 0 1
6
3
9
6
2 1 1 1
1 0 1 1
2 1 0 1
2 1 2
1 0 1
1 1 0
1 1 1− − −
−−
−
x
x
x
1
2
3
1 1
1 2
0 1
1
5 3
2 3
7 3
5 3
−
=
2 3
3 6
3
71
2
=
x
x
1 1 0
1 2 1
1 1
1 2
0 1
1
2
x
x==
=
1 1 0
1 2 1
2
1
3
Exercise Set 6.4 307
This system has solution x1 = 6, x2 = 3, x3 = 4, which is the least squares solution. Thedesired orthogonal projection is Ax, or
or (7, 2, 9, 5).
6. The solution of the given system can be expressed as x1 = –s + t, x2 = –s – t, x3 = 2s, x4 =2t. Thus the solution space is spanned by the vectors
Let A = [v1|v2]. We next find a least squares solution of Ax = u. The associated normalsystem is
or
This system has solution x1 = 1/2, x2 = 1/2, which is the least squares solution. The desiredorthogonal projection is Ax, or
or (0, –1, 1, 1).
−− −
=
−1 1
1 1
2 0
0 2
1 2
1 2
0
1
1
1
6 0
0 6
3
31
2
=
x
x
− −−
−− −
1 1 2 0
1 1 0 2
1 1
1 1
2 0
0 2
x
x
1
2
1 1 2 0
1 1 0 2
5
6
7
2
=
− −−
v and v1 2
1
1
2
0
1
1
0
2
=
−−
=−
2 1 2
1 0 1
1 1 0
1 1 1
6
3
4
−−
−
==
7
2
9
5
308 Exercise Set 6.4
7. (a) If we use the vector (1, 0) as a basis for the x-axis and let , then we have
[P] = A(AT A)–1 AT = [1] [1 0] =
8. (a) If we use the vectors (1, 0, 0) and (0, 0, 1) asa basis for the xz-plane and let
then we have
9. (a) By inspection, the vectors (1, 0, –5) and (0, 1, 3) lie in the plane W and they form alinearly independent set with the appropriate number of vectors. Hence they are abasis for W.
(b) Using the basis for W found in Part (a), we let
Thus the standard matrix for the orthogonal projection on W is
A =−
1 0
0 1
5 3
P[ ] =
1 0
0 0
0 1
1 0 0
0 0 1
1 0
0 0
0 11
1 0 0
0 0 1
1 0
0 0
0 1
1
=
−
=
1 0
0 1
1 0 0
0 0 1
1 0 0
0 0 00
0 0 1
A =
1 0
0 0
0 1
1 0
0 0
1
0
A =
1
0
Exercise Set 6.4 309
10. (c) By Part (b), the point (x0, y0, z0) projects to the point
on the plane W.
(d) By Part (c), the point (1, –2, 4) projects to the point (–8/7, –5/7, 25/7) on the plane.
The distance between these two points is 3 35/7.
11. (a) The vector v = (2, –1, 4) forms a basis for the line W.
(b) If we let A = [vT], then the standard matrix for the orthogonal projection on W is
1
35
10 15 5
15 26 3
5 3 34
0
0
0
−
−
x
y
z
=+ −
+ +−
( )
( )
(
2 3 7
15 26 2 35
5
0 0 0
0 0 0
x y z
x y z
x00 0 03 34 35+ +
y z )
[ ]P =−
−
−
1 0
0 1
5 3
1 0 5
0 1 3
1 0
0 1
5 3
−
=−
−1
1 0 5
0 1 3
1 0
0 1
5 3
−−
−
=
−26 15
15 10
1 0 5
0 1 3
1
1
00
0 1
5 3
2 7 3 7
3 7 26 35
1 0 5
0 1 3−
−
=−
−
=2 7 3 7 1 7
3 7 26 35 3 35
1 7 3 35 34 35
1
335
10 15 5
15 26 3
5 3 34
−
−
310 Exercise Set 6.4
(c) By Part (b), the point P0(x0, y0, z0) projects to the point on the line W given by
(d) By the result in Part (c), the point (2, 1, –3) projects to the point (–6/7, 3/7, –12/7).
The distance between these two points is 497/7.
12. We can see by inspection that when t = 1, we have the point P = (t, t, t) = (1, 1, 1), andwhen s = 1, we have the point Q = (s, 2s – 1, 1) = (1, 1, 1). Thus, when s = t = 1, we haveP – Q2 = (0, 0)2 = 0, which is the minimum possible value of P – Q2, since v ≥ 0 forany vector v.
13. (a) Using horizontal vector notation, we have b = (7, 0, –7) and Ax = (11/2, –9/2, –4).Therefore Ax – b = (–3/2, –9/2, 3), which is orthogonal to both of the vectors (1, –1,–1) and (1, 1, 2) which span the column space of A. Hence the error vector isorthogonal to the column space of A.
(c) In horizontal vector notation, b = (6, 0, 9, 3) and Ax = (3, 3, 9, 0). Hence Ax – b =(–3, 3, 0, –3), which is orthogonal to the three vectors (1, 2, 1, 1), (0, 1, 1, 1), and (–1,–2, 0, –1) which span the column space of A. Therefore Ax – b is orthogonal to thecolumn space of A.
14. Since A has linearly independent column vectors, Theorem 6.4.4 guarantees that there is aunique least squares solution x of the system Ax = b. Since x is a least squares solution, itminimizes Ax – b. But if there is a vector x for which Ax = b, then this minimum is zeroand occurs when x = x. Hence x = x by the uniqueness of the least squares solution.
1
21
4 2 8
2 1 4
8 4 16
0
0
0
−− −
−
x
y
z
=− +
− + −−
( )
( )
(
4 2 8 21
2 4 21
8 4
0 0 0
0 0 0
0
x y z
x y z
x yy z0 016 21+
)
=−
− −−
1
21
4 2 8
2 1 4
8 4 16
= −
−
2
1
4
1
212 1 4
P A A A AT T[ ] = = −
− −−( ) 12
1
4
2 1 4
2
11
4
2 1 4
1
−
−
Exercise Set 6.4 311
15. Recall that if b is orthogonal to the column space of A, then projW
b = 0.
16. (a) We have that [P] = A(AT A)–1 AT where A is any matrix formed using a set of basisvectors for W as its column vectors. Therefore
[P]2 = [A(AT A)–1 AT][A(AT A)–1 AT]
= A[(AT A)–1(AT A)](AT A)–1 AT
= A(AT A)–1 AT
= [P]
(c) We need to show that [P]T = [P]. Define A as in Part (a). Then
[P]T = [A(AT A)–1 AT]T
= (AT)T[(AT A)–1]T AT because (BC)T = CT BT
= A[(AT A)–1]T AT because (AT)T = A
We need only show that [(AT A)–1]T = (AT A)–1 to have the desired result. This followsfrom the fact that if B is any invertible matrix, then (B–1)T = (BT)–1, so that [(AT A)–1]T
= [(AT A)T]–1 = (AT A)–1. To see the above result, note that if B is invertible, then
BB–1 = (BB–1)T = I
so that
(B–1)T BT = I
or
(B–1)T = (BT)–1
17. If A is an m × n matrix with linearly independent row vectors, then AT is an n × m matrixwith linearly independent column vectors which span the row space of A. Therefore, byFormula (6) and the fact that (AT)T = A, the standard matrix for the orthogonal projection,S, of Rn on the row space of A is [S] = AT(AAT)–1 A.
312 Exercise Set 6.4
18. (a) If we assume a relationship V = IR, then the data points give the linear system
1 = 0.1 R
2.1 = 0.2 R
2.9 = 0.3 R
4.2 = 0.4 R
5.1 = 0.5 R
(b) No, this system is not consistent. There is no value of R that works for all 5 equations.
(c) The system in Part (a) can be written as Ax = b, where
Then, we have the least squares solution
x = (AT A)–1 ATb = [0,55]–1[5.62] = [5.62/0.55] = [10.2182].
This says the average resistance is approximately 10.2182 ohms.
19. If we assume a relationship V = IR + c, we have the linear system
1 = 0.1 R + c
2.1 = 0.2 R + c
2.9 = 0.3 R + c
4.2 = 0.4 R + c
5.1 = 0.5 R + c
A =
=
0 1
0 2
0 3
0 4
0 5
1
2 1
.
.
.
.
.
.
and bb 22 9
4 2
5 1
.
.
.
Exercise Set 6.4 313
This system can be written as Ax = b, where
Then, we have the least squares solution
.
Thus, we have the relationship V = 10.3 R – 0.03.
20. A relationship V = a4I4 + a3I
3 + a2I2 + a1I + a0, leads to the five equations
1 = a4(0.1)4 + a3(0.1)3 + a2(0.1)2 + a1(0.1) + a0
2.1 = a4(0.2)4 + a3(0.2)3 + a2(0.2)2 + a1(0.2) + a0
2.9 = a4(0.3)4 + a3(0.3)3 + a2(0.3)2 + a1(0.3) + a0
4.2 = a4(0.4)4 + a3(0.4)3 + a2(0.4)2 + a1(0.4) + a0
5.1 = a4(0.5)4 + a3(0.5)3 + a2(0.5)2 + a1(0.5) + a0
To solve this system, we need to row reduce the augmetned matrix
0 0001 0 001 0 01 0 1 1 1
0 0016 0 008 0 04 0 2 1 2 1
0
. . . .
. . . . .
.. . . . .
. . . . .
0081 0 027 0 09 0 3 1 2 9
0 0256 0 064 0 16 0 4 1 4 22
0 0625 0 125 0 25 0 5 1 5 1. . . . .
RREFF →
−
−
1 0 0 0 0 708 333
0 1 0 0 0 841 667
0 0 1 0 0 342 91
.
.
. 77
0 0 0 1 0 65 583
0 0 0 0 1 2 900
.
.−
x b= =
−−
( ). .
.
.
.A A A
T T11
0 55 1 55
1 5 5
5 62
15 3
=
−
.
..
10 3
0 03
A and
.
.
.
.
.
=
0 1 1
0 2 1
0 3 1
0 4 1
0 5 1
bbbb =
.
.
.
.
.
1
2 1
2 9
4 2
5 1
314 Exercise Set 6.4
Thus, keeping 3 decimal places of accuracy, we have the fourth-degree polynomial
V = –708.33I4 + 841.667I3 – 342.917I2 + 65.583I – 2.9.
This polynomial is shown in the graph below for 0.1 ≤ I ≤ 0.5, showing the fluctuation in theresistance during this experiment.
22. (a) Call the vector Ax. Then by Theorem 6.4.4, Ax = A(AT A)–1 ATb.
(c) The error is Ax – b.
(d) The standard matrix is A(AT A)–1 AT.
V, volts
I, amps
5
4
3
2
1
0.1 0.2 0.3 0.4 0.5
Exercise Set 6.4 315
EXERCISE SET 6.5
2. (b) T(x) = (14/3, –5/3, 11/3) and T(x) = x = 38.
3. (b) Since the row vectors form an orthonormal set, the matrix is orthogonal. Therefore itsinverse is its transpose,
(c) Since the Euclidean inner product of Column 2 and Column 3 is not zero, the columnvectors do not form an orthonormal set and the matrix is not orthogonal.
(f) Since the norm of Column 3 is not 1, the matrix is not orthogonal.
5. (b) We have (w)S
= (a, b) where w = au1 + bu2. Thus
2a + 3b = 1
–4a + 8b = 1
or and Hence (w)a b s.= = =5
28
3
14
5
28,
3
14
5
283
14
[ ] =
and
w s
1 2 1 2
1 2 1 2−
317
6. (a) Let v = av1
+ bv2
+ cv3. Then
a + 2b + 3c = 2
2b + 3c = –1
3c = 3
so that a = 3, b = –2, and c = 1. Thus (v)S
= (3, –2, 1) and
7. (b) Let p = ap1 + bp2 + cp3. Then
a + b = 2
a + c = –1
b + c = 1
or a = 0, b = 2, and c = –1. Thus (v)S
= (0, 2, –1) and
8. Let A = aA1 + bA2 + cA3 + dA4. Then
–a + b = 2
a + b = 0
c = –1
d = 3
so that a = –1, b = 1, c = –1, and d = 3. Thus (A)S
= (–1, 1, –1, 3) and
v[ ] =−
S
0
2
1
v[ ] = −
S
3
2
1
318 Exercise Set 6.5
9. (a) We have w = 6v1 – v2 + 4v3 = (16, 10, 12).
(c) We have B = –8A1 + 7A2 + 6A3 + 3A4 =
11. (a) Since v1 = 1310u1 – 25u2 and v2 = – 12u1 + 0u2, the transition matrix is
(b) Since u1 = 0v1 – 2v2 and u2 = – 52v1 – 132 v2, the transition matrix is
Note that P = Q–1.
(c) We find that w = – 1710 u1 + 85u2; that is
and hence
w[ ] =−
− −
−
B′
05
2
213
2
17
108
5
=−−
4
7
w[ ] =−
B
17
108
5
P =−
− −
05
2
213
2
Q =−
−
13
10
1
22
50
15 1
6 3
−
.
A S[ ] =
−
−
1
1
1
3
Exercise Set 6.5 319
11. (d) Verify that w = (–4)v1 + (–7)v2.
12. (a) We must find constants cij
such that
uj= c1j
v1 + c2jv2 + c3j
v3
for j = 1, 2, 3. For j = 1, this reduces to the system
–6c11 – 2c21 – 2c31 = –3
–6c11 – 6c21 – c31 = 0
4c21 + 7c31 = –3
which has the solution c11 = 3/4, c21 = –3/4, c31 = 0. For j = 2 and j = 3, we havesystems with the same coefficient matrix but with different right-hand sides. Thesehave solutions c12 = 3/4, c22 = –17/12, c32 = 2/3, c13 = 1/12, c23 = –17/12, and c33 = 2/3.Thus the transition matrix is
(b) We must find constants a, b, c such that w = au1 + bu2 + cu3. This means solving thesystem of equations
–3a – 3b + c = –5
2b + 6c = 8
–3a – b – c = –5
The solution is a = 1, b = 1, c = 1. Thus
and hence
w[ ]
/ / /
/ / /
/ /B′ = − − −
3 4 3 4 1 12
3 4 17 12 17 12
0 2 3 2 33
1
1
1
19 12
43 12
4 3
=
= −
/
/
/
[w]B =
1
1
1
P = − − −
3 4 3 4 1 12
3 4 17 12 17 12
0 2 3 2 3
320 Exercise Set 6.5
(c) Let w = av1 + bv2 + cv3 and solve the equations
–6a – 2b – 2c = –5
–6a – 6b – 3c = 8
4b + 7c = –5
for a, b, and c.
14. (a) The transition matrix from B′ to B is
(b) Since p1 = (3/4)q1 + (3/2)q2 and p2 = (7/2)q1 + q2, the transition matrix is
(c) Let p = ap1 + bp2. Thus
6a+ 10b = –4
3a + 2b = 1
so that a = 1 and b = –1. Therefore [p]B
= and hence
(d) Let p = aq1 + bq2 and solve for a and b.
15. (a) By hypothesis, f1 and f2 span V. Since neither is a multiple of the other, then f1, f2 is
a linearly independent set and hence is a basis for V. Now by inspection,
. Therefore, g1, g2 must also be a basis for V
because it is a spanning set which contains the correct number of vectors.
ff gg gg aanndd ff gg11 11 22 22 22= + −
=1
2
1
6
1
3
[ ]/ /
/
/
/p B =
=
−
=
−−3 4 7 2
3 2 1
1
1
11 4
1 2
1
1
1−
3 4 7 2
3 2 1
/ /
/
3 4 7 2
3 2 1
2 9 7 9
1 3 1 6
1/ /
/
/ /
/ /
=
−−
−
Exercise Set 6.5 321
15. (b) The transition matrix is
(c) From the observations in Part (a), we have
(d) Since h = 2f1 + (–5)f2, we have [h]B
= ; thus
16. (a) Equation (9) yields
(b) If we compute the inverse of the transition matrix in (a), we have
18. (a) Always true. The transition matrix is given in Equation (6).
(b) Always true. If the basis is B = v1, v2, …, vn, then the coordinate vectors are given by v
iB
= ei, so the transition matrix has columns e1, e2, …, e
n. This is the n × n identity matrix.
x
y
=
− −
−
=
−1 2 1 2
1 2 1 2
5
2
7/ /
/ /
22
3 2
x
y
′′
=
−cos( / ) sin( / )
sin( / ) cos(
3 4 3 4
3 4
π ππ 33 4
1 2 1 2
1 2 1 2
π / )
/ /
/ /
=−
− −
x
y
−
=
−
2
6
4 2
2 2
[ ]h B′ =−
−
=
−
1
20
1
6
1
3
2
5
1
2
2
5−
P =−
1
20
1
6
1
3
1
20
1
6
1
3
2 0
1 3
1
−
=
−
322 Exercise Set 6.5
(c) Always true. If A is invertible, then the column u1, u2, …, un
of A form a basis of Rn byTheorem 6.4.5. Denote this basis by B′, and denote the standard basis of Rn by B.Then we have [u
i]B
= ui
for i = 1, 2, …, n, so A = [[ui]B
| [ui]B
| … | [un]B] is the
transition matrix from the standard basis of Rn to the basis B′ consisting of the columnvectors of A.
19. The general transition matrix will be
In particular, if we rotate through θ = π3, then the transition matrix is
21. (a) See Exercise 19, above.
22. Use the result of Example 7 to find P1 where x′ = P1x. Use the result of Exercise 21 (a),above, to find P2 where x′′ = P2x′. Then x′′ = P2P1x, and hence A = P2P1.
23. Since the row vectors (and the column vectors) of the given matrix are orthogonal, thematrix will be orthogonal provided these vectors have norm 1. A necessary and sufficientcondition for this is that a2 + b2 = 1/2. Why?
24. Suppose that A = [aij] is a 2 × 2 orthogonal matrix. Since the column vectors form an
orthonormal set, we have the equations a112 + a21
2 = a122 + a22
2 = 1 and a11a12 + a21a22 = 0. Thus |aij|
≤ 1 for all aij. Since cos α takes on all values between –1 and + 1 for 0 ≤ α ≤ π, we can let a11
= cos θ for some value of θ where 0 ≤ θ ≤ π. Alternatively, we could let a11 = cos (2π – θ).Since a11
2 + a212 = 1, we have a21
2 = 1 – cos2 θ = sin2 θ, so that a21 = ±sin θ, or alternatively, a21 =sin(2π – θ). Thus the orthogonality equation becomes a12 cos θ ± a22 sin θ = 0, so that a12= k sin θ and a22 = k cos θ. Since a12
2 + a222 = 1, k = ±1. Thus we have four possible matrices:
cos sin
sin cos,
cos sin
sin cos
θ θθ θ
θ θθ θ
−
−− −
−
,
cos sin
sin cos,
cos sin
sin
θ θθ θ
θ θθθ θ−
cos
1
20
3
20 1 0
3
20
1
2
−
cos sinθ θ0
0 1 0
3
10
1
2
−
Exercise Set 6.5 323
If we substitute θ = 2π – θ for θ in the third and fourth matrices, they become the first andsecond matrices, respectively. Thus for some value of θ where 0 ≤ θ ≤ 2π, the first twomatrices represent all possible forms of a 2 × 2 orthogonal matrix. Since θ = 2π gives thesame value as θ = 0, we can restrict θ to the range 0 ≤ θ < 2π.
25. Multiplication by the first matrix A in Exercise 24 represents a rotation and det(A) = 1. Thesecond matrix has determinant –1 and can be written as
Thus it represents a rotation followed by a reflection about the x-axis.
28. Since a composition of rotations can be represented as a product of orthogonal matriceswith determinant 1 and, by Theorem 6.5.2, such a product is also orthogonal, it mustrepresent either a single rotation or a rotation followed by a reflection. However, if it is theproduct of matrices with determinant 1, its determinant is also 1, and hence it representsa rotation.
29. Note that A is orthogonal if and only if AT is orthogonal. Since the rows of AT are thecolumns of A, we need only apply the equivalence of Parts (a) and (b) to AT to obtain theequivalence of Parts (a) and (c).
30. (a) Rotations about the origin and reflections about any line through the origin are rigid.So is any combination of these transformations.
(b) Rotations about the origin, dilations and contractions, reflections about lines throughthe origin, and combinations of these are all angle preserving.
(c) Suppose that T is a linear operator in R2 and let A be the matrix associated with T. IfT is rigid, then Ax = x for all x in R2. Then Theorem 6.5.3 guarantees that Ax • Ay
= x • y or, since Ax = x, that
Therefore, the cosine of the angle between Ax and Ay is equal to the cosine of the anglebetween x and y and thus the angles themselves are the same. Hence, if T is rigid, then Tis angle preserving.
The converse is not true. Simply note that dilations in R2 by a factor k where k ≠ 1preserve the angle between two vectors but not their lengths.
Ax Ay
Ax Ay=
x y
x y
⋅ ⋅
cos sin
sin cos
cos sθ θθ θ
θ−− −
=
−
−1 0
0 1
iin
sin cos
θθ θ
324 Exercise Set 6.5
31. If A is the standard matrix associated with a rigid transformation, then Theorem 6.5.3guarantees that A must be orthogonal. But if A is orthogonal, then Theorem 6.5.2guarantees that det(A) = ±1.
32. If A is orthogonal, then its row vectors must form an orthonormal set r1, r2, r3. Since r1
= a2 + 1, then a = 0. This guarantees that r1 • r2 = r1 • r3 = 0. It remains to find b and c such
that b2 + 1/3 = 1, c2 + 2/3 = 1, and bc + 2/ 18 = 0. The first two equations imply that b =
± 2/ 3 and c = ±1/ 3, while the third equation tells us that b and c must have different
signs. Therefore a = 0, b = 2/ 3, c = 1/ 3 and a = 0, b = – 2/ 3, c = 1/ 3 are the
only two possibilities.
Exercise Set 6.5 325
EXERCISE SET 6.6
4. Let A be an orthogonal matrix. To establish that AT is also an orthogonal matrix, we needto show that (AT)–1 = (AT)T. This is easily done using properties of the transpose fromChapter 1. Since A is orthogonal, we know that
A–1 = AT.
Transposing both sides gives
(A–1)T = (AT)T.
and using Theorem 1.4.10, we have
(AT)–1 = (AT)T.
Thus, AT is also an orthogonal matrix.
327
SUPPLEMENTARY EXERCISES 6
1. (a) We must find a vector x = (x1, x2, x3, x4) such that
The first two conditions guarantee that x1 = x4 = 0. The third condition implies that x2= x3. Thus any vector of the form (0, a, a, 0) will satisfy the given conditions provideda ≠ 0.
(b) We must find a vector x = (x1, x2, x3, x4) such that x • u1 = x • u4 = 0. This implies
that x1 = x4 = 0. Moreover, since x = u2 = u3 = 1, the cosine of the angle between
x and u2 is x • u2 and the cosine of the angle between x and u3 is x • u3. Thus we are
looking for a vector x such that x • u2 = 2x • u3, or x2 = 2x3. Since x = 1, we have x
= (0, 2x3, x3, 0) where 4x32 + x3
2 = 1 or x3 = ±1/ 5. Therefore
2. We first show that A is symmetric, or that A = AT. Observe that
To show that A is orthogonal, it will suffice to show that A–1 = AT, or that AAT = In, or, in this
case, that A2 = In, or that A2 – I
n= O
n. To this end, observe that
A InT T2
2 424 4
= − +x
xxx
xx( )
A I I
I
Tn
T
T
nT T
T
n
= −
= − ( )= −
2 2
2
2 2
2
xxx
xxx
xxTT
TT
nT
I A( ) = − =xx
xx2
2
x = ±
02
5
1
50, , ,
x u =0, x u =0, andx u
x u=
x u
x u1 42
2
3
3⋅ ⋅ ⋅ ⋅
329
or that
Thus it will suffice to show that . To this end, let
x = [x1x2… x
n]T. Then
so that
This completes the proof.
4. Let . By the Cauchy-Schwarz Inequality,
or
n a aa an
21 1
1
1 1≤ + + + +
( )
( ) ( )u v u vterms
⋅ = + + ≤2 2 21 1L
n
u vand= =( ) ( ), , / , , /a a a an n1 11 1
(xx x x xTn
T
T
nT
x x x
x x
x x
x x
) [ ]21 2
1
2=
L
= + + +
= + + +
x x x
x x x
T Tn
T
n
12
22 2
12
22 2
xx xx xxL
L( )xxx x xxT T=
xx x x xT
T
T
nT
n
x x
x x
x x
x x x=
=
1
21 2 L[ ]]
xxx
xxT T( )=1
22
A I
x xn
T T22 2
24 1− = − −
xx xx( )
330 Supplementaary Exercises 6
6. If v = (a, b, c) where v = 1 and v • u1 = v • u2 = v • u3 = 0, then
a + b – c = 0
–2a – b + 2c = 0
–a + c = 0
where a2 + b2 + c2 = 12. The above system has the solution a = t, b = 0, and c = t where tis arbitrary. But a2 + b2 + c2 = 12 implies that 2t2 = 12 or t = ±1/ 2. The two vectors aretherefore v = ±(1/ 2, 0, 1/ 2).
7. Let
(*) ⟨u, v⟩ =w1u1v1 + w2u2v2 + … + wnu
nv
n
be the weighted Euclidean inner product. Since ⟨vi, v
j⟩ = 0 whenever i ≠ j, the vectors v1,
v2, …, vn form an orthogonal set with respect to (*) for any choice of the constants w1, w2,
…, wn. We must now choose the positive constants w1, w2, …, w
nso that v
k = 1 for all k.
But vk2 = kw
k. If we let w
k= 1/k for k = 1, 2, …, n, the given vectors will then form an
orthonormal set with respect to (*).
8. Suppose that ⟨u, v⟩ = w1u1v1 + w2u2v2 were such an inner product. Then
⟨(1, 2), (3, –1)⟩ = 0 ⇒ 3w1 – 2w2 = 0
(1, 2) = 1 ⇒ w1+ 4w2 = 1
(3, –1) = 1 ⇒ 9w1 + w2 = 1
But since the three equations for 1 and 2 above are inconsistent, there is no such innerproduct.
9. Let Q = [aij] be orthogonal. Then Q–1 = QT and det(Q) = ±1. If C
ijis the cofactor of a
ij, then
so that [aij
= det(Q)Cij.
Q a QQ
C QijT
ijT
T
= = =
=−[( )] ( )det( )
( ) det(1 1))( )Cij
Supplementaary Exercises 6 331
10. From the definition of the norm, we have
u – v2 = ⟨u – v, u – v⟩
= ⟨u, u⟩ + ⟨v, v⟩ – 2⟨u, v⟩
= u2 + v2 – 2u v cos θ
11. (a) The length of each “side” of this “cube” is |k|. The length of the “diagonal” is n|k|.The inner product of any “side” with the “diagonal” is k2. Therefore,
(b) As n → + ∞, cos θ → 0, so that θ → π/2.
12. (a) We have that u + v and u – v are orthogonal if and only if
⟨u + v, u – v⟩ = 0
But
⟨u + v, u – v⟩ = u – v
by the properties of an inner product (Why?). Thus u + v and u – v are orthogonal ifand only if u = v.
(b) Since u + v and u – v can be thought of as diagonals of a parallelogram with sides uand v, the above result says that the two diagonals are perpendicular if and only if theparallelogram has equal sides.
13. Recall that u can be expressed as the linear combination
u = a1v1 + … + anvn
where ai
= ⟨u, vi⟩ for i = 1, …, n. Thus
cosθ = =k
k n k n
2 1
332 Supplementaary Exercises 6
Therefore
14. We must show that ⟨u, v⟩ satisfies the four axioms which define an inner product. We show(2) and (4), and leave the others to you.
(2) Using Axiom 3 for ⟨u, v⟩1 and ⟨u, v⟩2, we have
⟨u + v, w⟩ = ⟨u + v, w⟩1 + ⟨u + v, w⟩2
= ⟨u, w⟩1 + ⟨v, w⟩1 + ⟨u, w⟩2 + ⟨v, w⟩2
= ⟨u, w⟩1 + ⟨u, w⟩2 + ⟨v, w⟩1 + ⟨v, w⟩2
= ⟨u, w⟩ + ⟨v, w⟩
(4) Using Axiom 4 for ⟨u, v⟩1 and ⟨u, v⟩2, we have
⟨v, v⟩ = ⟨v, v⟩1 + ⟨v, v⟩2 ≤ 0
since both ⟨v, v⟩1 and ⟨v, v⟩2 are nonnegative. Moreover, ⟨v, v⟩ = 0 if and only if ⟨v,v⟩1 = ⟨v, v⟩2 = 0, which is true if and only if v = 0.
15. Recall that A is orthogonal provided A–1 = AT. Hence
⟨u, v⟩ = vT AT Au
= vT A–1 Au = vT u
which is the Euclidean inner product.
cos cos21
2 12
22 2
12
22 2
α α+ + =+ + +
+ + +=L
L
Ln
n
n
a a a
a a a11
cos,2
2
2
2
12
22
α i
i
i
i
i
u v
a
u
a
a a
=
=
=+
u v
++ L an2
(Why?)
Supplementaary Exercises 6 333
16. To show that (W) = W, we first show that W ⊆ (W
). If w is in W, then w isorthogonal to every vector in W, so that w is in (W
) . Thus W ⊆ (W).
To show that (W) ⊆ W, let v be in (W
). Since v is in V, we have, by the Projection Theorem, that v = w1 + w2 where w1 is in W and w2 is in W . By definition, ⟨v, w2⟩ = ⟨w1, w2⟩ = 0. But
⟨v, w2⟩ = ⟨w1 + w2, w2⟩
= ⟨w1, w2⟩ + ⟨w2, w2⟩
= ⟨w2, w2⟩
so that ⟨w2, w2⟩ = 0. Hence w2 = 0 and therefore v = w1, so that v is in W. Thus (W) ⊆
W.
334 Supplementaary Exercises 6
EXERCISE SET 7.1
1. (a) Since
the characteristic equation is λ2 – 2λ – 3 = 0.
(e) Since
the characteristic equation is λ2 = 0.
3. (a) The equation (λI – A)x = 0 becomes
The eigenvalues are λ = 3 and λ = –1. Substituting λ = 3 into (λI – A)x = 0 yields
or
–8x1 + 4x2 = 0
0 0
8 4
0
01
2−
=
x
x
λλ
−− +
=
3 0
8 1
0
01
2
x
x
det( ) detλλ
λI A− =
0
8
det( ) det ( )( )λλ
λλ λI A− =
−− +
= − +
3 0
8 13 1
335
Thus x1 = 1–2s and x2 = s where s is arbitrary, so that a basis for the eigenspace
corresponding to λ = 3 is . Of course, and are also bases.
Substituting λ = –1 into (λI – A)x = 0 yields
or
–4x1 = 0
–8x1 = 0
Hence, x1 = 0 and x2 = s where s is arbitrary. In particular, if s = 1, then a basis for the
eigenspace corresponding to λ = –1 is .
3. (e) The equation (λI – A)x = 0 becomes
Clearly, λ = 0 is the only eigenvalue. Substituting λ = 0 into the above equation yieldsx1 = s and x2 = t where s and t are arbitrary. In particular, if s = t = 1, then we find
that and form a basis for the eigenspace associated with λ = 0.
4. (c) Since
= λ3 + 8λ2 + λ + 8
the characteristic equation of A is
λ3 + 8λ2 + λ + 8 = 0
det[ ] detλλ
λλ
I A− =+ −
+− − +
2 0 1
6 2 0
19 5 4
0
1
1
0
λλ0
0
0
01
2
=
x
x
0
1
−−
=
4 0
8 0
0
01
2
x
x
ππ2
1
2
1 2
1
336 Exercise Set 7.1
5. (c) From the solution to 4(c), we have
λ3 + 8λ2 + λ + 8 = (λ + 8)(λ2 + 1)
Since λ2 + 1 = 0 has no real solutions, then λ = –8 is the only (real) eigenvalue.
6. (c) The only eigenvalue is λ = –8. If we substitute the value λ = –8 into the equation (λI – A)x = 0, we obtain
The augmented matrix of the above system can be reduced to
Thus, x1 = – 1–6
s, x2 = – 1–6s, x3 = s is a solution where s is arbitrary. Therefore
forms a basis for the eigenspace.
(f) The eigenvalues are λ = –4 and λ = 3. If we substitute λ = –4 into the equation (λI – A)x = 0, we obtain
If we reduce the augmented matrix to row-echelon form, we obtain
This implies that x1 = –2 s, x2 = 8–3
s, x3 = s. If we set s = 1, we find that
1 0 2 0
0 1 8 3 0
0 0 0 0
−
− − −−
− −
9 6 2
0 3 8
1 0 2
1
2
3
x
x
x
=
0
0
0
−−
1 6
1 6
1
1 0 1 6 0
0 1 1 6 0
0 0 0 0
− −−
− − −
6 0 1
6 6 0
19 5 4
1
2
3
x
x
x
=
0
0
0
Exercise Set 7.1 337
is a basis for the eigenspace associated with λ = –4.
If we substitute λ = 3 into the equation (λI – A)x = 0, we obtain
If we reduce the augmented matrix to row-echelon form, we obtain
It then follows that
is a basis for the eigenspace associated with λ = 3.
7. (a) Since
= λ4 + λ3 – 3λ2 – λ + 2
= (λ – 1)2(λ + 2)(λ + 1)
det[ ] detλ
λλ
λI A− =
−− −
− +
0 2 0
1 1 0
0 1 2 0
0 0 0 λ −
1
5
2
1
−
1 0 5 0
0 1 2 0
0 0 0 0
−
− − −
−
2 6 2
0 4 8
1 0 5
1
2
3
x
x
x
=
0
0
0
−
2
8 3
1
338 Exercise Set 7.1
the characteristic equation is
(λ – 1)2 (λ + 2)(λ + 1) = 0
9. (a) The eigenvalues are λ = 1, λ = –2, and λ = –1. If we set λ = 1, then (λI – A)x = 0becomes
The augmented matrix can be reduced to
Thus, x1 = 2s, x2 = 3s, x3 = s, and x4 = t is a solution for all s and t. In particular, if welet s = t = 1, we see that
form a basis for the eigenspace associated with λ = 1.
If we set λ = –2, then (λI – A)x = 0 becomes
The augmented matrix can be reduced to
− −− − −
−−
2 0 2 0
1 2 1 0
0 1 0 0
0 0 0 3
=
x
x
x
x
1
2
3
4
0
0
0
0
2
3
1
0
0
0
0
1
and
1 0 2 0 0
0 1 3 0 0
0 0 0 0 0
0 0 0 0 0
−−
1 0 2 0
1 1 1 0
0 1 3 0
0 0 0 0
−− −
−
=
x
x
x
x
1
2
3
4
0
0
0
0
Exercise Set 7.1 339
This implies that x1 = –s, x2 = x4 = 0, and x3 = s. Therefore the vector
forms a basis for the eigenspace associated with λ = –2.
Finally, if we set λ = –1, then (λI – A)x = 0 becomes
The augmented matrix can be reduced to
Thus, x1 = –2s, x2 = s, x3 = s, and x4 = 0 is a solution. Therefore the vector
forms a basis for the eigenspace associated with λ = –1.
−
2
1
1
0
1 0 2 0 0
0 1 1 0 0
0 0 0 1 0
0 0 0 0 0
−
1 0 2 0
1 1 1 0
0 1 3 0
0 0 0 0
1
2
3
4
−− −
−
x
x
x
x
=
0
0
0
0
−
1
0
1
0
1 0 1 0 0
0 1 1 0 0
0 0 0 0 0
0 0 0 0 0
−
340 Exercise Set 7.1
11. By Theorem 7.1.1, the eigenvalues of A are 1, 1/2, 0, and 2. Thus by Theorem 7.1.3, theeigenvalues of A9 are 19 = 1, (1/2)9 = 1/512, 09 = 0, and 29 = 512.
12. Notice that A2 = I. Therefore A25 = A.
13. The vectors Ax and x will lie on the same line through the origin if and only if there existsa real number λ such that Ax = λx, that is, if and only if λ is a real eigenvalue for A and xis the associated eigenvector.
(a) In this case, the eigenvalues are λ = 3 and λ = 2, while associated eigenvectors are
respectively. Hence the lines y = x and y = 2x are the only lines which are invariantunder A.
(b) In this case, the characteristic equation for A is λ2 + 1 = 0. Since A has no realeigenvalues, there are no lines which are invariant under A.
14. (a) From the proof of Theorem 7.1.4, p(0) = det(–A) = 5. Since each signed elementaryproduct in det(–A) is just (–1)3 times the corresponding signed elementary product indet(A), we have that det(A) = –det(–A) = –5.
15. Let aij
denote the ijth entry of A. Then the characteristic polynomial of A is det(λI – A) or
This determinant is a sum each of whose terms is the product of n entries from the givenmatrix. Each of these entries is either a constant or is of the form λ – a
ij. The only term
with a λ in each factor of the product is
det
λ
λ
− −− −
− −
a a a
a a a
a a a
n
n
n n n
11 12 1
21 22 2
1 2
nn
1
1
1
2
and
Exercise Set 7.1 341
(λ – a11)(λ – a22) … (λ – ann
)
Therefore, this term must produce the highest power of λ in the characteristic polynomial.This power is clearly n and the coefficient of λn is 1.
17. The characteristic equation of A is
λ2 – (a + d)λ + ad – bc = 0
This is a quadratic equation whose discriminant is
(a + d)2 – 4ad + 4bc = a2 – 2ad + d2 + 4bc
= (a – d)2 + 4bc
The roots are
If the discriminant is positive, then the equation has two distinct real roots; if it is zero, thenthe equation has one real root (repeated); if it is negative, then the equation has no realroots. Since the eigenvalues are assumed to be real numbers, the result follows.
18. We are given the two distinct eigenvalues λ1 and λ2. To find the corresponding eigenvectors,we set
This yields the equations
(λi – a) x1 – bx2 = 0
–cx1 + (λi
– d)x2 = 0
This system of equations is guaranteed to have a nontrivial solution. Since b and a – λi
are
not both zero, then any nonzero multiple of will be an eignvector correspondingto λ
i. Therefore
−−
b
a iλ
λλ
i
i
a b
c d
x
x
− −− −
=
1
4
0
0
λ = + ± − +
1
24( ) (a )2
a d b bc
342 Exercise Set 7.1
are eigenvectors corresponding to λ1 and λ2, respectively.
19. As in Exercise 17, we have
Alternate Solution: Recall that if r1 and r2 are roots of the quadratic equation x2 + Bx + C = 0, then B = –(r1 + r2) and C = r1r2. The converse of this result is also true.Thus the result will follow if we can show that the system of equations
λ1 + λ2 = a + d
λ1λ2 = ad – bc
is satisfied by λ1 = a + b and λ2 = a– c. This is a straightforward computation and we leaveit to you.
20. Suppose that Ax = λx where A is invertible. Then
x = A–1 Ax = A–1 λx = λA–1x
Since A is invertible, we know that λ ≠ 0. Thus
A–1x = 1–λx
That is, 1/λ is an eigenvalue of A–1 and x is a corresponding eigenvector.
λ ( )
( )
=+ ± − +
=+ ± − +
a d a d bc
a d c b
2
2
4
2
44
2
2
2
bcd c b
a d c b
a
because a − = −
=+ ± +
=+
( )
dd c d a b c d
a b
+ + − − +
= + −2 2
or
or a c
−−
−−
b
a
b
aλ λ1 2and
Exercise Set 7.1 343
21. Suppose that Ax = λx. Then
(A – sI)x = Ax – sIx = λx – sx = (λ – s)x
That is, λ – s is an eigenvalue of A – sI and x is a corresponding eigenvector.
22. The characteristic equation of A is λ3 – 6λ2 + 11λ – 6 = 0. The roots of this equation are λ= 1, 2, 3, so that these are the eigenvalues of A. If λ = 1, then λI – A reduces to
Thus if x = [x1 x2 x3]T is an eigenvector of A corresponding to λ = 1, then x1 = x3 and x2 =
0, so that
If λ = 2, then λI – A reduces to
Thus any eigenvector x corresponding to λ = 2 is of the form
If λ = 3, then λI – A reduces to
1 0 1
0 1 1
0 0 0
−−
x =
=
s
s s
2
0
1 2
1
0
1 1 2 0
0 0 1
0 0 0
−
x =
=
r
r
r0
1
0
1
1 0 1
0 1 0
0 0 0
−
344 Exercise Set 7.1
Thus any eigenvector x corresponding to λ = 2 is of the form
(a) By Exercise 20, the eigenvalues of A–1 are 1, 1/2, and 1/3. The correspondingeigenspaces are spanned by the vectors
respectively.
23. (a) For any square matrix B, we know that det(B) = det(BT). Thus
det(λI – A) = det(λI – A)T
= det(λIT – AT)
= det(λI – AT)
from which it follows that A and AT have the same eigenvalues because they have thesame characteristic equation.
(b) Consider, for instance, the matrix which has λ = 1 as a (repeated) eigen-
value. Its eigenspace is spanned by the vector , while the eigenspace of its
transpose is spanned by the vector
24. (a) False, since x must be nonzero.
(b) True. The eigenvalues of A are the only values of λ for which the matrix λI – A issingular. So if λ is not an eigenvalue, then λI – A is invertible and the homogeneoussystem (λI – A)x = 0 has only the trivial solution.
1
1
−
1
1
2 1
1 0−
1
0
1
1 2
1
0
1
1
1
/
x =
=
t
t
t
t
1
1
1
Exercise Set 7.1 345
(c) True. By hypothesis, λ = 0 is an eigenvalue of A. Theorem 7.1.3 then guarantees thatλ2, which is also zero, is an eigenvalue for A2. Theorem 7.1.4 then guarantees that A2
is singular.
Alternatively, observe that if x is the (necessarily nonzero) eigenvector corre-sponding to λ, then Ax = 0. Thus, AAx = 0 for some x ≠ 0, so A2 is singular.
(d) True. Since λ = 0 is not a root of λn + 1, Theorem 7.1.4 guarantees that A is invertible.
25. (a) Since p(λ) has degree 6, A is 6 × 6.
(b) Yes, A is invertible because λ = 0 is not an eigenvalue.
(c) A will have 3 eigenspaces corresponding to the 3 eigenvalues.
346 Exercise Set 7.1
EXERCISE SET 7.2
1. The eigenspace corresponding to λ = 0 can have dimension 1 or 2. The eigenspacecorresponding to λ = 1 must have dimension 1. The eigenspace corresponding to λ = 2 canhave dimension 1, 2, or 3.
2. (a) The characteristic equation is
Hence the eigenvalues are λ = 3 and λ = 5.
(b) If λ = 3,
which has rank 1. If λ = 5,
which has rank 2.
(c) The nullities of the above matrices are 2 and 1, so the corresponding eigenspaces willhave dimensions 2 and 1. Therefore, the 3 × 3 matrix A has 3 linearly independenteigenvectors, and is diagonalizable.
λ I A− =−− −− −
→−1 0 1
2 2 2
1 0 1
1 0 1
0 1 0
0 0 00
λ I A− =−− −− −
→1 0 1
2 0 2
1 0 1
1 0 1
0 0 0
0 0 00
λλ
λλ λ
−− − −− −
= − − =4 0 1
2 3 2
1 0 4
3 5 02( ) ( )
347
5. Call the matrix A. Since A is triangular, the eigenvalues are λ = 3 and λ = 2. The matrices3I – A and 2I – A both have rank 2 and hence nullity 1. Thus A has only 2 linearlyindependent eigenvectors, so it is not diagonalizable.
6. Call the matrix A. The characteristic equation of A is (λ – 2)(λ2 + λ + 1) = 0, so λ = 2 is theonly (real) eigenvalue. Thus A has only one linearly independent eigenvector, so it is notdiagonalizable.
8. The characteristic equation is λ2 – 3λ + 2 = 0, the eigenvalues are λ = 2 and λ = 1, and theeigenspaces are spannedby the vectors
respectively. Thus, if we let
then
Clearly, there are other possibilities for P.
10. The characteristic equation is λ(λ – 1)(λ – 2) = 0, the eigenvalues are λ = 0, λ = 1, and λ= 2, and the eigenspaces are spannedby the vectors
Thus, if we let
P = −
0 1 0
1 0 1
1 0 1
0
1
1
1
0
0
0
1
1
−
P AP− =
1 2 0
0 1
P =
3 4
4 5
3 4
1
4 5
1
and
348 Exercise Set 7.2
then
12. The characteristic equation is λ3 – λ2 +5λ – 2 = 0, the eigenvalues are λ = 1 and λ = 2, andthe eigenspaces are spanned by the vectors
respectively. Since A is a 3 × 3 matrix with only 2 linearly independent eigenvectors, it isnot diagonalizable.
13. The characteristic equation is λ3 – 6λ2 + 11λ – 6 = 0, the eigenvalues are λ = 1, λ = 2, andλ = 3, and the eigenspaces are spannedby the vectors
Thus, one possibility is
and
P AP− =
11 0 0
0 2 0
0 0 3
P =
1 2 1
1 3 3
1 3 4
1
1
1
2 3
1
1
1 4
3 4
1
1
4 3
1
4 3
4 3
1
and
P AP− =
10 1 0
0 1 0
0 0 2
Exercise Set 7.2 349
15. The characteristic equation is λ2(λ – 1) = 0; thus λ = 0 and λ = 1 are the only eigenvalues.
The eigenspace associated with λ = 0 is spanned by the vectors and ; the
eigenspace associated with λ = 1 is spanned by . Thus, one possibility is
and hence
18. The eigenvalues of A are λ = 1 and λ = 2 and the eigenspaces are spanned by the vectors
Thus if we let
then
Now by Formula (10), we have
P AP− =
1 1 0
0 2
P =
1 0
1 1
1
1
and
0
1
P AP− =
10 0 0
0 0 0
0 0 1
P =−
1 0 0
0 1 0
3 0 1
0
0
1
0
1
0
1
0
3−
350 Exercise Set 7.2
20. By Theorem 7.1.1, the eigenvalues of A are λ1 = 1 and λ2 = λ3 = –1. Correspondingeigenvectors are
Since A has three linearly independent eigenvectors, it is diagonalizable. Let
Then
Thus, Ak = PDk P–1 for any positive integer k, or
APIP PP I k
PDP A k
k == =
=
− −
−
1 1
1
if is even
if is odd
P A P D− = = −
−
11 0 0
0 1 0
0 0 1
P =−
1 1 4
0 1 0
0 0 1
1
0
0
1
1
0
4
0
1
−
A P P10
1011 0
1 2
1 0
1 1
1 0
0 2
=
=
−
110
10 10
1 0
1 1
1 0
1 2 2
−
=−
Exercise Set 7.1 351
Finally, since D is its own inverse, we have
A–1 = (PDP–1)–1 = PD–1 P–1 = A
so that
A–k = (A–1)k = Ak
Of course, all of the above couldhave been greatly simplified if we had been cleverenough to notice at the beginning that A2 = I.
21. The characteristic equation of A is (λ – 1)(λ – 3)(λ – 4) = 0 so that the eigenvalues are λ = 1, 3, and 4. Corresponding eigenvectors are [1 2 1]T, [1 0 –1]T, and [1 –1 1]T,respectively, so we let
Hence
and therefore
22. (a) By Theorem 7.2.3, we can diagonalize A if it has two distinct eigenvalues. According toExercise 17 of Section 7.1, this will occur if and only if (a – d)2 + 4bc > 0.
(b) See Part (a). If the discriminant, (a – d)2 + 4bc, of the characteristic equation isnegative, then there are no real roots and hence no real eigenvalues or eigenvectors.Thus, by Theorem 7.2.1, A is not diagonalizable.
An
n
n
n
= −−
1 1 1
2 0 1
1 1 1
1 0 0
0 3 0
0 0 4
−−
/ / /
/ /
/ / /
1 6 1 3 1 6
1 2 0 1 2
1 3 1 3 1 3
P− = −
−
11 6 1 3 1 6
1 2 0 1 2
1 3 1 3 1 3
/ / /
/ /
/ / /
P = −−
1 1 1
2 0 1
1 1 1
352 Exercise Set 7.2
24. Since A is diagonalizable, we have A = PDP–1 where D is a diagonal matrix with theeigenvalues of A as its diagonal elements. Thus the rank of D is the number of nonzeroeigenvalues of A. Since P and P–1 are invertible, they are products of elementary matrices.But multiplication on the left or on the right by an elementary matrix does not change therank of the product. Therefore rank(A) = rank(D), and we are done.
25. (a) False. For instance the matrix , which has linearly independent column
vectors, has characteristic polynomial (λ – 1)2. Thus λ = 1 is the only eigenvalue.
The corresponding eigenvectors all have the form . Thus this 2 × 2 matrix has
only 1 linearly independent eigenvector, and hence is not diagonalizable.
(b) False. Any matrix Q which is obtained from P by multiplying each entry by a nonzeronumber k will also work. Why?
(c) True by Theorem 7.2.2.
(d) True. Suppose that A is invertible and diagonalizable. Then there is an invertiblematrix P such that P–1 AP = D where D is diagonal. Since D is the product of invertiblematrices, it is invertible, which means that each of its diagonal elements d
iis nonzero
and D–1 is the diagonal matrix with diagonal elements 1/di. Thus we have
(P–1 AP)–1 = D–1
or
P–1 A–1 P = D–1
That is, the same matrix P will diagonalize both A and A–1.
26. (a) The dimensions of the eigenspaces corresponding to λ = 1, λ = 3, and λ = 4 are atmost 1, 2, and 3 respectively. The dimension of the eigenspace corresponding to λ =1 must be exactly 1.
(b) If A is diagonalizable, then it follows from Theorem 7.2.1 that the dimensions of the 3spaces will be exactly 1, 2, and 3.
(c) The eigenvalue must be λ = 4.
27. (a) Since A is diagonalizable, there exists an invertible matrix P such that P–1 AP = Dwhere D is a diagonal matrix containing the eigenvalues of A along its diagonal.Moreover, it easily follows that P–1 AkP = Dk for k a positive integer. In addition,Theorem 7.1.3 guarantees that if λ is an eigenvalue for A, then λk is an eigenvalue forAk. In other words, Dk displays the eigenvalues of Ak along its diagonal.
t1
1
0 1
1 2−
Exercise Set 7.2 353
Therefore, the sequence
P–1 AP = D
P–1 A2 P = D2
...P–1 Ak P = Dk
...
will converge if and only if the sequence A, A2, . . ., Ak, . . . converges. Moreover, thiswill occur if and only if the sequences λ
i, λ
i2, . . ., λ i
k, . . . converges for each of the neigenvalues λ
iof A.
(b) In general, a given sequence of real numbers a, a2, a3, . . . will converge to 0 if and onlyif –1 < a < 1 and to 1 if a = 1. The sequence diverges for all other values of a.
Recall that P–1 Ak P = Dk where Dk is a diagonal matrix containing the eigenvalues
λ1k, λ 2
k, . . ., λnk on its diagonal. If |λ
i| < 1 for all i = 1, 2, . . ., n, then Dk = O
and hence Ak = O.
If λi
= 1 is an eigenvalue of A for one or more values of i and if all of the other
eigenvalues satisfy the inequality |λj| < 1, then Ak exists and equals PD
LP–1 where
DL
is a diagonal matrix with only 1’s and 0’s on the diagonal.
If A possesses one or more eigenvalues λ which do not satisfy the inequality
–1 < λ ≤ 1, then Ak does not exist.
28. (a) If Sn
= 1 + x + x2 + . . . + xn, then Sn
– xSn
= (1 – x)Sn
= 1 – xn+1 or Sn
= (1 – xn+1)/
(1 – x). If |x| < 1, then Sn
= 1/(1 – x). If |x| ≥ 1, then the limit doesn’t exist.
(b) and (c) If A is diagonalizable and λ1, λ2, . . ., λn
are the eigenvalues of A, then
P–1(I + A + A2 + . . . + Ak)P = I + P–1 AP + P–1 A2 P + . . . + P–1 Ak P
is a diagonal matrix with
di
= 1 + λi
+ λi–2 + . . . + λ
i–k
as its ith entry. If |λi| < 1 for all i = 1, 2, . . ., n, then the ith entry of that diagonal
matrix will converge to 1/(1 – λi).
If A possesses one or more eigenvalues λ such that |λ| ≥ 1, then the limit of thediagonal matrix will not exist.
limk→°
limk→°
limk→°
limk→°
limk→°
354 Exercise Set 7.2
29. The Jordan block matrix is
Since this is an upper triangular matrix, we can see that the only eigenvalue is λ = 1, withalgebraic multiplicity n. Solving for the eigenvectors leads to the system
( )λ I Jn− =
x
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
x.
Jn =
1 1 0 0 0
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1
.
Exercise Set 7.2 355
EXERCISE SET 7.3
1. (a) The characteristic equation is λ(λ – 5) = 0. Thus each eigenvalue is repeated onceand hence each eigenspace is 1-dimensional.
(c) The characteristic equation is λ2(λ – 3) = 0. Thus the eigenspace corresponding to λ= 0 is 2-dimensional and that corresponding to λ = 3 is 1-dimensional.
(e) The characteristic equation is λ3(λ – 8) = 0. Thus the eigenspace corresponding to λ= 0 is 3-dimensional and that corresponding to λ = 8 is 1-dimensional.
2. The eigenvalues of A are λ = 4 and λ = 2. The eigenspaces are spanned by the orthogonalvectors
respectively. If we normalize these vectors, we obtain
Thus
P =−
1
2
1
21
2
1
2
1
21
2
1
21
2
−
and
1
1
1
1
−
and
357
and
If we interchange the order of the vectors which span the eigenspaces, we obtain
and
6. The eigenvalues of A are λ = 2 and λ = 0. The eigenspace associated with λ = 2 is spanned
by ; the eigenspace associated with λ = 0 is spanned by and . If we
normalize these three orthogonal vectors, we obtain
Thus
P = −
1
2
1
20
1
2
1
20
0 0 0
1
21
20
1
21
20
−
0
0
1
0
0
1
1
1
0
−
1
1
0
P AP− =
1 2 0
0 4
P =−
1
2
1
21
2
1
2
P AP− =
1 4 0
0 2
358 Exercise Set 7.3
and
Another possibility (which results from changing the order of the spanning vectors) is
so that
8. The eigenvalues of A are λ = 0, λ = 4, and λ = 2. The eigenspace associated with λ = 0 is
spanned by and ; the eigenspaces associated with λ = 4 and λ = 2 are
spanned by and respectively. Hence, the eigenspaces of A are spanned by the
orthogonal vectors
0
0
1
0
0
0
0
1
1
1
0
00
1
1
0
0
−
1
1
0
0
−
1
1
0
0
0
0
0
1
0
0
1
0
P AP− =
10 0 0
0 0 0
0 0 2
P = −
1
20
1
21
20
1
21
21
1
2
P AP− =
12 0 0
0 0 0
0 0 0
Exercise Set 7.3 359
Normalizing each of these vectors yields
Thus
and
Changing the order of the spanning vectors may change the position of the eigenvalues onthe diagonal of P–1 AP.
10. The characteristic equation is λ2 – 2aλ + a2 – b2 = 0. By the quadratic formula, λ = a ± |b|,or λ = a ± b. Since b ≠ 0, there are 2 distinct eigenvalues with eigenspaces spanned by theorthogonal vectors
1
1
1
1
−
and
P AP− =
1
0 0 0 0
0 0 0 0
0 0 4 0
0 0 0 2
P
P
= −
−
0 0 1 2 1 2
0 0 1 2 1 2
1 0 0 0
0 1 0 0
11
0 0 0
0 0 0 1
1 2 1 2 0 0
1 2 1 2 0 0
=
−
1
0
0
1
0
0
0
0
1
1 2
1 2
0
0
−
1 2
1 2
0
0
360 Exercise Set 7.3
Thus
will orthogonally diagonalize the given matrix.
12. (a) By Theorem 7.3.1, it will suffice to show that I – v vT is symmetric. This follows fromproperties of the transpose.
(b) In this case,
The characteristic equation is (λ – 1) (λ – 1)(λ + 1) = 0, so the eigenvalues are λ = ±1.
Two linearly independent eigenvectors corresponding to λ = 1 are and .
An eigenvector corresponding to λ = –1 is . Thus the matrix
will orthogonally diagonalize I – vvT.
P =
−
1 2 0 1 2
0 1 0
1 2 0 1 2
1
0
1
0
1
0
1
0
1−
IT− =
−
v v
1 0 0
0 1 0
0 0 1
1
0
1
=−
−
1 0 1
0 0 1
0 1 0
1 0 0
1
2
1
21
2
1
2
−
Exercise Set 7.3 361
13. By the result of Exercise 17, Section 7.1, the eigenvalues of the symmetric 2 × 2 matrix
, are Since (a – d)2 + 4b2 cannot be negative,
the eigenvalues are real.
14. (a) True, since both are symmetric.
(b) True, since v1 and v2 must be orthogonal.
(c) False. For instance, let . Then A is orthogonal, but not
symmetric and hence not orthogonally diagonalizable.
(d) Since A is orthogonally diagonalizable, it follows from Theorem 7.3.1 that it must besymmetric, that is, A = AT. And since A is also invertible, we have
I = IT = (A–1 A)T = AT (A–1)T = A(A–1)T
This implies that A–1 = (A–1)T, so that A–1 is symmetric and hence is orthogonallydiagonalizable.
15. Yes. Notice that the given vectors are pairwise orthogonal, so we consider the equation
P–1 AP = D
or
A = PDP–1
where the columns of P consist of the given vectors each divided by its norm and where Dis the diagonal matrix with the eigenvalues of A along its diagonal. That is,
P D=
−
=−0 1 0
1 2 0 1 2
1 2 0 1 2
1 0 0
0 3 0
0
and
00 7
A =−
1 2 3 2
3 2 1 2
λ ( )= + ± −( ) +
1
24
2 2a d a d b
a b
b a
362 Exercise Set 7.3
From this, it follows that
Alternatively, we could just substitute the appropriate values for λ and x in the equationAx = λx and solve for the matrix A.
16. The converse of Theorem 7.3.2(b) is: “Eigenvectors from the same eigenspace are notorthogonal.” This is false. If we have an n-dimensional eigenspace, with n > 1, then we canuse the Gram-Schmidt process to convert any basis of eigenvectors to an orthogonal basis.Doing this, we have constructed a set of eigenvectors from the same eigenspace that areorthogonal.
A PDP= =
13 0 0
0 3 4
0 4 3
Exercise Set 7.3 363
SUPPLEMENTARY EXERCISES 7
1. (a) The characteristic equation of A is λ2 – 2 cos θ + 1 = 0. The discriminant of thisequation is 4 (cos2 θ – 1), which is negative unless cos2 θ = 1. Thus A can have no realeigenvalues or eigenvectors in case 0 < θ < π.
3. (a) If
then D = S2, where
Of course, this makes sense only if a21
≥ 0, . . ., an ≥ 0.
(b) If A is diagonalizable, then there are matrices P and D such that D is diagonal and D= P–1 AP. Moreover, if A has nonnegative eigenvalues, then the diagonal entries of Dare nonnegative since they are all eigenvalues. Thus there is a matrix T, by virtue ofPart (a), such that D = T2. Therefore,
A = PDP–1 = PT2 P–1 = PTP–1 PTP–1 = (PTP–1)2
That is, if we let S = PTP–1, then A = S2.
S
a
a
an
=
1
2
0 0
0 0
0 0
D
a
a
an
=
1
2
0 0
0 0
0 0
365
3. (c) The eigenvalues of A are λ = 9, λ = 1, and λ = 4. The eigenspaces are spanned by thevectors
Thus, we have
while
Therefore
4. (a) Since (λI – A)T = λI – AT, we have
det(λI – A) = det((λI – A)T)
= det(λI – AT)
Thus A and AT have the same characteristic polynomials, and hence the sameeigenvalues.
(b) By the result of Exercise 18, Section 7.1, if A has distinct eigenvalues λ1 and λ2, andif bc ≠ 0, then the eigenvectors of A are
−−
−−
b
a
b
aλ λ1 2and
S PTP= =
−11 1 0
0 2 1
0 0 3
D T=
=
9 0 0
0 1 0
0 0 4
3 0 0
0 1 0
0 0 2
and
P P=
= −−1 1 1
2 0 1
2 0 0
0 0 1 2
1 2 1 2
0 1
1and
−−
1
1
2
2
1
0
0
1
1
0
366 Supplementary Exercises 7
Corresponding eigenvectors of AT are
Thus we need only look for a matrix with two distinct eigenvalues, with bc ≠ 0, andwith b ≠ c. For instance, if we let
then the resulting pairs of eigenvectors will be different.
5. Since det(λI – A) is a sum of signed elementary products, we ask which terms involve λn–1.Obviously the signed elementary product
q= (λ – a11)(λ – a22
) …. (λ – ann)
= –n – (a11 + a22 + …. + ann
)λn–1
+ terms involving λr where r < n – 1
has a term – (trace of A)λn–1. Any elementary product containing fewer than n – 1 factorsof the form λ – a
iicannot have a term which contains λn–1. But there is no elementary
product which contains exactly n – 1 of these factors (Why?). Thus the coefficient of λn–1
is minus the trace of A.
6. The only eigenvalue is λ = a and if b ≠ 0, then the eigenspace is spanned by the vector
. Thus, by Theorem 7.2.1, A is not diagonalizable.
7. (b) The characteristic equation is
p(λ) = –1 + 3λ – 3λ2 + λ3 = 0
1
0
A =−
1 1
3 1
−−
−−
c
a
c
aλ λ1 2and
Supplementary Exercises 7 367
Moreover,
and
It then follows that
p(A) = –I + 3A – 3A2 + A3 = 0
8. (b) Let A be an n × n matrix, so that A = PDP–1, where D is a diagonalizable matrix whoseentries on the main diagonal are the eigenvalues of A. Then, for any positive integer m,we have
Am = (PDP–1)m = (PDP–1)(PDP–1) … (PDP–1) = PDn P–1
If the characteristic equation of A is
p(λ) = a0 + a1λ + a2λ2 + … + a
nλn = 0,
we then have
p(A) = a0I + a1A + a2A2 + … a
nAn
= a0I + a1(PDP–1) + a2(PDP–1)2 + … + an(PDP–1)n
= a0I + a1(PDP–1) + a2)PDP–1)2 + … + an(P⟨nP–1)
=P(a0I)P–1 + P(a1D)P–1 + P(a2D2)P–1 + … P(a
nDn)P–1
= P[a0I + a1D + a2S2 + … + a
nDn]P–1
A3
1 3 3
3 8 6
6 15 10
=−−
−
A2
0 0 1
1 3 3
3 8 6
= −−
368 Supplementary Exercises 7
However, each λi
is a root of the characteristic polynomial, so p(λi) = 0 for i = 1, . . ., n.
Then,
= 0.
Thus, a diagonalizable matrix satisfies its characteristic polynomial.
9. Since c0 = 0 and c1 = –5, we have A2 = 5A, and, in general, An = 5n–1 A.
p A P P( ) =
−
0 0 0
0 0 0
0 0 0
1
=
+P
a
a
a
a0
0
0
10 0
0 0
0 0
λ11
1 2
20 0
0 0
0 01
a
a
a
n
λ
λ
λ
+
112
2 22
22
0 0
0 0
0 0
a
a n
λ
λ
+ +
a
a
a
nn
nn
n nn
λ
λ
λ
1
2
0 0
0 0
0 0
=
+ + +
−P
P
a a a
a
nn
0 1 1 1
0
0 0
0
λ λ
++ + +
+ + +
a a
a a a
nn
n n nn
1 2 2
0 1
0
0 0
λ λ
λ λ
=
( )
( )
(
p
P
p
p
p
1
1
2
0 0
0 0
0 0
λλ
λ
nn
P
)
−1
Supplementary Exercises 7 369
11. Call the matrix A and show that A2 = (c1 + c2 + . . . + cn)A = [tr(A)]A. Thus Ak =
[tr(A)]k–1 A for k = 2, 3, . . . . Now if λ is an eigenvalue of A, then λk is an eigenvalue of Ak,so that in case tr(A) ≠ 0, we have that λk[tr(A)]k–1 = [λ/tr(A)]k–1 λ is an eigenvalue of Afor k = 2, 3, . . . . Why? We know that A has at most n eigenvalues, so that this expressioncan take on only finitely many values. This means that either λ = 0 or λ = tr(A). Why? In case tr(A) = 0, then all of the eigenvalues of A are 0. Why? Thus the only possibleeigenvalues of A are zero and tr(A). It is easy to check that each of these is, in fact, aneigenvalue of A.
Alternatively, we could evaluate det(Iλ – A) by brute force. If we add Column 1 toColumn 2, the new Column 2 to Column 3, the new Column 3 to Column 4, and so on, weobtain the equation
If we subtract Row 1 from each of the other rows and then expand by cofactors along thenth column, we have
= (–1)n+1 (λ – tr(A))(–λ)n–1 because the above matrix is triangular
= (–1)2n(λ – tr(A))λn–1
= λn–1(λ – tr(A))
Thus λ = tr(A) and λ = 0 are the eigenvalues, with λ = 0 repeated n – 1 times.
det( ) det
( )
I A
c c c c c c A
− =
− − − − − − −λ λ λ λ1 1 2 1 2 3 tr
−−− −
− − −
=
λλ λ
λ λ λ
0 0 0
0 0
0
( ) ( ( )) det− −
−− −− − −+1
0 0 0
0 0
01ntr Aλ
λλ λλ λ λ
− − − −
λ λ λ λ
det( ) detI A
c c c c c c c
λ
λ λ λ λ
− =
− − − − − − − −1 1 2 1 2 3 1 cc c
c c c c c c c c c
c
n
n
2
1 1 2 1 2 3 1 2
− −− − − − − − − − − −−
λ λ λ
11 1 2 1 2 3 1 2
1 1
− − − − − − − − −
− − −
c c c c c c c c
c c c
nλ λ
22 1 2 3 1 2− − − − − − −
c c c c c cn λ
370 Supplementary Exercises 7
12. (a) The characteristic polynomial of the given matrix is
If we add λ times the second row to the first row, this becomes
Expanding by cofactors along the first column gives
Adding λ2 times the second row to the first row yields
p
c c c
c
c( ) detλ
λ λ λ
λ
λ=
+ +
−
−
0 0 0
1 0 0
0 1 0
31 2 3
2
3
44
10 0 0 1
− +
−λ cn
p
c c
c
c( ) detλ
λ λ
λ
λ=
+
−
−
21 2
3
4
0 0 0
1 0 0
0 1 0
00 0 0 1 1 − +
−λ cn
p
c
( ) detλ
λ λ
λ
λ=
+
−
−
0 22
1
2
3
0 0 c
1 0 0 c
0 1 0 c
0 0 1 cn 10 −
−λ +
p
c
c
c( ) detλ
λλ
λ=−
−
0 0 0 0
1 0 0 0
0 1 0 0
0 0
1
2
3
−− +
−1 1λ cn
Supplementary Exercises 7 371
Expanding by cofactors along the first column gives
Eventually this procedure will yield
Adding λn–2 times the second row to the first row gives
or
p(λ) = c1 + c2λ + … + cn–1λ
n–2 + λn–1
which is the desired result.
14. The characteristic equation of an n × n matrix has degree n and therefore n roots. Sincethe complex roots of a polynomial equation with real coefficients occur in conjugate pairs,any such equation of odd degree must have at least one real root. Hence A has at least onereal eigenvalue.
16. The characteristic equation of the matrix is
(λ – 1)(λ + 2 (λ – 3)(λ + 3) = λ4 + λ3 – 11λ2 – 9λ + 18 = 0
(a) By Exercise 14 of Section 7.1, det(A) = 18.
(b) By Exercise 5, above, tr(A) = –1.
17. Since every odd power of A is again A, we have that every odd power of an eigenvalue ofA is again an eigenvalue of A. Thus the only possible eigenvalues of A are λ = 0, ±1.
pc c c
c
nn n
n
( ) detλ λ λ λλ
= + + + +− +
−− −
−
0
11 2 1
2 1
1
pc c c
c
nn
n
n
( ) detλ λ λ λλ
= + + +− +
−−
−
−
21 2 2
3
11
p
c c c
c
c( ) detλ
λ λ λ
λ
λ=
+ +
−
−
0 0 0
1 0 0
0 1 0
1 2 32
4
55
10 0 0 1
− +
−λ cn
372 Supplementary Exercises 7
EXERCISE SET 8.1
2. Using the usual notation, we have
T(u + v)= (2(u1 + v1) – (u2 + v2) + (u3 + v3), (u2 + v2) – 4(u3 + v3))
= (2u1 – u2 + u3, u2 – 4u3) + (2v1 –v2 + v3, v2 – 4v3)
= T(u) + T(v)
and
T(cu) = (2cu1 – cu2 + cu3, cu2 – 4cu3) = cT(u)
3 Since T (–u) = –u = u = T(u) ≠ –T(u) unless u = 0, the function is not linear.
5. We observe that
T(A1 + A2) = (A1 + A2)B = A1B + A2B = T(A1) + T(A2)
and
T(cA) = (cA)B = c(AB) = cT(A)
Hence, T is linear.
6. Since T(A + B) = tr(A + B) = tr(A) + tr(B) = T(A) + T(B) and T(cA) = tr(cA) = c(tr(A))= cT(A), T is linear.
373
8. (b) Since both properties fail, T is nonlinear. For instance,
10. (a) T is not linear because it sends the zero function to f(x) = 1.
(b) Since
T(f(x) + g(x))= T((f + g)(x)) = (f + g)(x + 1)
= f(x + 1) + g(x + 1) = T(f(x)) + T(g(x))
and
T(cf(x)) = cf(x + 1) = cT(f(x))
T is linear.
12. Let v = (x1, x2) = av1 + bv2 = a(1, 1) + b(1, 0). This yields the equations x1 = a + b and x2 = a, so that v = x2v1 + (x1 – x2)v2. Thus
T(v) = T(x1, x2) = x2T(v1) + (x1 – x2)T(v2)
= x2(1, –2) + (x1 – x2)(–4, 1)
= (–4x1 + 5x2, x1 – 3x2)
so that
T(2, –3) = (–8 – 15, 2 + 9) = (–23, 11)
14. Let v = (x1, x2, x3) = av1 + bv2 + cv3 = a(1, 1, 1) + b(1, 1, 0) + c(1, 0, 0). Thus x1 = a + b+ c, x2 = a + b, and x3 = a, so that v = x3v1 + (x2 – x3)v2 + (x1 – x2)v3. Therefore
T ka b
c dk a k b
kTa b
c d
= +
≠
2 2 2 2
in general.
374 Exercise Set 8.1
T(v)= T(x1, x2, x3) = x3T(v1) + (x2 – x3)T(v2) + (x1 – x2)T(v3)
= x3(2, –1, 4) + (x2 – x3)(3, 0, 1) + (x1 – x2)(–1, 5, 1)
= (–x1 + 4x2 – x3, 5x1 – 5x2 – x3, x1 + 3x3)
and
T(2, 4, –1) = (–2 + 16 + 1, 10 –20 + 1, 2 – 3) = (15, –9, –1)
16. We have
T(2v1 – 3v2 + 4v3)= 2T(v1) – 3T(v2) + 4T(v3)
= (2, –2, 4) + (0, –9, –6) + (–12, 4, 8)
= (–10, –7, 6)
17. (a) Since T1 is defined on all of R2, the domain of T2 T1 is R2. We have T2
T1(x, y) =T2(T1(x, y)) = T2(2x, 3y) = (2x – 3y, 2x + 3y). Since the system of equations
2x – 3y = a
2x + 3y = b
can be solved for all values of a and b, the codomain is also all of R2.
(d) Since T1 is defined on all of R2, the domain of T2 T1 is R2. We have
T2(T1(x, y)) = T2(x – y, y + z, x – z) = (0, 2x)
Thus the codomain of T2 T1 is the y-axis.
18. (a) Since T1 is defined on all of R2, the domain of T3 T2
T1 is R2. We have
T3 T2
T1(x, y) = T3 T2 (–2y, 3x, x – 2y)
= T3(3x, x – 2y, –2y)
= (3x – 2y, x)
This vector can assume all possible values in R2, so the codomain is R2.
Exercise Set 8.1 375
19. (a) We have
(b) Since the range of T1 is not contained in the domain of T2, T2 T1 is not well defined.
20. We have
(T1 T2)(p(x)) = T1(p(x + 1)) = p((x – 1) + 1) = p(x)
and
(T2 T1)(p(x)) = T2(p(x – 1)) = p((x + 1) –1) = p(x)
22. We have
T2 T1 (a0 + a1x + a2x
2) = T2 (a0 + a1(x + 1) + a2(x + 1)2)
= T2((a0 + a1 + a2) + (a1 + 2a2)x + a2 x2)
= (a0 + a1 + a2)x + (a1 + 2a2)x2 + a2x3
24. (b) We have
(T3 T2) T1(v) = (T3
T2)(T1(v))
= T3(T2(T1(v)))
= T3 T2
T1(v)
25. Since (1, 0, 0) and (0, 1, 0) form an orthonormal basis for the xy-plane, we have T(x, y, z)= (x, 0, 0) + (0, y, 0) = (x, y, 0), which can also be arrived at by inspection. Then T(T(x,y, z)) = T(x, y, 0) = (x, y, 0) = T(x, y, z). This says that T leaves every point in the x-y
plane unchanged.
( )( ) ( )T T A Aa c
b da d
T1 2 = =
= +tr tr
376 Exercise Set 8.1
26. (a) From the linearity of T and the definition of kT, we have
(kT)(u + v) = k(T(u + v)) = k(T(u) + T(v))
= kT(u) + kT(v) = (kT)(u) + (kT)(v)
and
(kT)(cu) = k(T(cu)) = kcT(u) = c(kT)(u)
Therefore, kT is linear.
28. (a) F is linear since
F((x1, y1) + (x2, y2)) = F(x1 + x2, y1 + y2)
= (a1(x1 + x2) + b1(y1 + y2), a2 (x1 + x2) + b2 (y1 + y2))
= (a1x1 + b1y1, a2x1 + b2y1) + (a1x2 + b1y2, a2x2 + b2y2)
= F(x1, y1) + F(x2, y2)
and
F(c(x, y)) = F(cx, cy)
= (a1cx + b1cy, a2cx + b2cy)
= c(a1x + b1y, a2x + b2y)
= cF(x, y)
(b) Since F(c(x, y)) = F(cx, cy) = c2F(x, y), F is not a linear operator.
30. Let v = a1v1 + … + anv
nbe any vector in V. Then
T(v)= a1T(v1) + … + anT(v
n)
= a1v1 + … + anv
n
= v
Hence T is the identity transformation on V.
Exercise Set 8.1 377
31. (b) We have
(c) We have
33. (a) True. Let c1 = c2 = 1 to establish Part (a) of the definition and let c2 = 0 to establishPart (b).
(b) False. All linear transformations have this property, and, for instance there is morethan one linear transformation from R2 to R2.
(c) True. If we let u = 0, then we have T(v) = T(–v) = –T(v). That is, T(v) = 0 for allvectors v in V. But there is only one linear transformation which maps every vector tothe zero vector.
(d) False. For this operator T, we have
T(v + v) = T(2v) = v0 + 2 v
But
T(v) + T(v) = 2T(v) = 2v0 + 2v
Since v0 ≠ 0, these two expressions cannot be equal.
34. Since B is a basis for the vector space V, we can write any vector v in V in one and only oneway as a linear combination v = c1v1 + c2v2 + … + c
nv
nof vectors from B. Thus for any
linear operator T, we have
T(v) = c1T(v1) + c2T(v2) + … + cnT(v
n)
That is, to define T, we need only specify where it maps each of the basis vectors in B. IfT is to map B into itself, then we have n choices for each of the n vectors T(v
i), and hence
a total of nn possible operators T.
35. Yes. Let T Pn → Pm be the given transformation, and let TR
Tn+1 → Rm+1 be thecorresponding linear transformation in the sense of Section 4.4. Let
n P
n→ Rn+1 be the
function that maps a polynomial in Pn
to its coordinate vector in Rn+1, and let m
Pm
→Rm+1 be the function that maps a polynomial in P
nto its coordinate vector in Rm+1.
( )( ) ( )J D e e dt ex x xx
+ = + ′ = −∫3 3 10
( )(sin ) (sin ) sin( ) – sin( ) sin( )J D x t dt x x= ′ = =00
xx
∫
378 Exercise Set 8.1
By Example 7, both n
and m
are linear transformations. Theorem 5.4.1 implies that acoordinate map is invertible, so
m–1 is also a linear transformation.
We have T=m–1
TR
n, so T is a composition of linear transformations. Refer to the
diagram below:
Thus, by Theorem 8.1.2., T is itself a linear transformation.
T
TR
Pn
nϕ ϕ m–1
Pm
Rm+1Rn+1
Exercise Set 8.1 379
EXERCISE SET 8.2
1. (a) If (1, –4) is in R(T), then there must be a vector (x, y) such that T(x, y) = (2x – y,–8x + 4y) = (1, –4). If we equate components, we find that 2x – y = 1 or y = t and x= (1 + t)/2. Thus T maps in finitely many vectors into (1, –4).
(b) Proceeding as above, we obtain the system of equations
2x – y = 5
–8x + 4y = 0
Since 2x – y = 5 implies that –8x + 4y = –20, this system has no solution. Hence (5, 0)is not in R(T).
2. (a) The vector (5, 10) is in ker(T) since
T(5, 10) = (2(5) – 10, –8(5) + 4(10)) = (0, 0)
(b) The vector (3, 2) is not in ker(T)since T(3, 2) = (4, –16) ≠ (0, 0).
3. (b) The vector (1, 3, 0) is in R(T) if and only if the following system of equations has asolution:
4x + y – 2z – 3w = 1
2x + y + z – 4w = 3
6x –9z + 9w = 0
This system has infinitely many solutions x = (3/2)(t – 1), y = 10 – 4t, z = t, w = 1where t is arbitrary. Thus (1, 3, 0) is in R(T).
381
4. (b) Since T(0, 0, 0, 1) = (–3, –4, 9) ≠ (0, 0, 0), the vector (0, 0, 0, 1) is not in ker(T).
5. (a) Since T(x2) = x3 ≠ 0, the polynomial x2 is not in ker(T).
6. (a) Since T(1 + x) = x + x2, the polynomial x + x2 is in R(T).
(c) Since there doesn’t exist a polynomial p(x) such that xp(x) = 3 – x2, then 3 – x2 is notin R(T).
7. (a) We look for conditions on x and y such that 2x – y = and –8x + 4y = 0. Since theseequations are satisfied if and only if y = 2x, the kernel will be spanned by the vector(1, 2), which is then a basis.
(c) Since the only vector which is mapped to zero is the zero vector, the kernel is 0 andhas dimension zero so the basis is the empty set.
8. (a) Since –8x + 4y = –4(2x – y) and 2x – y can assume any real value, the range of T isthe set of vectors (x, –4x) or the line y = –4x. The vector (1, –4) is a basis for thisspace.
(c) The range of T is just the set of all polynomials in P3 with constant term zero. The setx, x2, x3 is a basis for this space.
9. (a) Here n = dim(R2) = 2, rank(T) = 1 by the result of Exercise 8(a), and nullity(T) = 1by Exercise 7(a). Recall that 1 + 1 = 2.
(c) Here n = dim(P2) = 3, rank(T) = 3 by virtue of Exercise 8(c), and nullity(T) = 0 byExercise 7(c). Thus we have 3 = 3 + 0.
10. (a) We know from Example 1 that the range of T is the column space of. Using elementarycolumn operations, we reduce to the matrix
Thus and as well as and form a basis for the column space
and therefore also for the range of T.
0
1
1
1
0
2
0
1
1
1
5
7
1 0 0
5 1 0
7 1 0
1 0 0
0 1 0
2 1 0
or
382 Exercise Set 8.2
(b) To investigate the solution space of A, we look for conditions on x, y, and z such that
The solution of the resulting system of equations is x = –14t, y = 19t, and z = 11t
where t is arbitrary. Thus one basis for ker(T)is the vector .
(c) By Parts (a) and (b), we have rank(T) = 2 and nullity(T) = 1.
(d) Since rank(A) = 2 and has 3 columns, nullity (A) = 3 – 2 = 1. This also follows fromTheorem 8.2.2 and Part (c) above.
12. (a) Since the range of T is the column space of A, we reduce using column operations.This yields
Thus and form a basis for the range of T.
(b) If
then x = –s – 4r, y = –s + 2r, z = s, and w = 7r where s and r are arbitrary. Thus,
That is, the set [–1, –1, 1, 0]T, [–4, 2, 0, 7]T is one basis for ker(T).
x
y
z
w
s
=
−−
+
1
1
1
0
−−
4
2
0
7
r
4 1 5 2
1 2 3 0
0
0
=
x
y
z
w
0
1
1
0
1 0 0 0
0 1 0 0
−
14
19
11
1 1 3
5 6 4
7 4 2
−−
x
y
z =
0
0
0
Exercise Set 8.2 383
12. (c) By Parts (a) and (b), rank(T) = nullity(T) = 2.
(d) Since A has 4 columns and rank(A) = 2, we have nullity(A) = 4 – 2 = 2. This alsofollows from Theorem 8.2.2 and Part (c) above.
14. (a) The orthogonal projection on the xz-plane maps R3 to the entire xz-plane. Hence itsrange is the xz-plane. It maps only the y-axis to (0, 0, 0), so that its nullspace is they-axis.
16. (a) The nullity of T is 5 – 3 = 2.
(c) The nullity of T is 6 – 3 = 3.
18. (a) The dimension of the solution space of Ax = 0 is, by virtue of the Dimension Theorem,7 – 4 = 3.
(b) Since the range space of A has dimension 4, the range cannot be R5, which hasdimension 5.
19. By Theorem 8.2.1, the kernel of T is a subspace of R3. Since the only subspaces of R3 arethe origin, a line through the origin, a plane through the origin, or R3 itself, the resultfollows. It is clear that all of these possibilities can actually occur.
21. (a) If
then x = –t, y = –t, z = t. These are parametric equations for a line through the origin.
(b) Using elementary column operations, we reduce the given matrix to
1 0 0
3 5 0
2 8 0
−−
1 3 4
3 4 7
2 2 0
0
0
0−
=
x
y
z
384 Exercise Set 8.2
Thus, (1, 3, –2)T and (0, –5, 8)T form a basis for the range. That range, which we caninterpret as a subspace of R3, is a plane through the origin. To find a normal to thatplane, we compute
(1, 3, –2) × (0, –5, 8) = (14, –8, –5)
Therefore, an equation for the plane is
14x – 8y – 5z = 0
Alternatively, but more painfully, we can use elementary row operations to reducethe matrix
to the matrix
Thus the vector (x, y, z)T is in the range of T if and only if 14x – 8y – 5z = 0.
22. Suppose that v is any vector in V and write
(*) v = a1v1 + a2 v2 + … + anvn
Now define a transformation T V | W by
T(v) = a1w1 + a2w2 + … + anw
n
Note that T is well-definied because, by Theorem 5.4.1, the constants a1, a2, … , an
in (*)are unique.
In order to complete the problem, you must show (i) that T is linear and (ii) that T(v
i) = w
ifor i = 1, 2, … , n.
23. The rank of T is at most 1, since dimR = 1 and the image of T is a subspace of R. So, weknow that either rank(T) = 0 or rank(T) = 1. If rank(T) = 0, then every matrix A is in thekernel of T, so every n × n matrix A has diagonal entries that sum to zero. This is clearlyfalse, so we must have that rank(T) = 1. Thus, by the Dimension Theorem (Theorem 8.2.3),dim (kerT) = n2 – 1.
1 0 1 4 3 5
0 1 1 3 5
0 0 0 14 8 5
− +( )−( )
− −
x y
x y
x y z
1 3 4
3 4 7
2 2 0
x
y
z−
Exercise Set 8.2 385
24. (a) Suppose that v1, v2, … , vn is a basis for V and let
(*) v = a1v1 + a2v2 + … + a
nv
n
be an arbitrary vector in V. Since T is linear, we have
(**) T(v) = a1T(v1) + … + anT(v
n)
The hypothesis that dim(ker(T)) = 0 implies that T(v) = 0 if and only if v = 0. Sincev1, … , v
n is a linearly independent set, by (*), v = 0 if and only if
a1 = a2 = … = an
= 0
Thus, by (**), T(v) = 0 if and only if
a1 = a2 = … = an
= 0
That is, the vectors T(v1), T(v2), … , T(vn) form a linearly independent set in R(T).
Since they also span R(T), then dim(R(T)) = n.
26. If J(p) -2b, then the kernel of J is the set of all
p(x) = ax + b for which b = 0 except the y-axis.
27. If f(x)is in the kernel of D D, then f′′(x) = 0 or f(x) = ax + b. Since these are the onlyeligible functions f(x)for which f′′(x) = 0 (Why?), the kernel of D D is the set of allfunctions f(x) = ax + b, or all straight lines in the plane. Similarly, the kernel of D D D
is the set of all functions f(x) = ax2 + bx + c, or all straight lines except the y-axis andcertain parabolas in the plane.
28. (a) kernel, range
(b) (1, 1, 1)
(c) the dimension of V
(d) 3
29. (a) Since the range of T has dimension 3 minus the nullity of T, then the range of T hasdimension 2. Therefore it is a plane through the origin.
( ) ( / )ax b dx ax bx+ = + − −
∫ 21
1
1
1
2
386 Exercise Set 8.2
(b) As in Part (a), if the range of T has dimension 2, then the kernel must have dimension1. Hence, it is a line through the origin.
30. (a) If T(f(x)) = f′′′′(x), then T(ax3 + bx2 + cx + d) = 0 for all polynomials in P3. Moreover,these are the only functions which are mapped to the zero vector, that is, the functionwhich is identically zero.
(b) If T transforms f(x) to its (n + 1)st derivative, then kernel (T) = Pn.
Exercise Set 8.2 387
EXERCISE SET 8.3
1. (a) Clearly ker(T) = (0, 0), so T is one-to-one.
(c) Since T(x, y) = (0, 0) if and only if x = y and x = –y, the kernel is (0, 0) and T is one-to-one.
(e) Here T(x, y) = (0, 0, 0) if and only if x and y satisfy the equations x – y = 0, –x + y= 0, and 2x – 2y = 0. That is, (x, y) is in ker(T) if and only if x = y, so the kernel ofT is this line and T is not one-to-one.
2. (a) Since det(A) = 1 ≠ 0, then T has an inverse. By direct calculation
and so
3. (a) Since det(A) = 0, or equivalently, rank(A) < 3, T has no inverse.
Tx
xA
x
x
x x
x
− −
=
=
−− +
1 1
2
1 1
2
1 2
1
2
2 55 2x
A− =
−−
1 1 2
2 5
389
3. (c) Since A is invertible, we have
4. (a) Since the kernel of this transformation is the line x = 2y, the transformation is not one-to-one.
(c) Since the kernel of the transformation is (0, 0), the transformation is one-to-one.
5. (a) The kernel of T is the line y = –x since all points on this line (and only those points)map to the origin.
(b) Since the kernel is not (0, 0), the transformation is not one-to-one.
7. (b) Since nullity(T) = n – rank(T) = 1, T is not one-to-one.
(c) Here T cannot be one-to-one since rank(T) ≤ n < m, so nullity(T) ≥ 1.
8. (b) If p(x) = a0 + a1x + a2x2, then
T(p(x)) = a0 + a1 (x + 1) + a2(x + 1)2
= (a0 + a1 + a2) + (a1 + 2a2)x + a2x2
Thus T(p(x)) = 0 if and only if a2 = a1 + 2 a2 = a0 + a1 + a2 = 0, or a2 = a1 = a0 = 0.That is, the nullity of T is zero so T is one-to-one.
T
x
x
x
A
x
x
x
− −
=
=11
2
3
11
2
3
1
2
1
2
1
21
2
1
2
1
21
2
1
2
1
2
−
−
−
x
x
x
1
2
3
=
− +1
2 1 2x x x33
1 2 3
1 2 3
1
21
2
( )
− + +( )
+ −( )
x x x
x x x
390 Exercise Set 8.3
10. (a) Since infinitely many vectors (0, 0, … , 0, xn) map to the zero vector, T is not one-to-
one.
(c) Here T is one-to-one and T–1 (x1, x2, … , xn) = (x
n, x1, … , x
n–1).
11. (a) We know that T will have an inverse if and only if its kernel is the zero vector, whichmeans if and only if none of the numbers a
i= 0.
12. Note that T1 and T2 can be represented by the matrices
that is,
(a) Since det(AT1) = –2 ≠ 0 and det(AT2
) = –5 ≠ 0, it follows that both T1 and T2 are one-to-one.
(b) The transformations T1–1 and T2
–1 are represented by AT1–1 and AT2
–1, respectively, where
Then T2 T1 can be represented by
and (T2 T1)–1 can be represented by
A AT T2 1
1 1
10
3 1
1 3( ) =
−
−
A AT T2 1
3 1
1 3=
−
A AT T1 2
1 11
2
1 1
1 1
1
5
2 1
1 2− −=
−
=
−and
Tx
yA
x
yT
x
yA
x
yT T1 21 2
=
=and
A AT T1 2
1 1
1 1
2 1
1 2=
−
=
−
and
Exercise Set 8.3 391
that is
12. (c) It is easy to verify that
AT1–1 AT2
–1 = (AT2AT1
)–1
Alternative Solution: If you want to do this the hard way, notice that if T1(x, y) =
(x + y, x – y), then T2–1(x, y) = and if T2(x, y) = (2x + y, x – 2y)
then T2–1(x, y) = . Also T2 T1(x, y) = T2(x + y, x – y) =
(3x + y, –x + 3y) and (T1 T2)–1(x, y) = .
13. (a) By inspection, T1–1(p(x)) = p(x)/x, where p(x) must, of course, be in the range of T1
and hence have constant term zero. Similarly T2–1(p(x)) = p(x – 1), where, again, p(x)
must be in the range of T2. Therefore (T2 T1)–1(p(x)) = p(x – 1)/x for appropriate
polynomials p(x).
16. If there were such a transformation T, it would have nullity zero (because it would be one-to-one) and rank equal to the dimension of W (because its range would be W). But theDimension Theorem guarantees that rank(T) + nullity(T) = dim(V), which is, byhypothesis, greater than dim(V). Thus there can be no such transformation.
17. (a) Since T sends the nonzero matrix to the zero matrix, it is not one-to-one.
(c) Since T sends only the zero matrix to the zero matrix, it is one-to-one. By inspection,T–1(A) = T(A).
Alternative Solution: T can be represented by the matrix
0 1
0 0
3
10
3
10
x y x y− +
,
2
5
2
5
x y x y+ −
,
x y x y+ −
2 2
,
( )T Tx
y
x
y2 1
1 1
10
3 1
1 3 −
=
−
=
−+
1
10
3
3
x y
x y
392 Exercise Set 8.3
By direct calculation, TB
= (TB)–1, so T = T–1.
18. If T(x, y) = (x + ky, –y) = (0, 0), then x = y = 0. Hence ker(T) = (0, 0) and therefore Tis one-to-one. Since T(T(x, y)) = T(x + ky, –y) = (x + ky + k(–y), –(–y)) = (x, y), we haveT–1 = T.
19. Suppose that w1 and w2 are in R(T). We must show that
T–1(w1 + w2) = T–1(w1) + T–1(w2)
and
T–1(kw1) = kT–1(w1)
Because T is one-to-one, the above equalities will hold if and only if the results of applyingT to both sides are indeed valid equalities. This follows immediately from the linearity of thetransformation T.
21. It is easy to show that T is linear. However, T is not one-to-one, since, for instance, it sendsthe function f(x) = x – 5 to the zero vector.
24. (a) False. Since T is not one-to-one, it has no inverse. For T–1 to exist, T must map eachpoint in its domain to a unique point in its range.
(b) True. If T1(v1) = T1(v2) where v1 ≠ v2, then T2 T1(v1) = T2
T1(v2) where v1 ≠ v2.
(c) True, since such rotations and reflections are one-to-one linear transformations.
25. Yes. The transformation is linear and only (0, 0, 0) maps to the zero polynomial. Clearlydistinct triples in R3 map to distinct polynomials in P2.
TB =−
−
0 0 0 1
0 1 0 0
0 0 1 0
1 0 0 0
Exercise Set 8.3 393
26. It does. T is a linear operator on M22. If EA = O for any matrix A, then A = E–1O = O. ThusT is one-to-one. Alternatively, if EA1 = EA2, then E(A, –A2) = O, so A1 = A2 = E–1O = O. Thatis, A1 = A2.
27. No. T is a linear operator by Theorem 3.4.2. However, it is not one-to-one since T(a) = a× a = 0 = T(0). That is, T maps a to the zero vector, so if T is one-to-one, a must be thezero vector. But then T would be the zero transformation, which is certainly not one-to-one.
394 Exercise Set 8.3
EXERCISE SET 8.4
2. (a) Let B = 1, x, x2 = u1, u2, u3 and B′ = 1, x = v1, v2. Observe that
T(u1) = T(1) = 1 = 1v1
T(u2) = T(x) = 1 – 2x = 1v1 – 2v2
T(u3) = T(x2) = – 3x = –3v2
Hence the matrix of T with respect to B and B′ is
(b) We have
and
[T(c0 + c1x + c2x2)]
B′ = [(c0 + c1) – (2c1 + 3c2)x]B′
=+
− −
c c
c c
0 1
1 22 3
[ ] [ ]',T
c
c
c
B B Bx =− −
=1 1 0
0 2 3
0
1
2
c c
c c
0 1
1 22 3
+− −
1 1 0
0 2 3− −
395
4. (a) Since
and
we have
(b) Since
and
Formula (5a) does, indeed, hold.
6. (a) Observe that
T(v1) = (1, –1, 0) = v1 – v2
T
T
v v v v
v
2 1 2 3
3
1 1 13
2
1
2
1
2
0 0 1
( ) = − −( ) = − + +
( ) = (
, ,
, , )) = + −1
2
1
2
1
21 2 3v v v
Tx x
x xx x x
B
B
x( )[ ] =−+
= +( ) +1 2
1 21 2 1 2 22uu uu
[ ] [ ]Tx
xB B
B
x =−
=−
2 1
2 0
2 1
2 0
1
2
−( ) + +( )
=−
x xB1 2 2 1 2
2 1
2 0
u u u
−
=
+
x
x x
x x
x
2
2 1
1 2
22
[ ]T B =−
2 1
2 0
T( )u2 11
1=
−−
= −uu
T( )uu uu uu1 1 20
22 2=
= +
396 Exercise Set 8.4
Thus the matrix of T with respect to B is
(b) If we solve for (x1, x2, x3)B= a1 v1 + a2v2 + a3v3, we find that
Thus we have
and
[T(x)]B
= [(x1 – x2, x2 – x1, x1 – x3)]B
=
− − − + −
− − + − +
( ) ( ) ( )
( ) ( )
x x x x x x
x x x x
1 2 2 1 1 3
1 2 2 1
2(( )
( ) ( ) ( )
x x
x x x x x x
1 3
1 2 2 1 1 3
2
2
−
− + − − −
=
− −
− + −
− +
3 2
22
2
2
1 2 3
1 2 3
1 3
x x x
x x x
x x
[ ] [ ]T B Bx =
−
−
−
13
2
1
2
11
2
1
2
01
2
1
2
− +
− + +
+ −
x x x
x x x
x x x
1 2 3
1 2 3
1 2
2
2
33
1 2 3
1 2 3
2
3 2
22
=
− +
− + +
x x x
x x x
22
21 3x x−
x =− + − + + − +
B
x x x x x x x x x1 2 3 1 2 3 1 2 3
2 2 2
T
13
2
1
2
11
2
1
2
01
2
1
2
−
−
−
Exercise Set 8.4 397
6. (c) No. The transformation T is not invertible because x1 – x2 = –(x2 – x1) and thusrank(T) = 2 ≠ 3. We can see this because the reduced row-echelon form of the matrix[T]
Bof the transformation is
8. (a) Since
T(1) = x
T(x) = x(x – 3) = –3x + x2
T(x2) = x(x – 3)2 = 9x – 6x2 + x3
we have
(b) Since we are working with standard bases, we have
so that T(1 + x – x2) = –11x + 7x2 – x3.
(c) We have T(1 + x – x2) = x(1 + (x – 3) – (x – 3)2) = –11x + 7x2 – x3.
9. (a) Since A is the matrix of T with respect to B, then we know that the first and secondcolumns of A must be [T(v1)]
Band [T(v2)]
B, respectively. That is
T
T
B
B
( )
( )
v
v
1
1
1
2
3
5
=−
=
[ ] [ ]',T B B Bx =−
−
0 0 0
1 3 9
0 1 6
0 0 1
1
1
1
0
11
7
1−
=−
−
[ ] ,T B B′ =−
−
0 0 0
1 3 9
0 1 6
0 0 1
1 0 1
0 1 1
0 0 0
−−
≠ .I
398 Exercise Set 8.4
Alternatively, since v1 = 1v1 + 0v2 and v2 = 0v1 + 1v2, we have
and
(b) From Part (a),
and
(c) Since we already know T(v1) and T(v2), all we have to do is express [x1 x2]T in terms
of v1 and v2. If
then
x1 = a – b
x2 = 3a + 4b
or
a = (4x1 + x2)/7
b = (–3x1 + x2)/7
Thus
x
x
x x1
2
1 24
7
3
5
3
=
+−
+
− xx x
x x
x x
1 2
1 2
1 2
7
2
29
18
7107 24
7
+ −
=
+
− +
x
xa b a b
1
21 2
1
3
1
4
= + =
+
−
v v
T( )v v v2 1 23 52
29= + =
−
T( )v v v1 1 223
5= − =
−
( )T AB
v20
1
3
5 =
=
T AB
( )v1 =
=
−
1
0
1
2
Exercise Set 8.4 399
9. (d) By the above formula,
11. (a) The columns of A, by definition, are [T(v1)]B, [T(v2)]
B, and [T(v3)]
B, respectively.
(b) From Part (a),
T(v1) = v1 + 2v2 + 6v3 = 16 + 51x + 19x2
T(v2) = 3v1 –2v3 = –6 – 5x + 5x2
T(v3) = –v1 + 5v2 + 4v3 = 7 + 40x + 15x2
(c) Let a0 + a1x + a2x2 = b0v1 + b1v2 + b2v3. Then
a0 = – b1 + 3b2
a1 = 3b0 + 3b1 + 7b2
a2 = 3b0 + 2b1 + 2b2
This system of equations has the solution
b0 = (a0 – a1 + 2a2)/3
b1 = (–5a0 + 3a1 – 3a2)/8
b2 = (a0 + a1 – a2)/8
Thus
T(a0 + a1x + a2x2) = b0T(v1) + b1T(v2) + b2T(v3)
=− +
+− +
+
239 161 247
24
201 111 247
8
0 1 2
0 1 2
a a a
a a ax
61 31 107
120 1 2 2a a a
x− +
T1
1
19 7
83 7
=
−
400 Exercise Set 8.4
(d) By the above formula,
T(1 + x2) = 2 + 56x + 14x2
13. (a) Since
T1(1) =2 and T1(x) = –3x2
T2(1) = 3x T2(x) = 3x2 and T2(x2) = 3x3
T2 T1(1) = 6x and T2 T1(x) = –9x3
we have
and
(b) We observe that here
[T2 T1]B′,B = [T2]B′,B′′ [T1]B′′,B
T TB2 1
0 0
6 0
0 0
0 9
ο′, =
−
Β
[ ] [ ], ,T TB B B B1 2
2 0
0 0
0 3
0 0 0
3 0′′ ′ ′′=
−
=00
0 3 0
0 0 3
Exercise Set 8.4 401
15. If T is a contraction or a dilation of V, then T maps any basis B = v1, … , vn of V to
kv1, … , kvn where k is a nonzero constant. Therefore the matrix of T with respect to B is
17. The standard matrix for T is just the m × n matrix whose columns are the transforms of thestandard basis vectors. But since B is indeed the standard basis for Rn, the matrices are thesame. Moreover, since B′ is the standard basis for Rm, the resulting transformation will yieldvector components relative to the standard basis, rather than to some other basis.
18. (a) Since D(1) = 0, D(x) = 1, and D(x2) = 2x, then
is the matrix of D with respect to B.
(b) Since D (2) = 0, D(2 – 3x) = –3 = – 3–2
p1 and D(2 – 3x + 8x2) = – 23—6
p1 – 16—6
p2, the
matrix of D with respect to B is
03
2
23
6
0 016
3
0 0 0
−
−
0 1 0
0 0 2
0 0 0
k
k
k
k
0 0 0
0 0 0
0 0 0
0 0 0
402 Exercise Set 8.4
(c) Using the matrix of Part (a), we obtain
(d) Since 6 – 6x + 24x2 = p1 – p2 + 3p3, we have
or
D(6 – 6x + 24x2) = 13(2) – 16(2 – 3x) = –6 + 48x
19. (c) Since D(f1) = 2f1, D(f2) = f1 + 2f2, and D(f3) = 2f2 + 2f3, we have the matrix
20. The upper left-hand corner is all vectors in the space V. The upper right-hand corner is allvectors in the range of T. Thus it is a subspace of W. Since V is a real vector space, thelower left-hand corner is all of R4. The lower right-hand corner is the range of thetransformation obtained by multiplying every element of R4 by the matrix [T]
B′,B. It is thusa subspace of R7.
2 1 0
0 2 2
0 0 2
D x xB
( )6 6 24
03
2
23
6
0 016
30 0 0
2− +
=
−
−
−
= −1
1
3
13
166
0
B
D x x( )6 6 24
0 1 0
0 0 2
0 0 0
6
6
24
2− + =
−
=−
= − +6
48
0
6 48x
Exercise Set 8.4 403
EXERCISE SET 8.5
1. First, we find the matrix of T with respect to B. Since
and
then
In order to find P, we note that v1 = 2u1 + u2 and v2 = –3u1 + 4u2. Hence the transitionmatrix from B′ to B is
Thus
P− =
1
4
11
3
111
11
2
11
P =−
2 3
1 4
A TB
= =−−
1 2
0 1
T( )u2
12 =−−
T( )u11
0=
405
and therefore
2. In order to compute A = [T]B, we note that
and
Hence
In order to find P, we note that
v1 = (1.3)u1 + (–0.4)u2
v2 = (–0.5)u1
Hence
P =−
−
. .
.
1 3 0 5
0 4 0
A TB
= =−
0 8 6 1
3 6 3 8
. .
. .
A TB
= =−
= +
3
166 1 3 81 2( . ) ( . )u u
T( ) ( . ) ( . )u u u1 1 216
20 8 3 6=
−
= +
′ = = =−
−−′
−A T P T PB B[ ] [ ]1 1
11
4 3
1 2
1 2
0 1
−
=− −
−
2 3
1 4
3
11
56
112
11
3
11
406 Exercise Set 8.5
and
It then follows that
3. Since T(u1) = (1/ 2, 1/ 2) and T(u2) = (–1/ 2, 1/ 2), then the matrix of T withrespect to B is
From Exercise 1, we know that
Thus
5. Since T(e1) = (1, 0, 0), T(e2) = (0, 1, 0), and T(e3) = (0, 0, 0), we have
In order to compute P, we note that v1 = e1, v2 = e1 + e2, and v3 = e1 + e2 + e3. Hence,
A TB
= =
1 0 0
0 1 0
0 0 0
A T P APB
′ = = =−
−'
1 1
11 2
13 25
5 9
P P=−
=
−
−2 3
1 4
1
11
4 3
1 21and
A TB
= =−
1 2 1 2
1 2 1 2
A T P APB
′ ′= = =−−
−1 15 5 4 5
37 5 12 5
. .
. .
P− =
−− −
1 0 2 5
2 6 5
.
.
Exercise Set 8.5 407
and
Thus
7. Since
and
we have
We note that q p p and q p p Hence1 1 2 2 1 22
9
1
3
7
9
1
6= − + = − .
TB
=−
2
3
2
91
2
4
3
T x( )p p p2 1 212 22
9
4
3= + = − +
T x( )p p p1 1 29 32
3
1
2= + = +
TB
=−
−
′
1 1 0
0 1 1
0 0 1
1 0 0
0 11 0
0 0 0
1 1 1
0 1 1
0 0 1
1 0 0
0
= 11 1
0 0 0
P− =
−−
11 1 0
0 1 1
0 0 1
P =
1 1 1
0 1 1
0 0 1
408 Exercise Set 8.5
and
Therefore
8. (a) Since T(1, 0) = (3, –1) and T(0, 1) = (–4, 7), we have
(c) Since T(1) = 1, T(x) = x – 1, and T(x2) = (x – 1)2, we have
9. (a) If A and C are similar n × n matrices, then there exists an invertible n × n matrix Psuch that A = P–1CP. We can interpret P as being the transition matrix from a basis B′for Rn to a basis B. Moreover, C induces a linear transformation T Rn → Rn where C= [T]
B. Hence A = [T]
B′. Thus A and C are matrices for the same transformation withrespect to different bases. But from Theorem 8.2.2, we know that the rank of T isequal to the rank of C and hence to the rank of A.
det( )T =−
− =1 1 1
0 1 2
0 0 1
1
det( )T =−
−=
3 4
1 717
TB
=
−
′
3
4
7
23
21
2
3
2
91
2
4
3
−
−
=
2
9
7
91
3
1
6
1 1
0 1
P− =
1
3
4
7
23
21
P =−
−
2
9
7
91
3
1
6
Exercise Set 8.5 409
Alternate Solution: We observe that if P is an invertible n × n matrix, then P2represents a linear transformation of Rn onto Rn. Thus the rank of the transformationrepresented by the matrix CP is the same as that of C. Since P–1 is also invertible, itsnull space contains only the zero vector, and hence the rank of the transformationrepresented by the matrix P–1 CP is also the same as that of C. Thus the ranks of Aand C are equal. Again we use the result of Theorem 8.2.2 to equate the rank of alinear transformation with the rank of a matrix which represents it.
Second Alternative: Since the assertion that similar matrices have the same rank dealsonly with matrices and not with transformations, we outline a proof which involvesonly matrices. If A = P–1 CP, then P–1 and P can be expressed as products ofelementary matrices. But multiplication of the matrix C by an elementary matrix isequivalent to performing an elementary row or column operation on C. From Section5.5, we know that such operations do not change the rank of C. Thus A and C musthave the same rank.
10. (a) We use the standard basis for P4. Since T(1) = 1, T(x) = 2x + 1, T(x2) = (2x + 1)2,T(x3) = (2x + 1)3, and T(x4) = (2x + 1)4, we have
(b) Thus rank(T) = 5, or [T] is invertible, so, by Theorem 8.4.3, T is one-to-one.
11. (a) The matrix for T relative to the standard basis B is
The eigenvalues of [T]B
are λ = 2 and λ = 3, while corresponding eigenvectors are(1, –1) and (1, –2), respectively. If we let
P =− −
=
− −
1 1
1 2
2 1
1 1then P-1
TB
=−
1 1
2 4
det( )T =
1 1 1 1 1
0 2 4 6 8
0 0 4 12 24
0 0 0 8 32
0 0 0 0 16
1024=
410 Exercise Set 8.5
and
is diagonal. Since P represents the transition matrix from the basis B′ to the standardbasis B, we have
as a basis which produces a diagonal matrix for [T]B′.
12. (b) The matrix for T relative to the standard basis B is
The eigenvalues of [T]B
are λ = 1 and λ = –2. The eigenspace corresponding to λ = 1
is spanned by the vectors and and the eigenspace correspond to λ = –2
is spanned by the vector . If we let
then P–1[T]BP will be diagonal. Hence the basis
will produce a diagonal matrix for [T]B′.
B′ =−
1
1
0
1
0
1
1
, , 11
1−
P =−
−
1 1 1
1 0 1
0 1 1
1
1
1−
1
0
1
−
1
1
0
TB
=−
−
0 1 1
1 0 1
1 1 0
B′ = ,1
1
1
2−
−
P T PB− =
1 2 0
0 3[ ]
Exercise Set 8.5 411
13. (a) The matrix of T with respect to the standard basis for P2 is
The characteristic equation of A is
λ3 – 2λ2 – 15λ + 36 = (λ – 3)2(λ + 4) = 0
and the eigenvalues are therefore λ = –4 and λ = 3.
(b) If we set λ = –4, then (λI – A)x = 0 becomes
The augmented matrix reduces to
and hence x1 = –2s, x2 = 8–3s, and x3 = s. Therefore the vector
is a basis for the eigenspace associated with λ = –4. In P2, this vector represents thepolynomial –2 + 8–
3x + x2.
If we set λ = 3 and carry out the above procedure, we find that x1 = 5 s, x2 = –2s,and x3 = s. Thus the polynomial 5 – 2x + x2 is a basis for the eigenspace associatedwith λ = 3.
−
2
8 3
1
1 0 2 0
0 1 8 3 0
0 0 0 0
−
− − −−
− −
9 6 2
0 3 8
1 0 2
1
2
3
x
x
x
=
0
0
0
A = − −−
5 6 2
0 1 8
1 0 2
412 Exercise Set 8.5
14. (a) We look for values of λ such that
or
If we equate corresponding entries, we find that
λa –2c = 0
–a + λb – c = 0(*)
– b + (λ + 2)c = 0
(λ – 1)d= 0
This system of equations has a nontrivial solution for a, b, c, and d only if
or
(λ – 1)(λ + 2)(λ2 – 1) = 0
Therefore the eigenvalues are λ = 1, λ = –2, and λ = –1.
(b) We find a basis only for the eigenspace of T associated with λ = 1. The basesassociated with the other eigenvalues are found in a similar way. If λ = 1, the equationT(x) = λx becomes T(x) = x and the augmented matrix for the system of equations(*) above is
1 0 2 0 0
1 1 1 0 0
0 1 3 0 0
0 0 0 0 0
−− −
−
det
λλ
λλ
0 2 0
1 1 0
0 1 2 0
0 0 0 1
−− −
− +−
= 0
2
2
c a c
b c d
a b
c d
+−
=
λ λλ λ
Ta b
c d
a b
c d
=
λ
Exercise Set 8.5 413
This reduces to
and hence a = 2t, b = 3t, c = t, and d = s. Therefore the matrices
form a basis for the eigenspace associated with λ = 1.
15. If v is an eigenvector of T corresponding to λ, then v is a nonzero vector which satisfiesthe equation T(v) = λv or (λI – T)v = 0. Thus λI – T maps v to 0, or v is in the kernel ofλI – T.
17. Since C[x]B
= D[x]B
for all x in V, we can, in particular, let x = vi
for each of the basisvectors v1, … , v
nof V. Since [v
i]B
= e3i
for each i where e1, … , e
n is the standard basis
for Rn, this yields Cei
= Dei
for i = 1, … , n. But Cei
and Dei
are just the ith columns of Cand D, respectively. Since corresponding columns of C and D are all equal, we have C = D.
18. Let B be the standard basis for R2, and let
B′ = (cos θ, sin θ), (–sin θ, cos θ) = v1, v2
be the basis consisting of the unit vector v1 lying along and the unit vector v2perpendicular to . Note that to change basis from B to B′, we rotate through an angle –θ.Now relative to B′, T is just the orthogonal projection on the v1-axis, which is accomplishedby setting the v2-coordinate equal to zero. Then to return to the basis B, we rotate throughan angle θ. Thus
2 3
1 0
0 0
0 1
and
1 0 2 0 0
0 1 3 0 0
0 0 0 0 0
0 0 0 0 0
−−
414 Exercise Set 8.5
19. (a) False. Every matrix is similar to itself, since A = I–1 AI.
(b) True. Suppose that A = P–1 BP and B = Q–1 CQ. Then
A = P–1(Q–1 CQ)P = (P–1Q–1)C(QP) = (QP)–1C(QP)
Therefore A and C are similar.
(c) True. By Table 1, A is invertible if and only if B is invertible, which guarantees that Ais singular if and only if B is singular.
Alternatively, if A = P–1 BP, then B = PAP–1. Thus, if B is singular, then so is A.Otherwise, B would be the product of 3 invertible matrices.
(d) True. If A = P–1 BP, then A–1(P–1 BP)–1 = P–1 B–1(P–1)–1 = P–1B–1P, so A–1 and B–1 aresimilar.
20. If A is invertible and B is singular, then it follows from Exercise 19(c)that A and B cannotbe similar. Hence
are not similar.
Moreover, if A and B are both invertible but have different determinants, then it followsfrom Table 1 that they cannot be similar. Hence.
are not similar.
A B=
=
1 0
0 1
2 0
0 1and
A B=
=
1 0
0 1
1 0
0 0and
Tx
yT T I
x
yB B B B B
=
′ ′ ′[ ] [ ] [ ], ,
=−
cos sin
sin cos
coθ θθ θ
1 0
0 0
ss sin
sin cos
cos sin c
θ θθ θ
θ θ
−
=
x
y
2 oos
sin cos sin
θ
θ θ θ2
x
y
Exercise Set 8.5 415
22. Let B = P–1 AP so that A = PBP–1. Let λ be an eigenvalue common to A and B and let x bean eigenvector of A corresponding to λ. Then Ax = λx, or
Ax = (PBP–1)x = λx
so that
BP–1x = P–1 λx
Thus
B(P–1x) = λ(P–1x)
That is, P–1x is an eigenvector of B corresponding to λ.
24. For every f in V, T(f) = f(x – 2π). Thus, if f is an eigenvector with eigenvalue λ = 1, wehave T(f) = 1 • f, so f(x – 2π) = f(x) for all x. Thus, the eigenspace of λ = 1 is precisely theset of all functions f that are periodic with periods that evenly divide 2π.
25. First, we need to prove that for any square matrices A and B, the trace satisfies tr(A) =tr(B). Let
Then,
[ ]AB a b a b a b a b an n j
j
11 11 11 12 21 13 31 1 1 11
= + + + =+=
nn
j
n n
b
AB a b a b a b a b
∑
= + + + =+
1
22 21 21 22 22 23 32 2 2[ ] aa b
AB a b a b a b
j
j
n
j
nn n n n n n n
21
2
1 1 2 2 3 3
=
+
∑
= + + +[ ] aa b a bnn nn nj
j
n
jn==∑
1
.
A
a a a
a a a
a a a
n
n
n n nn
=
11 12 1
21 22 2
1 2
=and
b11
B
b b
b b b
b b
n
n
n n
12 1
21 22 2
1 2 bnn
416 Exercise Set 8.5
Thus,
tr(AB) = [AB]11 + [AB]22 + … + [AB]nn
Reversing the order of summation and the order of multiplication, we have
Now, we show that the trace is a similarity invariant. Let B = P–1 AP. Then
tr(B) = tr(P–1 AP)
= tr((P–1 A)P)
= tr(P(P–1 A))
= tr(PP–1)A)
= tr(I A)
= tr(A).
tr( )AB b a
b a b
jk
k
n
jk
j
n
k
k
n
k k
k
n
=
= +
==
= =
∑∑
∑
11
11
1 21
∑∑ ∑+ +
= + + +=
=a b a
BA BA BA
k nk
k
n
kn
nn
21
11 22[ ] [ ] [ ]
ttr BA( ).
= + + +
=
= = =∑ ∑ ∑a b a b a b
a
j
j
n
j j j
j
n
nj jn
j
n
kj
j
21
1 2 21 1
===∑∑
11
n
kj
k
n
b .
Exercise Set 8.5 417
EXERCISE SET 8.6
1. (a) This transformation is onto because for any ordered pair (a, b) in R2, T(b, a) = (a, b).
(b) We use a counterexample to show that this transformation is not onto. Since there isno pair (x, y) that satisfies T(x, y) = (1, 0), T is not onto.
(c) This transformation is onto. For any ordered pair (a, b) in R2, T = (a, b).
(d) This is not an onto transformation. For example, there is nopair (x, y) that satisfies T(x, y) = (1, 1, 0).
(e) The image of this transformation is all vectors in R3 of the form (a, –a, 2a). Thus, theimage of T is a one-dimensional subspace of R3 and cannot be all of R3. In particular,there is no vector (x, y) that satisfies T(x, y) = (1, 1, 0), and this transformation is notonto.
(f) This is an onto transformation. For any point (a, b) in R2, there are an infinite number
of points that map to it. One such example is T = (a, b).
2. The transformation TA
is not onto if the image of TA
is not all of Rn. Thus, rank(TA) < n, or
equivalently, null(T) > 0. This is equivalent to the matrix A being singular.
3. (a) We find that rank(A) = 2, so the image of T is a two-dimensional subspace of R3. Thus,T is not onto.
(b) We find that rank(A) = 3, so the image of T is all of R3. Thus, T is not onto.
(c) We find that rank(A) = 3, so the image of T is all of R3. Thus, T is onto.
(d) We find that rank(A) = 3, so the image of T is all of R3. Thus, T is onto.
4. (a) We find that rank(A) = 1 and the codomain is R3, so this is not onto.
(b) We find that rank(A) = 3 and the codomain is R3, so this is an onto transformation.
(c) We find that rank(A) = 2 and the codomain is R3, so this is not onto.
a b a b+ −
2 2
0, ,
a b a b+ −
2 2
,
419
5. (a) The transformation T is not a bijection because it is not onto. There is no p(x) inP2(x) so that xp(x) = 1.
(b) The transformation T(A) = AT is one-to-one, onto, and linear, so it is a bijection.
(c) By Theorem 8.6.1, there is no bijection between R4 and R3, so T cannot be a bijection. Inparticular, it fails being one-to-one. As an example, T(1, 1, 2, 2) = T(1, 1, 0, 0) = (1, 1, 0).
(d) Because dim P3 = 4 and R3 = 3, Theorem 8.6.1 states that there is no bijection betweenP3 and R3, so T cannot be a bijection. In particular, it fails being one-to-one. As anexample, T(x + x2 + x3) = T(1 + x + x2 + x3) = (1, 1, 1).
6. Let T V → W be a bijection. Then T is one-to-one, onto, and invertible.First, we show that T–1 W → V is one-to-one. If T–1(w1) = T–1(w2), then T(T–1(w1)) =
T(T–1(w2)) so w1 = w2, and T–1 is one-to-one. Second, we show that T–1 is onto. Let v beany vector in V, and let w = T(v). Then T–1(w = T(w) = T–1(T(v)) = v, so v is in the imageof T–1 for any v in V, and T–1 is onto. Thus, the inverse of a bijection is again a bijection.
Now, we want to show that if T V → W is a bijective linear transformation, thenT–1 W → V is also a bijective linear transformation.
Because T is onto, for any w in W, there is a vector v in V so that w = T(v). Then,
T–1(cw) = T–1(cT(v)).
Since T is a linear transformation, cT(v) = T(cv), so
T–1(cw) = T–1(T(cv))
= cv
= cT–1(w).
Also, for any two vectors w1 and w2 in W, there exist vectors v1 and v2 in V so that w1 =T(v1) and w2 = T(v2), since T is onto. Thus,
T–1(w1 + w2) = T–1(v1) + T(v2)).
Since T is a linear transformation, T(v1 + T(v2) = T(v1 + v2), and we have
T–1(w1 + w2) = T–1(v1 + v2))
= v1 + v2
= T–1(w1) + T–1(w2).
Thus, T–1 is a linear transformation.
7. Assume there exists a subjective (onto) linear transformation T V → W, where dim W >dim V. Let m = dim V and n = dim W, with m < n. Then, the matrix A
Tof the
transformation is an n × m matrix, with m < n. The maximal rank of AT
is m, so thedimension of the image of T is at most m. Since the dimension of the image of T is smallerthan the dimension of the codomain Rn, T is not onto. Thus, there cannot be a subjectivetransformation from V onto W if dim V < dim W.
420 Exercise Set 8.6
If n = dim W ≤ dim V = m, then the matrix AT
of the transformation is an n × m matrixwith maximim possible rank n. If rank(A
T) = n, then T is a subjective transformation.
Thus, it is only possible for T V → W to be a subjective linear transformation if dim W≤ dim V.
8. Let V be the vector of all symmetric 3 × 3 matrices, and let T V → R6 be defined by
Clearly, T is one-to-one and onto, so T is a bijection. Because T is the coordinate mapfrom V to R6, it is a linear transformation (refer to Example 7 in Section 8.1). Thus, T is abijective linear transformation.
9. Let T V → Rn be defined by T(v) = (v)S, where S = u1, u2, … , u
n is a basis of V. We
know from Example 7 in Section 8.1 that the coordinate map is a linear transformation.Let (a1, a2, … , a
n) be any point in Rn. Then, for the vector v = a1u1 + a2u2 + … + a
nu
n, we
have
T(v) = T(a1u1 + a2u2 + … + anu
n) = (a1, a2, … , a
n)
so T is onto.
Also, let v1 = a1u1 + a2u2 + … + anu
nand v2 = b1u1 + b2u2 + … + b
nu
n. If T(v1) =
T(v2), then (v1_S= (v2)S
, and thus a1, a2, … , an) = (b1, b2, … , b
n). It follows that a1 = b1,
a2 = b2, … , an
= bn, and thus
v1 = a1u1 + a2u2 + … + anu
n) = v2.
So, T is one-to-one and is thus an isomorphism.
10. T1 and T2 are both bijective linear transformations, we know from Theorem 8.1.2 thatT1
T2 is also a linear transformation. Let T1 V → W and T2 U → V. Because T1 is onto,we know that for any w in W, there exists a v in V so that T1(v) = w. Similarly, because T2is onto, we know that for any v in V, there exists a u so that T2(u) = v. Then
(T1 T2)(u) = T1(T2(u)) = T1(v) = w,
and T1 T2 is onto.
If there are two vectors u1 and u2 so that (T1 T2)(u1) = (T1
T2)(u2), then T1(T2(u1))= T1(T2(u2)). Since T1 is one-to-one, this means that T2(u1) = T2(u2), and since T2 is one-to-one, this means that u1 = u2. Thus, the composition T1
T2 is one-to-one.
Thus, the composition of two bijective linear transformations is again a bijective lineartransformation.
T
a b c
b d e
c e f
a
b
c
d
e
f
=
Exercise Set 8.6 421
11. Let V = Span1, sin x, cos x, sin 2x, cos 2x. Differentiation is a linear transformation (seeExample 11, Section 8.1). In this case, D maps functions in V into other functions in V. Toconstruct the matrix of the linear transformation with respect to the basis B = 1, sin x,cos x, sin 2x, cos 2x, we look at coordinate vectors of the derivatives of the basis vectors:
D(1) = 0 D(sin x) = cos x D(cos x = –sin x D(sin 2x) = 2 cos 2x
D(cos 2x) = 2 sin 2x
The coordinate matrices are:
Thus, the matrix of the transformation is
Then, differentiation of a function in V can be accomplished by matrix multiplication by theformula
[D(f)]B
= SD[f]
B.
The final vector, once transformed back to V from coordinates in R5, will be the desiredderivative.
For example,
Thus, D(3 – 4 sin x + sin 2 x + 5 cos 2x) = –4 cos x – 10 sin 2x + 2 cos 2x.
12. (a) We have
uV
= ⟨u,u⟩ 1/2—V
= ⟨TPu, T(u)⟩1/2—W
= T(u)W
[ ( sin sin cos )]D x x x AB D3 4 2 5 2
3
4
− + + =−
0
1
5
0 0 0 0 0
0 0 1 0
=− 0
0 1 0 0 0
0 0 0 0 2
0 0 0 2 0
−
−
=
3
4
0
1
5
0
.
0
4
10
2
−−
AD =−
−
0 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 0 0 0 2
0 0 00 2 0
[ ( )] [ (sin )]D D xB B1
0
0
0
0
0
0
0
1=
=00
0
0
1
0
0
=−
[ (cos )]D x B
00
2
0
0
0
0
2
=
[ (sin )]D x B
=−
[ (cos )]D x B2
0
0
0
2
0
.
422 Exercise Set 8.6
so an inner product space isomorphism preserves lengths. The angle θV
between uand v in V is given by
and the angle θW
between T(u) and T(v) in W is given by
Thus, because the inner product space isomorphism preserves lengths,
and the inner product space isomorphism also preserves angles.
(b) If B = v1, v2, … , v2 is an orthonormal set in V, then
Thus, the set T(v1, Y(v2), … , T(vn) = is an orthonormal set in W.
However, in general this is not true. Consider the inner transformation T R2 → R2
defined by
This transformation is a change of basis transformation, and is thus one-to-one andonto, but the vectors T(e1) and T(e2) are not orthonormal. In fact, T*e2) = 2, and⟨T(e1), T(e2)⟩ = 1 ≠ 0. Thus, length and angles are not preserved with thistransformation.
(c) The set e1, e2, … , en is an orthonormal basis of W. By the Gram-Schmidt process, we
can construct an orthonormal basis v1, v2, … , vn of V. Define a linear transformation
T V | W by
T(vj) = e
j, j = 1, 2, … , n.
T e T e( ) ( )1 21
0
1
0=
=
and
( ( ), ( )) ( ,T Ti j
i ji j jv v v vif
1 if= =
≠=
10
cos, ,
θWw v
v
T T
T w T w=
( ) ( )( ) ( )
=( )
=u v
u v
u v
u v vcos .θV
cos,
.θWw
T T
T w T w=
( ) ( )( ) ( )
u v
u v
cos,
θ =( )u v
u vv
v v
Exercise Set 8.6 423
Let u = a1v1 + a2v2 + … + anv
nand v = b1v1 + b2v2 + … + b
nv
n. Then
T(u) = a1T(v1) + a2T(v2) + … + anT(v
n)
= a1e1 + a2e2 + … + ane
n
Similarly,
T(v) = b1T(v1) + v2T(v2) + … + bnT(v
n)
= b1e1 + v2e2 + … + bne
n
So, if T(u) = T(v), then and a1 = b1, a2 = b2, … , an
= bn. Thus, u = v
and T is one-to-one.
For any vector in W, we have
T(c1v1 + c2v2 + … + cnv
n) = = w.
so T is onto.
c
c
cn
1
2
w =
c
c
cn
1
2
a
a
a
b
b
bn n
1
2
1
2
=
=
b
b
bn
1
2
=
a
b
an
1
2
424 Exercise Set 8.6
We now need to show that T preserves the inner product. We have
⟨u, v⟩V
= ⟨a1v1 + a2v2 + … + anv
n, b1v1 + b2v2 + … + b
nv
n⟩
= ajb
k⟨v
j, v
k⟩.
Recall that if j ≠ k, then ⟨vj, v
k⟩ = 0. We then have
⟨u, v⟩V
= a1b1 ⟨v1, v1⟩ + a2b2 ⟨v2, v2⟩ + … + anb
n⟨v
n, v
n⟩.
Now recall that ⟨vj, v
j⟩ = 1. Then
⟨u, v⟩ = ⟨a1b1 + a2b2 + … + anb
n
We also have
⟨T(u), T(v)⟩W
= ⟨a1e1 + a2e2 + … + ane
n, b1e1 + b2e2 + … + b
ne
n⟩
= a1b1 + a2b2 + … + anb
n.
Thus, ⟨T(u), T(v)⟩W
= ⟨u, v⟩V
and T is an inner product space isomorphism.
(d) Let dim V = dim W = n. Then by Part (c), there is an inner product isomorphism T1
V → Rn and another inner product isomorphism T2 W → Rn. Then T2–1 Rn → W is
an isomorphism by Problem 6. And, since
⟨u, v⟩ = ⟨T2(u), T2(v)⟩Rn,
we rename x = T2(u) and y = T2(v) and we have
⟨T2–1(x), T2
–1(y)⟩W
= ⟨x, y⟩Rn.
Thus, T2–1 Rn → W is another inner product space isomorphism. Then, the
composition (T2–1 T1) V → W is an isomorphism by Problem 10. We have
⟨u, v⟩V
= ⟨T1(u), T1(v(⟩Rn
= ⟨T2–1(T1(u)), T2
–1(T1(v))⟩W
.
= ⟨(T2–1 T
1)(u), (T2
–1 T1)(v)⟩W
.
so T2–1 T
1is an inner product isomorphism from V to W.
=
.
a
a
a
b
b
bn n
1
2
1
2
j
n
j
n
==∑∑
11
Exercise Set 8.6 425
(e) An orthonormal basis for P5 is 1, x, x2, x3, x4, x5, using the inner product
⟨a0 + a1x + a1x2 + … + a5x
5, b0 =+ b1x + b2x2 + … + b5x
5⟩P
5
= a0b
0+ a
1b
1+ … + a
5b
5.
An orthonormal basis for M22 is
using the inner product
⟨A, B⟩M
23= tr(AT B).
We can define a linear transformation T P5 → M23 by
The transformation T is one-to-one and onto. To see that is an inner productisomorphism, we need to compute ⟨u, v⟩
P5and ⟨T(u), T(v)⟩
M23. Let u = a0 + a1x +
a2x2 + … + a5x
5, and let v = b0 + b1x + b2x2 + … + b5x
5. Then we have
⟨u, v⟩P
5= a0b0 + a1b1 + … a5b5
and
Thus, ⟨u, v⟩P
5= ⟨T(u),T(v)⟩
M23
, and T is an inner product space isomorphism.
T Ta a a
a a a
b b b
bM( ), ( ) ,u v
23
0 1 2
3 4 5
0 1 2=
33 4 5
0 0 1 1 3 3
b b
a b a b a b
= + + + + .
T T x T11 0 0
0 0 0
0 1 0
0 0 0( ) =
( ) =
, , xx
T x T
2
3
0 0 1
0 0 0
0 0 0
1 0 0
( ) =
( ) =
,
, xx T x4 50 0 0
0 1 0
0 0 0
0 0 1( ) =
( ) =
, .
1 0 0
0 0 0
0 1 0
0 0 0
0 0 1
0 0 0
0
, , ,
00 0
1 0 0
0 0 0
0 1 0
0 0 0
0 0 1
, ,
,
426 Exercise Set 8.6
SUPPLEMENTARY EXERCISES 8
3. By the properties ofan inner product, we have
T(v + w) = ⟨v + w, v0⟩v0
= (⟨v, v0J + ⟨w, v0⟩)v0
= ⟨v, v0⟩v0 + ⟨w, v0⟩v0
= T(v) + T(w)
and
T(kv) = ⟨kv, v0⟩v0 = k⟨v, v0⟩v0 = kT(v)
Thus T is a linear operator on V.
4. (a) By direct computation, we have
T(x + y) = ((x + y) • v1, … , (x + y) • vm
)
= (x • v1 + y • v1, … , x • vm
+ y • vm
)
= (x • v1, … , x • vm
) + (y • v1, … , y • vm
)
= T(x) + T(y)
and
T(kx) = ((kx) • v1, … , (kx) • vm
)
= (k(x • v1), … , k(x • vm
))
= kT(x)
427
4. (b) If e1, … , en is the standard basis for Rn, and if v
i= (a1i
, a2i, … , a
ni) for i = 1, … , m,
then
T(ej) = (a
j1, aj2, … , ajm
)
But T(ej), interpreted as a column vector, is just the jth column of the standard matrix
for T. Thus the ith row of this matrix is (a1i, a2i
, … , ani
), which is just vi.
5. (a) The matrix for T with respect to the standard basis is
We first look for a basis for the range of T; that is, for the space of vectors B such thatAx = b. If we solve the system ofequations
x + z + w = b1
2x + y + 3x + w = b2
x + w = b3
we find that = z = b1 – b3 and that any one of x, y, or w will determine the other two.Thus, T(e3) and any two of the remaining three columns of A is a basis for R(T).
Alternate Solution :We can use the method of Section 5.5 to find a basis for thecolumn space of A by reducing AT to row-echelon form. This yields
so that the three vectors
form a basis for the column space of T and hence for its range.
1
2
1
0
1
0
0
0
0
1 1
0 1 0
0 0 1
0 0 0
2
A =
1 0 1 1
2 1 3 1
1 0 0 1
428 Supplementary Exercises 8
Second Alternative: Note that since rank(A) = 3, then R(T) is a 3-dimensionalsubspace of R3 and hence is all of R3. Thus the standard basis for R3 is also a basis forR(T).
To find a basis for the kernel of T, we consider the solution space of Ax = 0. If we setb1 = b2 = b3 = 0 in the above system of equations, we find that z = 0, x = –w, and y =w. Thus the vector (–1, 1, 0, 1) forms a basis for the kernel.
7. (a) We know that T can be thought of as multiplication by the matrix
where reduction to row-echelon form easily shows that rank([T]B) = 2. Therefore the
rank of T is 2 and the nullity of T is 4 – 2 = 2.
(b) Since [T]B
is not invertible, T is not one-to-one.
9. (a) If A = P–1 BP, then
AT = (P–1 BP)T
= PT BT (P–1)T
= ((PT)–1)–1BT (P–1)T
= ((P–1)T)–1B–1 (PT)T
Therefore AT and BT are similar. You should verify that if P is invertible, then so is PT
and that (PT)–1 = (P–1)T.
10. If statement (i) holds, then the range of T is V and hence rank(T) = n. Therefore, by theDimension Theorem, the nullity of T is zero, so that statement (ii) cannot hold. Thus (i)and (ii) cannot hold simultaneously.
Now if statement (i) does not hold, then the range of T is a proper subspace of V andrank(T) < n. The Dimension Theorem then implies that the nullity of T is greater thanzero. Thus statement (ii) must hold. Hence exactly one of the two statements must alwayshold.
[ ]T B =
−− −
−−
1 1 2 2
1 1 4 6
1 2 5 6
3 2 3 2
Supplementary Exercises 8 429
11. If we let then we have
The matrix X is in the kernel of T if and only if T(X) = 0, i.e., if and only if
a + b + c = 0
2b + d = 0
d = 0
Hence
The space of all such matrices X is spanned by the matrix and therefore has
dimension 1. Thus the nullity is 1. Since the dimension of M22 is 4, the rank of T musttherefore be 3.
Alternate Solution. Using the computations done above, we have that the matrix for thistransformation with respect to the standard basis in M22 is
Since this matrix has rank 3, the rank of T is 3, and therefore the nullity must be 1.
1 1 1 0
0 2 0 1
0 0 0 1
0 0 0 1
,1 0
1 0−
Xa
a=
−
0
0
Ta b
c d
a c b d b b
d d
=
+ +
+
0 0
=+ + +
a b c b d
d d
2
Xa b
c d=
,
430 Supplementary Exercises 8
12. We are given that there exist invertible matrices P and Q such that A = P–1 BP and B = Q–1
CQ. Therefore
A = P–1(Q–1 CQ)P = (QP)–1 C(QP)
That is, A and C are similar.
13. The standard basis for M22 is the set of matrices
If we think of the above matrices as the vectors
[1 0 0 0]T, [0 1 0 0]T, [0 0 1 0]T, [0 0 0 1]T
then L takes these vectors to
[1 0 0 0]T, [0 0 1 0]T, [0 1 0 0]T, [0 0 0 1]T
Therefore the desired matrix for L is
14. (a) Reading directly from P, we have
v1 = 2u1 + u2
v2 = –u1 + u2 + u3
v3 = 3u1 + 4u2 + 2u3
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 1
1 0
0 0
0 1
0 0
0 0
1 0
1 0
0 1
, , ,
Supplementary Exercises 8 431
14. (b) Since
by direct calculation we have
u1 = –v1 – v2 + v3
u2 = 5v1 + 4v2 – 2v3
u3 = – 7v1 – 5v2 + 3v3
15. The transition matrix P from B′ to B is
Therefore, by Theorem 8.5.2, we have
Alternate Solution: We compute the above result more directly. It is easy to show that u1= v1, u2 = –v1 + v2, and u3 = –v2 + v3. So
T(v1) = T(u1) = –3u1 + u2 = –4v1 + v2
T(v2) = T(u1 + u2) = T(u1) + T(u2)
= u1 + u2 + u3 = v3
T(v3) = T(u1 + u2 + u3) = T(u1) + T(u2) + T(u3)
= 8u1 – u2 + u3
= 9v1 – 2v2 + v3
[ ] [ ]T P T PB B′−= =
−−
14 0 9
1 0 2
0 1 1
P =
1 1 1
0 1 1
0 0 1
P− =
− −− −
−
12 5 7
2 4 5
1 2 3
432 Supplementary Exercises 8
16. Let and solve for a, b, c, and d so that
This yields the system of equations
a – b = 2a + c
a + 4b = 2b + d
c – d = a + 3c
c + 4d = b + 3c
which has solution b = –a, c = 0, and d = –a. Thus, for instance,
will work.
However, if we try the same procedure on the other pair of matrices, we find that a =3d, b = d, and c = 3d, so that det(P) = 3d2 – 3d2 = 0. Thus P is not invertible.
Alternatively, if there exists an invertible matrix P such that
then we have a matrix with zero determinant on the left, and one with nonzero determinanton the right. Why? Thus the two matrices cannot be similar.
17. Since
T ,
1
0
0
1
0
1
=
T
0
1
0
1
1
0
=−
, and T
0
0
1
=−
1
0
1
3 1
6 2
1 2
1 01
− −
−
−= P P
P P=−−
=
−−
−1 1
0 1
1 1
0 11and
1 1
1 4
2 1
1 3
1 1
1 41
−
=
−
−P P Por
=
2 1
1 3P
Pa b
c d=
Supplementary Exercises 8 433
we have
In fact, this result can be read directly from [T(X)]B.
18. We know that det(T) ≠ 0 if and only if the matrix of T relative to any basis B has nonzerodeterminant; that is, if and only if every such matrix is invertible. Choose a particular basisB for V and let V have dimension n. Then the matrix [T]
Brepresents T as a matrix
transformation from Rn to Rn. By Theorem 4.3.1, such a transformation is one-to-one if andonly if [T]
Bis invertible. Thus T is one-to-one if and only if det(T) ≠ 0.
19. (a) Recall that D (f + g) = (f(x) + g(x))′′ = f′′(x) + g′′(x) and D(cf) = (cf(x))′′ = cf ′′(x).
(b) Recall that D(f) = 0 if and only if f′(x) = a for some constant a if and only if f(x) = ax
+ b for constants a and b. Since the functions f(x) = x and g(x) = 1 are linearlyindependent, they form a basis for the kernel of D.
(c) Since D(f) = f(x)if and only if f′′(x) = f(x) if and only if f(x) = aex + be–x for a and barbitrary constants, the functions f(x) = ex and g(x) = e–x span the set of all suchfunctions. This is clearly a subspace of C2 (–∞, ∞) (Why?), and to show that it hasdimension 2, we need only check that ex and e–x are linearly independent functions.To this end, suppose that there exist constants c1 and c2 such that c1e
x + c2e–x = 0. If
we let x = 0 and x = 1, we obtain the equations c1 + c2 = 0 and c1e + c2e–1 = 0. These
imply that c1 = c2 = 0, so ex and e–x are linearly independent.
21. (a) We have
and
T kp x
kp x
kp x
kp x
k( ( ))
( )
( )
( )
=
=1
2
3
( )
( )
( )
( ( ))
p x
p x
p x
kT p x
1
2
3
=
T p x q x
p x q x
p x q x( ( ) ( ))
( ) ( )
( ) ( )+ =++
1 1
2 2
pp x q x
p x
p x
p x( ) ( )
( )
( )
( )3 3
1
2
3+
=
+
=( )
( )
( )
(
q x
q x
q x
T
1
2
3
pp x T q x( )) ( ( ))+
[ ]T B =−
−
1 1 1
0 1 0
1 0 1
434 Supplementary Exercises 8
(b) Since T is defined for quadratic polynomials only, and the numbers x1, x2, and x3 aredistinct, we can have p(x1) = p(x2) = p(x3) = 0 if and only if p is the zero polynomial.(Why?) Thus ker(T) = 0, so T is one-to-one.
(c) We have
T(a1P1(x) + a2P2(x) + a3P3(x)) = a1T(P1(x)) + a2T(P2(x)) + a3T(P3(x))
(d) From the above calculations, we see that the points must lie on the curve.
23. Since
then
where the above vectors all have n + 1 components. Thus the matrix of D with respect toB is
[ ( )]( , , )
( , , , , ) , ,D x
k
k k
kB =
==
0 0 0
0 0 1 2
if
if , n
↑
kth component
D xk
kx k n
k
k( )
, , ,=
=
=
−
0 0
1 21
if
if
=
+
+
a a a1 2 3
1
0
0
0
1
0
0
0
1
=
a
a
a
1
2
3
Supplementary Exercises 8 435
24. Call the basis B and the vectors v0, v1, … , vn. Notice that D(v
i) = v
i–1 for i = 1, … , n,while D(v0) = 0. That is,
where the above vectors all have n + 1 components. Thus the matrix of D with respect toB is
In fact, the differentiation operator maps Pn
to Pn–1.
25. Let Bn
and Bn+1 denote the bases for P
nand P
n+1, respectively. Since
J(xk) = xk+1
——k + 1
for k = 0, … , n
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
0 0 0 0 0
[ ( )]( , , )
( , , , , )D
k
kk Bv
if
if=
==
0 0 0
0 1 0 1,, , ,2 n
↑
kth componeent
0 1 0 0 0
0 0 2 0 0
0 0 0 3 0
0 0 0 0
0 0 0 0 0
n
436 Supplementary Exercises 8
we have
where [xk]Bn
= [0, … , 1, … , 0]T with the entry 1 as the (k + 1)st component out of a totalof n + 1 components. Thus the matrix of J with respect to B
n+1 is
with n + 2 rows and n + 1 columns.
0 0 0 0
1 0 0 0
0 1 2 0 0
0 0 1 3 0
0 0 0 1 1( )n +
[ ( )] , , , , (J x Bk
nk
n+ =+
+1 01
10 2 )components
↑
+k 2nnd component
Exercise Set 8.1 437
EXERCISE SET 9.1
1. (a) The system is of the form y′ = Ay where
The eigenvalues of A are λ = 5 and λ = –1 and the corresponding eigenspaces arespanned by the vectors
respectively. Thus if we let
we have
Let y = Pu and hence y′ = Pu′. Then
or
u′1 = 5u1
u′2 = –u2
′−
u = u
5 0
0 1
D P AP= =−
−1 5 0
0 1
P =−
1 2
1 1
1
1
2
1
−
and
A =
1 4
2 3
439
Therefore
u1 = c1e5x
u2 = c2e–x
Thus the equation y = Pu is
or
y1 = c1e5x– 2c2e
–x
y2 = c1e5x + c2e
–x
1. (b) If y1(0) = y2(0) = , then
c1 – 2c2 = 0
c1 + c2 = 0
so that c1 = c2 = 0. Thus y1 = and y2 = 0.
3. (a) The system is of the form y′ = Ay where
The eigenvalues of A are λ = 1, λ = 2, and λ = 3 and the corresponding eigenspaces arespanned by the vectors
respectively. Thus, if we let
P
/
=− −
0 1 2 1
1 1 1
0 1 1
0
1
0
1 2
1
1
1
1
1
−
−
/
A = −−
4 0 1
2 1 0
2 0 1
y
y
c e
c e
cx
x
1
1
15
2
1 2
1 1
=
−
=−
115
2
15
2
2e c e
c e c e
x x
x x
−
+
−
−
440 Exercise Set 9.1
then
Let y = Pu and hence y′ = Pu′. Then
so that
u′1 = u1
u′2 = 2u2
u′3 = 3u3
Therefore
u1 = c1ex
u2 = c2e2x
u3 = c3e3x
Thus the equation y = Pu is
or
y1 = – 12c2e2x – c3e
3x
y2 = c1ex + c2e
2x + c3e3x
y3 = c2e2x + c3e
3x
y
y
y
1
2
3
=− −
/0 1 2 1
1 1 1
0 1 1
c e
c e
c e
x
x
x
1
22
33
′
u u
1 0 0
0 2 0
0 0 3
D P AP=
−11 0 0
0 2 0
0 0 3
Exercise Set 9.1 441
Note: If we use
as basis vectors for the eigenspaces, then
and
y1 = –c2e2x + c
3e3x
y2 = c1ex + 2c2e
2x – c3e3x
y3 = 2c2e2x – c3e
3x
There are, of course, infinitely many other ways of writing the answer, depending uponwhat bases you choose for the eigenspaces. Since the numbers c1, c2, and c3 arearbitrary, the “different” answers do, in fact, represent the same functions.
3. (b) If we set x = 0, then the initial conditions imply that
– 12c2 – c3 = –1
c1 + c2 + c3 = 1
c2 + c3 = 0
or, equivalently, that c1 = 1, c2 = –2, and c3 = 2. If we had used the “different” solutionwe found in Part (a), then we would have found that c1 = 1, c2 = –1, and c3 = –2. Ineither case, when we substitute these values into the appropriate equations, we findthat
y1 = e2x – 2e3x
y2 = ex – 2e2x + 2e3x
y3 = –2e2x + 2e3x
P =−
−−
0 1 1
1 2 1
0 2 1
0
1
0
1
2
2
1
1
1
−
−−
442 Exercise Set 9.1
5. Following the hint, let y = f(x) be a solution to y′ = ay, so that f′(x) = af(x). Now considerthe function g(x) = f(x)e–ax. Observe that
g′(x) = f′(x)e–ax – af(x)e–ax
= af(x)e–ax – af(x)e–ax
= 0
Thus g(x) must be a constant; say g(x) = c. Therefore,
f(x)e–ax = c
or
f(x) = ceax
That is, every solution of y′ = ay has the form y = ceax.
6. Suppose that y′ = Ay where A is diagonalizable. Then there is an invertible matrix P suchthat P–1 AP = D where D is a diagonal matrix whose diagonal entries are the eigenvalues ofα. If we let y = Pu so that y′ = Pu′, then y′ = Ay becomes Pu′ = APu or
u′ = P–1 APu
= Du
That is,
u1′ = λ
1u
1. .. .. .
u′n
= λnu
n
and hence
u1 = c1eλ1x
. .. .. .
un
= cneλnx
Exercise Set 9.1 443
Therefore y = Pu can be written
y1 = p11c1eλ1x + p12c2e
λ2x + … + p1nc
neλnx
. . . .. . . .. . . .
yn
= pn1c1e
λ1x + pn2c2e
λ2x + … + pnn
cneλnx
where pij
is the ijth entry in the matrix P. That is, each of the functions yi
is a linearcombination of eλ1x, … , eλnx.
7. If y1 = y and y2 = y′, then y1′ = y2 and y2′ = y′′ = y′ + 6y = y2 + 6y1. That is,
y1′ = y2
y2′ = 6y1+ y2
or y′ = Ay where
The eigenvalues of A are λ = –2 and λ = 3 and the corresponding eigenspaces are spannedby the vectors
respectively. Thus, if we let
then
Let y = Pu and hence y′ = Pu′. Then
P AP− =
−
1 2 0
0 3
P =−
1 1
2 3
−
1
2and
1
3
A =
0 1
6 1
444 Exercise Set 9.1
or
y1 = –c1e–2x + c2e
3x
y2 = 2c1e–2x + 3c2e
3x
Therefore
u1 = c1e–2x
u2 = c2e3x
Thus the equation y = Pu is
or
y1 = –c1e–2x + c2e
3x
y2 = 2c1e–2x + 3c2e
3x
Note that y1′ = y2, as required, and, since y1 = y, then
y = –c1e–2x + c2e
3x
Since c1 and c2 are arbitrary, any answer of the form y = ae–2x + be3x is correct.
8. Letting y = y1, y′ = y2, y′′ = y3 leads to the system
y1′ = y2
y2′ = y3
y3′ = 64 – 11y2 + 6y3
y
y
c e
c e
x
x
1
2
12
23
1 1
2 3
=
−
−
′−
u u
2 0
0 3
Exercise Set 9.1 445
or
The eigenvalues and corresponding eigenvectors of A are:
The solution to the system is:
Then y= y1 = c1et + c2e
2t + c3e3t .
9. If we let y1 = y, y2 = y′, and y3 = y′′, then we obtain the system
y1′′ = y2
y2′ = y3
y3′ = 6y1 –11y2 + 6y3
The associated matrix is therefore
y
y
y
1
2
3
1
1
1
=
+c e c1t
222t
33te c e
1
2
4
1
3
9
λ
λ
1 1
2 2
1
1
1
1
2
,
,
= =
=
X
X
,
=
= =
1
2
4
3
1
3
93 3λ X
′′′
=−
y
y
y
1
2
3
0 1 0
0 0 1
6 11 6
′ =y
y
y
AY
1
2
3
, or y
446 Exercise Set 9.1
The eigenvalues of A are λ = 1, λ = 2, and λ = 3 and the corresponding eigenvectors are
The solution is, after some computation,
y = c1ex + c2e
2x + c3e3x
10. (b) Suppose that the coefficient matrix A of the system y = Ay is diagonalizable and let D= P–1 AP where D and P are as given in the problem. Let y = Pu and y′ = Pu′ so thatu′ = Du. Thus u
i′ = λ
iu
ifrom which it follows that u
i= c
ieλix for i = 1, … , n. Since y
= Pu, we have
which is the desired result.
11. Consider y′ = Ay, where , with aij
REAL. Solving the system
yields the quadratic equation
λ2 – (a11 + a22) λ + a11a22 – a21, a12 = 0, or
λ2 – (TrA)λ + det A = 0.
det ( )λλ
λI-A =
− −− −
=a a
a a
11 12
12 220
Aa a
a a=
11 12
21 22
y
c e
c e
c e
x
x
nnx
=
X X L X1 2 n
11
22
λ
λ
λ
1
1
1
1
2
4
1
3
9
, and
A =−
0 1 0
0 0 1
6 11 6
Exercise Set 9.1 447
Let λ1, λ2 be the solutions of the characteristic equation. Using the quadratic formula yields
Now the solutions y1(t) and y2(t) to the system y′ = Ay will approach zero as t → ∞ ifand only if Re(λ1, λ2) < 0. (Both are < 0)
Case I: Tn2A – 4 det A < 0.
In this case Rn
(λ1) = Re
(λ2) = TrA2 . Thus y1(t), y2(t) → 0 if and only if TrA < 0.
Case II: Tr2A – 4 det A = 0. Then λ1 = λ2, Reλ1 = TrA/2, so y1, y2 → 0 (if and only if) ↔ TrA
< 0. Tr2A – 4 det A = 0. Then λ1, λ2, and Re(λ1, λ2) = TrA2 , so y1, y2 → 0 if and only if TrA
< 0.
Case III: Tr2A – 4 det A > 0. Then λ1, λ2 are both real.
Subcase 1: det A > 0.
If TrA > 0, then both (λ1, λ2) > 0, so y1, y2 0.
If TrA < 0, then both (λ1, λ2) < 0, so y1, y2 → 0. TrA = 0 is not possible in this case.
Subcase 2: det A < 0
If TrA > 0, then one root (say λ1) is positive, the other is negative, so y1, y2 0.
If TrA = 0, then again λ1 > 0, λ2 < 0, so y1, y2 0.
Subcase 3: det A = 0. Then λ1 = 0 or λ2 = 0, so y1, y2 0.→
→
→
Then Tn Tn A2A A− > ≥4 0det
→
Then Tn A > Tn2A A− >4 0det
λ1 2
2 4
2,det
=± −Tr A Tr A A
448 Exercise Set 9.1
12. From
we get yθ2 = et.
Thus, consider the differential equation
y′ = y = rt.
Solving, y = c1et + tet
13. The system
can be put into the form
The eigenvalues and eigenvectors of A are:
Solving:
14. The system
can be put into the form
y
y=
2
1
y
y+ t1
1
21
1
2
1
2
11
2= Ay+f
′ = + +
′ = − +
−
y y y e
y y y e
k
k
1 1 2
2 1 2
y
yc e c e t
k k1
22
31
1
1
1
0
1
=
−
+
+
−,
+
−
13
23
/
/
λ λ1 1 211
13, ,= =
−
= =x x2
11
1
′′
=
+
y
y
y
yt
1
2
1
2
2 1
1 2
1
22
′ = + +′ = +
y y y t
y y y t
1 1 2
2 1 2
2
2 2_
′ = +′ =
y y y
y y
1 1 2
2 2
Exercise Set 9.1 449
The eigenvalues and eigenvectors of A are
Solving
=
− +
+
−+
−y
yc e c e
t t1
2
22
21
1 2
1
1 2,
+
−−
−
−e te
t
t
t2
1
1
2
λ λ1 1 1 2 221
1 22
1
1 2= =
− +
= − =
−+
x x
, ,
450 Exercise Set 9.1
EXERCISE SET 9.2
1. (a) Since T(x, y) = (–y, –x), the standard matrix is
(b)
(c) Since T(x, y) = (x, 0), the standard matrix is
(d)
3. (b) Since T(x, y, z) = (x, –y, z), the standard matrix is
0 1 0
1 0 0
0 0 1
−
A =
0 0
0 1
1 0
0 0
A =−
−
1 0
0 1
A =−
−
1 0
0 1
451
5. (a) This transformation leaves the z-coordinate of every point fixed. However it sends (1,0, 0) to (0, 1, 0) and (0, 1, 0) to (–1, 0, 0). The standard matrix is therefore
(c) This transformation leaves the y-coordinate of every point fixed. However it sends (1,0, 0) to (0, 0, –1) and (0, 0, 1) to (1, 0, 0). The standard matrix is therefore
12. (b) To reduce this matrix to the identity matrix, we add –2 times Row 1 to Row 2 and thenadd –4 times Row 2 to Row 1. These operations when performed on I yield
respectively. The inverses of the above two matrices are
respectively. Thus,
and therefore the transformation represents a shear by a factor of 4 in the x-directionfollowed by a shear by a factor of 2 in the y-direction.
(c) To reduce this matrix to the identity matrix, we multiply Row 1 by –1/2, multiply Row2 by 1/4, and interchange Rows 1 and 2. These operations when performed on I yield
−
1 2 0
0 1
1 0
0 1 4
0 1
1 0
/
/
1 4
2 9
1 0
2 1
1 4
0 1
=
1 4
0 1
1 4
0 1
and
1 0
2 1
1 4
0 1−
−
and
0 0 1
0 1 0
1 0 0−
0 1 0
1 0 0
0 0 1
−
452 Exercise Set 9.2
respectively. The inverses of the above matrices are
respectively. Thus,
Therefore, the transformation represents a reflection about the line y = x followed byexpansions by factors of 4 and –2 in the y- and x- directions, respectively. Note that theorder in which the two expansion matrices occur is immaterial. However the position ofthe reflection matrix cannot be changed.
We return to the original matrix and note that we could, of course, interchange Rows 1and 2 first and then multiply Row 1 by 1/4 and Row 2 by –1/2. This yields the factorization
Here the order of the two expansion matrices can also be interchanged. Note that there aretwo other factorizations, for a grand total of six.
13. (a)
(c)
14. (a)
15. (b) The matrices which represent compressions along the x- and y- axes are
and , respectively, where 0 < k < 1. But
k k0
0 1
1 0
0 1
1
=
−/
1 0
0 k
k 0
0 1
−−
=
−
=
1 0
0 1
5 0
0 1
1 0
0 1
0 1
−−
5 0
−−
=
−−
1 0
0 1
0 1
1 0
0 1
1 0
1 0
0 5
1 2 0
0 1
1 2 0
0 5
=
/ /
0 2
4 0
0 1
1 0
4 0
0 1
1 0
0 2
−
=
−
0 2
4 0
2 0
0 1
1 0
0 4
0 1
1 0
−
=
−
−
2 0
0 1
1 0
0 4
0 1
1 0
Exercise Set 9.2 453
and
Since 0 < k < 1 implies that 1/k > 1, the result follows.
15. (c) The matrices which represent reflections about the x- and y- axes are and
, respectively. Since these matrices are their own inverses, the
result follows.
16. Since , it follows that points (x′, y′) on the image line must satisfy the
equations
x = –2x′ + 3y′
y = –3x′ + 4y′
where y = –4x + 3. Hence –3x′ + 4y′ = 8x′ – 12y′ + 3, or 11x′ – 16y′ + 3 = 0. That is, theimage of y = –4x + 3 has the equation 11x – 16y + 3 = 0.
17. (a) The matrix which represents this shear is ; its inverse is . Thus,
points(x′, y′) on the image line must satisfy the equations
x = x – 3y′
y = y′
where y = 2x. Hence y′ = 2x′ – 6y′, or 2x′ – 7y′ = 0. That is, the equation of the imageline is 2x – 7y = 0.
Alternatively, we could note that the transformation leaves (0, 0) fixed and sends(1, 2) to (7, 2). Thus (0, 0) and (7, 2) determine the image line which has theequation 2x – 7y = 0.
(c) The reflection and its inverse are both represented by the matrix . Thus the
point (x′, y′) on the image line must satisfy the equations
x = y′
y = x′
where y = 2x. Hence x′ = 2y′, so the image line has the equation x – 2y = 0.
0 1
1 0
1 3
0 1
−
1 3
0 1
A− = −
−
1 2 3
3 4
−
1 0
0 1
1 0
0 1−
1 0
0
1 0
0 1
1
k k
=
−
/
454 Exercise Set 9.2
(e) The rotation can be represented by the matrix . This sends the
origin to itself and the point (1, 2) to the point ((1 – 2 3 )/2, (2 + 3)/2). Since both
(0, 0) and (1, 2) lie on the line y = 2x, their images determine the image of the line
under the required rotation. Thus, the image line is represented by the equation (2 +
3)x + (2 3 – 1)y = 0.
Alternatively, we could find the inverse of the matrix, , and
proceed as we did in Parts (a) and (c).
18. Since the shear is represented by the matrix , it will send (2, 1) to (2 + k, k) and (3,
0) to (3, 0). The origin, of course, remains fixed. Thus we must find a value of k for whichthe vectors (2 + k, k) and (3, 0) are orthogonal; that is,
(2 + k, k) • (3, 0) = 3(2 + k) = 0
Clearly k = –2.
20. The equation of a line in the plane is Ax + By + C = 0 where not both A and B are zero.Moreover
where ad – bc ≠ 0. Thus (x, y) is transformed into (x′, y′) where
Therefore, the line Ax + By + C = is transformed into
or
dx by
ad bcB
cx ay
ad bc
−−
+ =− ′ + ′
−+ =
−−
+ =− +
C
da cB
ad bcB
bA a
0
BB
ad bcC
−+ = 0
xdx by
ad bc
ycx ay
ad bc
=′ − ′
−
=′ + ′
−
a b
c d ad bc
d b
c a
=
−−
−
−11
1
0 1
k
1 2 3 2
3 2 1 2
/ /
/ /−
1 2 3 2
3 2 1 2
/ /
/ /
−
Exercise Set 9.2 455
This is the equation of a line provided dA – cB and –bA + aB are not both zero. But if bothnumbers are zero, then we have
bdA – bcB = 0
–bdA + adB = 0
or
(ad – bc) = 0
This implies that B = 0 since ad – bc ≠ 0. However, B = 0 implies that dA = bA = 0. Thuseither A = 0 or d = b = 0. But since ad – bc ≠ 0, b and d cannot both be zero. Hence A mustequal zero. Finally, since not both A and B can equal zero, then not both dA – cB and –bA
+ aB can equal zero.
21. We use the notation and the calculations of Exercise 20. If the line Ax + By + C = 0 passesthrough the origin, then C = 0, and the equation of the image line reduces to (dA – cB)x +(–bA + aB)y = 0. Thus it also must pass through the origin.
The two lines A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 are parallel if and only if A1B2 = A2B1. Their image lines are parallel if and only if
(dA1 – cB1)(–bA2 + aB2) = (dA2 – cB2)(–bA1 + aB1)
or
bcA2B1 + adA1B2 = bcA1B2 + adA2B1
or
(ad – bc)(A1B2 – A2B1) = 0
or
A1B2 – A2B1 = 0
Thus the image lines are parallel if and only if the given lines are parallel.
22. (a) Since x is fixed, but y and z are interchanged, we have
1 0 0
0 0 1
0 1 0
456 Exercise Set 9.2
23. (a) The matrix which transforms (x, y, z) to (x + kz, y + kz, z) is
24. (a) A reflection about the x-axis will not change the length of any vector. Thus the onlypossible eigenvalues are 1 and –1. The only two unit vectors which are transformed tovectors with the same or opposite direction are the standard basis vectors, e1 and e2.The vector e1 goes to itself and e2 goes to –e2. Therefore e1 and e2 are linearlyindependent eigenvectors with corresponding eigenvalues 1 and –1.
This is easily checked using the matrix
with characteristic equation (λ – 1)(λ + 1) = 0. The eigenspaces will be the x and yaxes.
(c) This is similar to Part (a) above only the two lines through the origin which remainfixed will be y = x and y = –x. Thus 2 linearly independent eigenvectors will be (1, 1)and (–1, 1). The corresponding eigenvalues will be 1 and –1.
Again, this is easily checked using the matrix
with characteristic equation λ2 – 1 = 0.
(e) The only line through the origin which remains fixed when k ≠ 0 is the y-axis. Sincedistance along the y-axis is not changed, the only possible eigenvalue is λ = 1. Acorresponding eigenvector ise2 = (0, 1).
(f) The only angles through which we can rotate R2 and keep a line through the originfixed are integer multiples of 0, π, and 2π. Since rotations by integer multiples of 0 and2π leave the plane unchanged, we consider only rotations by odd multiples of π. Theseleave all lines through the origin fixed but reverse their directions. Thus such rotationshave –1 as an eigenvalue, and (1, 0) and (0, 1) form a basis for the eigenspace.
A =
0 1
1 0
A =−
1 0
0 1
1 0
0 1
0 0 1
k
k
Exercise Set 9.2 457
EXERCISE SET 9.3
1. We have
Thus the desired line is y = –1/2 + (7/2)x.
3. Here
M =
1 2 4
1 3 9
1 5 25
1 6 36
a
b
T
=
1 0
1 1
1 2
1 0
1 1
1 2
=
−11 0
1 1
1 2
0
2
7
T
3 3
3 5
9
16
5
6
1
21
2
1
2
1
=
−
−
−
=−
/
9
16
1 2
7 2
459
and
Thus the desired quadratic is y = 2 + 5x – 3x2.
5. The two column vectors of M are linearly independent if and only if neither is a nonzeromultiple of the other. Since all of the entries in the first column are equal, the columns arelinearly independent if and only if the second column has at least two different entries, orif and only if at least two of the numbers x
iare distinct.
6. Since we have
M
x x x
x x x
m
n n nm
=
1
1
1 12
1
2
a
b
c
M M MTT
1
2
3
1
0
10
48
76
=−−−
−( )
=
4 16 74
16 74 376
74 376 2018
−−−
=
−1134
726
4026
221
10
62
5
3
2262
5
649
90
8
93
2
8
9
1
9
134
−−−−
=−
726
4026
2
5
3
460 Exercise Set 9.3
the columns of M will be linearly independent if and only if a0 = a1 = … = am
= is the onlysolution to the equation
a0 [1 … 1]T + a1 [x1… x
n]T + a2 [x1
2 … xn2]T + … + a
m[xm
1… x
nm]T = [0 … 0]T
That is, if and only if the system of equations
has only the trivial solution for the numbers a0, … , am
. But if any of the numbers ai
≠ 0,then there are at least m + 1 distinct numbers x
iwhich must satisfy the equation
a0 + a1x + a2x2 + … + a
mxm = 0
This is a polynomial of degree at most m and can therefore have at most m distinct realroots. Thus, indeed, a0 = … = a
m= 0.
8. Let xi
denote the month, starting with January = 1, and yi
denote the sales for that month.Then we have
where
M M MT=
1 1 1
1 2 2
1 3 3
1 4 4
1 5 5
2
2
2
2
2
==
5 15 55
15 55 225
55 225 979
a
a
a
M M MT T
0
1
2
1
4 0
4 4
5 2
6 4
8 0
=
−( )
.
.
.
.
.
a a x a x a xmm
0 1 1 2 12
1 0+ + + + =
a a x a x a xn n m nm
0 1 22+ + + + = 00
Exercise Set 9.3 461
so that the desired quadratic is y = 4 – (.2)x + (.2)x2. When x = 12, y = 30.4, so theprojected sales for December are $30,400.
( )M MT − =
−
− −
−
1
23
5
33
10
1
233
10
187
70
3
71
2
3
7
1
14
=and M yT
28 0
94 0
3
.
.
770 8.
462 Exercise Set 9.3
EXERCISE SET 9.4
1. (a) Since f(x) = 1 + x, we have
Using Example 1 and some simple integration, we obtain
k = 1, 2, . . .
Thus, the least squares approximation to 1 + x on [0, 2π] by a trigonometric polynomialof order ≤2 is
1 + x (1 + π) – 2 sin x – sin 2x
2. (a) Since f(x) = x2,
Using Example 1 and integration by parts or a table of integrals, we find that
k = 1, 2, …
b x kx dxk k
xk sin( )( )
cos(= = − +1 2 22
2
πππ π
kkx dxk
) = −∫∫4
0
2
0
2 ππ
a x kx dxk
x kx dxk
k cos( ) sin( )= = − =1 2 42
π π 220
2
0
2 ππ∫∫
a x dx02 2
0
21 8
3= =∫π
ππ
b x kx dxk
k ( )sin( )= + = −∫1
12
0
2
ππ
a x kx dxk ( )cos( )= + =∫1
1 00
2
ππ
a x dx0 0
211 2 2( )= + = +∫π
ππ
463
Thus, the least squares approximation to x2 on [0, 2π] by a trigonometric polynomialof order at most 3 is
3. (a) The space W of continuous functions of the form a + bex over [0, 1] is spanned by thefunctions u1 = 1 and u2 = ex. First we use the Gram-Schmidt process to find anorthonormal basis v1, v2 for W.
Since ⟨f, g⟩ = f(x)g(x)dx, then u1 = 1 and hence
v1 = 1
Thus
v2 =
where α is the constant
Therefore the orthogonal projection of x on W is
projW
x x
x x xe e e e
xd
, ,= +− + − +
=
1 11 1
α α
xxe e x e e
dx
e e
x x
x
+− + − +
= +− +
∫ ∫1 1
1
2
1
0
1
0
1
α α
α
( )
( )
3
2
1
2
1
11
−
= +−
− +
= −
e
ee e
x
α
11
2
1
1+
−
ee
x
α ( )
(
/
= − + = − +
=
∫e e e e dxx x1 1 2
0
1 1 2
33 1
2
1 2− −
e e)( )/
e e
e e
e ex x
x x
x−
−=
− +( , )
( , )
1 1
1 1
1
α
0
1
∫
x x x x2 24
34 2
4
93 4cos cos cos sin π π+ + + − xx x xsin sin− −2
4
33π π
464 Exercise Set 9.4
(b) The mean square error is
The answer above is deceptively short since a great many calculations are involved.
To shortcut some of the work, we derive a different expression for the mean squareerror (m.s.e.). By definition,
m.s.e. = [f(x) – g(x)]2 dx
= f – g2
= ⟨f – g, f – g⟩
= ⟨f, f – g⟩ – ⟨g, f – g⟩
Recall that g = projW
f, so that g and f – g are orthogonal (see Figure 3). Therefore,
m.s.e. = ⟨f, f – g⟩
= ⟨f, f⟩ – ⟨fg⟩
But g = ⟨f, v1⟩v1 + ⟨f, v2⟩v2, so that
(*) m.s.e. = ⟨f, f⟩ – ⟨f, v1⟩2 – ⟨f, v2⟩
2
Now back to the problem at hand. We know ⟨f, v1⟩ and ⟨f, v2⟩ from Part (a). Thus,in this case,
Clearly the formula (*) above can be generalized. If W is an n-dimensional space withorthonormal basis v1, v2, . . . , v
n, then
(**) m.s.e. = f2 – ⟨f, v1⟩2 – • • • – ⟨f, v
n⟩2
m.s.e. = −
−−
∫ x dx
e22
0
12
1
2
3
2α
== −−−
≈( )
.1
12
3
2 10014
e
e
xe
e dxe
e
x− − +−
= +
+−
(( )
1
2
1
1
13
12
1
2 1
2
== +−−∫
1
2
3
2 10
1 e
e( )
Exercise Set 9.4 465
4. (a) The space W of polynomials of the form a0 + a1x over [0, 1] has the basis 1, x. TheGram-Schmidt process yields the orthonormal basis 1, ƒ3(2x – 1). Therefore, theorthogonal projection of ex on W is
(b) From Part (a) and (*) in the solution to 3(b), we have
5. (a) The space W of polynomials of the form a0 + a1x + a2x2 over [–1, 1] has the basis 1,
x, x2. Using the inner product ⟨u, v⟩ = u(x)v(x)dx and the Gram-Schmidt
process, we obtain the orthonormal basis
(See Exercise 29, Section 6.3.) Thus
Therefore,
sinππ
x x3
sin , sin( )
sin , s
π π
π
x x dx
x x
v
v
1 1
1
2
1
20
3
2
= =
=
−∫
iin( )
sin , ( )sin
ππ
π
x dx
x x
−∫ =
= −
1
1
32
2 3
2
1
2
5
23 1v (( )π x dx
−∫ =1
10
1
2
3
2
1
2
5
23 12, , ( )x x −
−∫ 1
1
m.s.e = − − − −
=− −
∫ e dx e e
e e
x2 2 2
0
11 3 3
3 7
( ) ( )
( )( 119
2004
).≈
projWx x x
e e e x x
e
= + − −
= − +
, , ( ) ( )
(
1 1 3 2 1 3 2 1
1 3 3 −− −
= −( ) + −( )e x
e e x
) ( )3 2 1
4 10 18 6
466 Exercise Set 9.4
(b) From Part (a) and (**) in the solution to 3(b), we have
8. The Fourier series of π – x is
Thus the required Fourier series is
where
and
k = 1, 2, . . .
Thus the required Fourier series is . Notice that this is consistent with the result
of Example 1, Part (b).
9.
(note a slight correction to definition of f(x) as stated on pg. 466.)
Let ( ),
,f x
x
x=
≤ <≤ ≤
1 0
0 2
ππ π
sin2
1 kkx
k=
∞
∑
b x kx dxk
k = − =∫1 2
0
2
ππ
π( )sin( )
a x kx dxk = − =∫1
00
2
ππ
π( )cos( )
a x dx0 0
210= − =∫π
ππ
( )
aa kx b kxk k
k
0
12( cos sin )+ +
=
∞
∑
m.s.e. = − = − ≈−∫ sin ( ) .2
2 21
1 61
639π
π πx dx
Exercise Set 9.4 467
Then
So Fourier Series is
10.
Thus, the Fourier Series for f(x) = sin 3x is sin 3x.
a x dx
a x kxk
0 0
213 0
13
sin
sin cos
= =
=
∫π
π
π
sin sin ,
dx
b x kx dx kk
=
= = ≠
∫ 0
13 0 3
0
2π
π 00
2
32
0
213
1
21 6
π
π
π
π
∫
∫=
= − =
a x dx
x dx
sin
cos 110
2π∫
1
2
11 1
1
+ − −
=
∞
∑ ( ) sink
kxk
kπ
a f x dx dx
a f x kxdxk
0 00
21 11
1
= = =
= =
∫∫π π
π
ππ( )
( )cos11
0
1
00
2
π
π
ππcos
( )sin
kxdx
k
b f x kxk
=
=
=
∫∫1, 2,
ddx kxdx
k
k
=
= − −
∫∫1
11 1
00
2
π
π
ππsin
( ) .
468 Exercise Set 9.4
EXERCISE SET 9.5
1. (d) Since one of the terms in this expression is the product of 3 rather than 2 variables,the expression is not a quadratic form.
5. (b) The quadratic form can be written as
The characteristic equation of A is (λ – 7)(λ – 4) – 1/4 = 0 or
4λ2 – 44λ + 111 = 0
which gives us
If we solve for the eigenvectors, we find that for
x1 = (3 + 10)x2
and
x1 = (3 - 10)x2
Therefore the normalized eigenvectors are
3 10
20 6 10
1
20 6 10
3 10
20 6 1
+
+
+
−
−and 00
1
20 6 10−
λ =−11 10
2
λ =+11 10
2
λ =±11 10
2
[ ]x xx
xx Ax
T1 2
1
2
7 1 2
1 2 4
=
469
or, if we simplify,
Thus the maximum value of the given form with its constraint is
The minimum value is
6. (b) We write the quadratic form as
The characteristic equation of a is
λ3 – 4λ2 + 3λ = λ(λ – 3)(λ – 1) = 0
with roots λ = 0, 1, 3. The corresponding normalized eigenvectors are
respectively. Thus the maximum value of the given form with its constraint is
1 3
1 3
1 3
0
1 2
1 2
−
−
−
2 6
1 6
1 6
x x x
x
x
x
1 2 3
1
2
3
2 1 1
1 1 0
1 0 1
= x xTA
11 10
2
1
20 6 10
1
20 6 101 2
−=
−
+=
−at andx x
11 10
2
1
20 6 10
1
20 6 101 2
+=
−=
+at andx x
1
20 6 10
1
20 6 10
1
20 6 10
1
20
−
−
−
+
−
and
66 10
470 Exercise Set 9.5
3 at x1 = 2/ 6, x2 = 1/ 6, x3 = 1/ 6
The minimum value is
0 at x1 = 1/ 3, x2 = –1/ 3, x3 = –1/ 3
7. (b) The eigenvalues of this matrix are the roots of the equation λ2 – 10λ + 24 = 0. Theyare λ = 6 and λ = 4 which are both positive. Therefore the matrix is positive definite.
8. (b) The principal submatrices are
Their determinants are 5 and 24, so the matrix is positive definite.
9. (b) The characteristic equation of this matrix is λ3 – 3λ + 2 = (λ + 1)2(λ – 2). Since twoof the eigenvalues are negative, the matrix is not positive definite.
11. (a) Since x12 + x2
2 > unless x1 = x2 =, the form is positive definite.
(c) Since (x1 ~ x2)2 ≥ 0, the form is positive semidefinite. It is not positive definite because
it can equal zero whenever x1 = x2 even when x1 = x2 ≠ 0.
(e) If |x1| > |x2|, then the form has a positive value, but if |x1| < |x2|, then the form has anegative value. Thus it is indefinite.
12. (a) The related quadratic form is 3x12 – 2x2
2 + x32 which is indefinite since its values can be
both positive and negative. Hence the matrix is indefinite.
(c) The related quadratic form is 6x12 + 9x2
2 + x32 + 14x1x2 + 2x1x3 + 4x2x3. The possible
signs of the values which this form can assume are not obvious and its characteristicequation is rather messy. However, the determinant of the matrix is zero while thedeterminants of the remaining principal submatrices are positive. Hence the matrix ispositive semidefinite.
55 1
1 5[ ] −
−
and
Exercise Set 9.5 471
13. (a) By definition,
T(x + y) = (x + y)T A(x + y)
= (xT + yT)A(x + y)
= xTAx + xT Ay + yT Ax + yT A y
= T(x) + xT Ay + (xT A Ty)T + T(y)
= T(x) + xT αy + xT ATy + T(y)
(The transpose of a 1 × 1 matrix is itself.)
= T(x) + 2xT Ay + T(y)
(Assuming that A is symmetric, AT = A.)
(b) We have
T(kx) = (kx)T A(kx)
9 = k2xT Ax (Every term has a factor of k2.)
= k2T(x)
(c) The transformation is not linear because T(kx) ≠ kT(x) unless k = or 1 by Part (b).
14. (b) The symmetric matrix associated with the quadratic form is
The determinants of the principal submatrices are 5, 1, and k – 2. These are positiveprovided k > 2. Thus the quadratic form is positive definite if and only if k > 2.
15. If we expand the quadratic form, it becomes
c12x1
2 + c22x2
2 + … + cn2x
n2 + 2c1c2x1x2 + 2c1c3x1x3 +
… + 2c1cnx1xn
+ 2c2c3x2x3 + … + 2cn–1cnx
n–1xn
5 2 1
2 1 1
1 1
−−
− −
k
472 Exercise Set 9.5
Thus we have
and the quadratic form is given by xT Ax where x = [x1 x2… x
n]T.
16. (a) We have
Thus sx2 = xT Ax where
(b) Since sx2 is 1/(n –1) times a sum of squares, it cannot be negative. In fact, it can only
be zero if (xi
– x) = 0 for i = 1, … , n, which can only be true in case x1 = x2 = … =x
n. If we let all the values of x
ibe equal but not zero, we have that s
x2 = while x ≠ 0.
Thus the form is positive semidefinite.
1 1
1
1
1
1
1
1
1
1 1
n n n n n n n
n n n n n
−−
−−
−−
−−
−−
( ) ( ) ( )
( ) (
11
1
1
1
1
1
1
1
1
1
) ( )
( ) ( ) ( )
−−
−−
−−
−−
n n
n n n n n n n
sn
x x x x x x nxn nx2
12
22 2
121
12=
−+ + − + +( ) +
=−
+ + − + +( ) + + +( )1
1
2 112 2
12
12
nx x
nx x
nx xn n n
=−
+ + − + + + +1
1
121
2 212 2
1 2n
x xn
x x x xn n ++( )( )
=−
−+ +( ) −
−x x
n
n
nx x
nx
n n
n
1
12 21
1
1 2 11 2 1x x xn n+ +( )
−
A
c c c c c c c
c c c c c c c
c c c c
n
n
=
12
1 2 1 3 1
1 2 22
1 3 2
1 3 2 33 32
3
1 2 32
c c c
c c c c c c c
n
n n n n
Exercise Set 9.5 473
17. To show that λn
≤ xT Ax if x = 1, we use the equation from the proof dealing with λ1 andthe fact that λ
nis the smallest eigenvalue. This gives
xT Ax = ⟨x, Ax⟩ = λ1 ⟨x, v1⟩2 + λ2 ⟨x, v2⟩
2 + … + λn
⟨x, vn⟩2
≥ λn
⟨x, v1⟩2 + λ
n⟨x, v2⟩
2 + … + λn
⟨x, vn⟩2
= λn
(⟨x, v1⟩2 + … + ⟨x, v
n⟩2)
= λn
Now suppose that x is an eigenvector of A corresponding to λn. As in the proof dealing
with λ1, we have
xT Ax = ⟨x, Ax⟩ = ⟨x, λnx⟩ = λ
n⟨x, x⟩ = λ
nx2 = λ
n
474 Exercise Set 9.5
EXERCISE SET 9.6
1. (a) The quadratic form xT Ax can be written as
The characteristic equation of A is λ2 – 4λ + 3 = 0. The eigenvalues are λ = 3 and λ =1. The corresponding eigenspaces are spanned by the vectors
respectively. These vectors are orthogonal. If we normalize them, we can use the resultto obtain a matrix P such that the substitution x = Py or
will eliminate the cross-product term. This yields the new quadratic form
or 3y12 + y2
2.
2. (a) The quadratic form can be written as xT Ax where
xT = [x1 x2 x3]
y yy
y1 2
1
2
3 0
0 1
x
x
y
y
1
2
1
2
1 2 1 2
1 2 1 2
=
−
1
1
1
1−
and
x xx
x1 2
1
2
2 1
1 2
−−
475
and
The characteristic equation of A is
λ3 – 12λ2 + 39λ – 28 = (λ – 7)(λ – 4)(λ – 1) = 0
The eigenspaces corresponding to λ = 7, λ = 4, and λ = 1 are spanned by the vectors
respectively. If we normalize these vectors, we obtain the matrix
By direct calculation, we have
so the substitution x = Py will yield the quadratic form
yT By or 7y12 + 4y2
2 + y32
7. (a) If we complete the squares, then the equation 9x2 + 4y2 – 36x – 24y + 36 = 0 becomes
9(x2 – 4x + 4) + 4(y2 – 6y + 9) = –36 + 9(4) + 4(9)
or
9(x – 2)2 + 4(y – 3)2 = 36
P AP BT =
=7 0 0
0 4 0
0 0 1
P = −− −
1 3 2 3 2 3
2 3 1 3 2 3
2 3 2 3 1 3
1
2
2
2
1
2
2
2
−
−−11
A = −−
3 2 0
2 4 2
0 2 5
476 Exercise Set 9.6
or
This is an ellipse.
(c) If we complete the square, then y2 – 8x – 14y + 49 = becomes
y2 – 14y + 49 = 8x
or
(y – 7)2 = 8x
This is the parabola (y′)2 = 8x′.
(e) If we complete the squares, then 2x2 – 3y2 + 6x + 2y = –41 becomes
or
or
12(x′)2 – 18(y′)2 = –419
8. (a) The matrix form for the conic 2x2 – 4xy – y2 + 8 = 0 is
(*) xT A x + 8 = 0
where
A =−
− −
2 2
2 1
23
23
10
3
419
6
2 2
x y+
− −
= −
2 39
43
20
3
100
941
92 2x x y y+ +
− − +
= − +22
100
3−
′( )+
′( )=
x y2 2
4 91
Exercise Set 9.6 477
The eigenvalues of A are λ1 = 3 and λ2 = –2 and the eigenspaces are spanned by thevectors
respectively. Normalizing these vectors yields
We choose
Note that det(P) = 1. (Interchanging columns yields a matrix with determinant –1.) Ifwe substitute x = Px′ into (*), we have
(Px′)T A(Px′) + 8 = 0
or
(x′)T (PT AP)x′ + 8 = 0
Since
this yields the hyperbola
–2(x′)2 + 3(y′)2 + 8 = 0
Note that if we had let
P APT =
−
2 0
0 3
P =
−
1
5
2
52
5
1
5
−
2
51
5
1
52
5
and
−
2
1
1
2and
478 Exercise Set 9.6
then det(P) = 1 and the rotation would have produced the diagonal matrix
and hence the hyperbola
3(x′)2 – 2(y′)2 + 8 = 0
That is, there are two different rotations which will put the hyperbola into the twodifferent standard positions.
8. (b) The matrix form for the conic 5x2 + 4xy + 5y2 = 9 is
xT Ax = 9
where
The eigenvalues of A are λ1 = 3 and λ2 = 7 and the eigenspaces are spanned by thevectors
If we normalize these vectors, we obtain
−
1
21
2
1
21
2
and
−
1
1
1
1and
A =
5 2
2 5
P APT =
−
3 0
0 2
P =−
2
5
1
51
5
2
5
Exercise Set 9.6 479
Hence we can let
Note that det(P) = 1. Thus if we let x = Px′, then we have
(x′)T (PT AP)x′ = 9
where
This yields the ellipse,
7(x′)2 + 3(y′)2 = 9
Had we let
then we would have obtained
3(x)2 + 7(y′) 2 = 9
which is the same ellipse rotated 90°.
9. The matrix form for the conic 9x2 – 4xy + 6y2 – 10x – 20y = 5 is
xT Ax + Kx = 5
where
and K = [ – 10 – 20]A =−
−
9 2
2 6
P =
−
1
2
1
21
2
1
2
P APT =
7 0
0 3
P =
−
1
2
1
21
2
1
5
480 Exercise Set 9.6
The eigenvalues of A are λ1 = 5 and λ2 = 1 and the eigenspaces are spanned by the vectors
Thus we can let
Note that det(P) = 1. If we let x = Px′, then
(x′)T (PT AP)x′ + KPx′ = 5
where
Thus we have the equation
5(x′)2 + 10(y′)2 – 10 5x′ = 5
If we complete the square, we obtain the equation
5(x′)2 – 2 5x′ + 5 + 10 (y) 2 = 5 + 25
or
(x′′)2 + 2(y′′)2 = 6
where x′′ = x′ – 5 and y′′ = y′. This is the ellipse
′′( )+
′′( )=
x y2 2
6 31
P AP KPT =
= −
5 0
0 1010 5 0and
P =
−
1
5
2
52
5
1
5
1
2
2
1
−
and
Exercise Set 9.6 481
Of course we could also rotate to obtain the same ellipse in the form 2(x′′)2 + (y′′)2 = 6,which is just the other standard position.
11. The matrix form for the conic 2x2 – 4xy – y2 – 4x – 8y = –14 is
xT Ax + Kx = –14
where
The eigenvalues of A are λ1 = 3, λ2 = –2 and the eigenspaces are spanned by the vectors
Thus we can let
Note that det(P) = 1. If we let x = Px′, then
(x′)T (PT AP)x′ + KPx′ = –14
where
Thus we have the equation
3(x′)2 – 2(y′)2 – 4 5y′ = –14
If we complete the square, then we obtain
P AP KPT =
−
= −
3 0
0 20 4 5and
P =−
2
5
1
51
5
2
5
−
2
1
1
2and
A K=−
− −
= − −[ ]2 2
2 14 8and
482 Exercise Set 9.6
3(x′)2 – 2((y′)2 + 2 5y′ + 5)= –14 – 10
or
3(x′′)2 – 2(y′′)2 = –24
where x′′ = x′ and y′′ = y′ + 5. This is the hyperbola
We could also rotate to obtain the same hyperbola in the form 2(x′′)2 – 3(y′′)2 = 24.
14. The matrix form for the conic 4x2 – 20xy + 25y2 – 15x – 6y = 0 is
xT Ax + Kx = 0
where
The eigenvalues of A are λ1 = 29 and λ2 = 0 and the eigenspaces are spanned by the vectors
Thus we can let
Note that det(P) = 1. If we let x = Px′, then
(x′)T (PT AP)x′ + KPx′ = 0
P =−
2
29
5
295
29
2
29
−
2
5
5
2and
A K=−
−
= − −[ ]4 10
10 2515 6and
′′( )−
′′( )=
x y2 2
12 81
Exercise Set 9.6 483
where
Thus we have the equation
which is the parobola . The other standard positions for this parabola are
represented by the equations
15. (a) The equation x2 – y2 = 0 can be written as (x – y)(x + y) = 0. Thus it represents thetwo intersecting lines x ± y = 0.
(b) The equation x2 + 3y2 + 7 = 0 can be written as x2 + 3y2 = –7. Since the left side ofthis equation cannot be negative, then there are no points (x, y) which satisfy theequation.
(c) If 8x2 + 7y2 = 0, then x = y = 0. Thus the graph consists ofthe single point (0, 0).
(d) This equation can be rewritten as (x – y)2 = 0. Thus it represents the single line y =x.
(e) The equation 9x2 + 12xy + 4y2 – 52 = 0 can be written as (3x + 2y)2 = 52 or 3x + 2y
= ± 52. Thus its graph is the two parallel lines 3x + 2y ± 2 13 = 0.
(f) The equation x2 + y2 – 2x – 4y = –5 can be written as x2 – 2x + 1 + y2 – 4y + 4 = 0 or(x – 1)2 + (y – 2)2 = 0. Thus it represents the point (1, 2).
and′( ) = − ′ ′( ) = ± ′x y y x2 23
29
3
29
2′( ) =x33
29′y
2987
290
2′( ) − ′ =x y
P AP KPT =
= −
29 0
0 00 87 29and
484 Exercise Set 9.6
EXERCISE SET 9.7
5. (a) ellipse 36x2 + 9y2 = 32
(b) ellipse 2x2 + 6y2 = 21
(c) hyperbola 6x2 – –3y2 = 8
(d) ellipse 9x2 + 4y2 = 1
(e) ellipse 16x2 + y2 = 16
(f) hyperbola 3y2 – 7x2 = 1
(g) circle x2 + y2 = 24
6. (a)
(b)
(c)
(d)
−
=9 0 0
0 4 0
0 0 1
0[ ]x y z
x
y
z
6 0 0
0 3 0
0 0 2
6[ ]x y z
x
y
z
−−
− == 0
[[ ]x y z
x
y
z
2 0 0
0 6 0
0 0 3
18
−
− = 00
[ ]x y z
x
y
z
36 0 0
0 9 0
0 0 4
36 0
− =
485
(e)
(f)
(g)
7. (a) If we complete the squares, the quadratic becomes
9(x2 – 2x + 1) + 36(y2 – 4y + 4) + 4(x2 – 6z + 9)
= –153 + 9 + 144 + 36
or
9(x – 1) 2 + 36(y – 2) 2 + 4z – 3) 2 = 36
or
This is an ellipsoid.
(c) If we complete the square, the quadratic becomes
3(x2 + 14z + 49) – 3y2 – z2 = 144 + 147
or
3(x + 7) 2 – 3y2 – z2 = 3
( ) ( ) ( )′+
′+
′=
x y z2 2 2
4 1 91
x y[ zz
x
y
z
]
1 0 0
0 1 0
0 0 1
25 0
− =
7 0 0
0 3 0
0 0 1
−
x y z[ ]
+
=x
y
z
x
y
z
0 0 1 0
[[ ]x y z
x
y
z
16 0 0
0 1 0
0 0 0
0 0
+ −66 0
=
x
y
z
486 Exercise Set 9.7
or
This is a hyperboloid of two sheets.
7. (e) If we complete the squares, the quadric becomes
(x2 + 2x + 1) + 16(y2 – 2y + 1) – 16z = 15 + 1 + 16
or
(x + 1)2 + 16(y – 1)2 – 16(z + 2) = 0
or
This is an elliptic paraboloid.
(g) If we complete the squares, the quadric becomes
(x2 – 2x + 1) + (y2 + 4y + 4) + (z2 – 6z + 9) = 11 + 1 + 4 + 9
or
(x – 1)2 + (y + 2)2 + (z – 3)2 = 25
or
This is a sphere.
( ) ( ) ( )′+
′+
′=
x y z2 2 2
1 1 31
( ) ( ) ( )′+
′+
′=
x y z2 2 2
4 1 91
( ) ( ) ( )′+
′+
′=
x y z2 2 2
1 1 31
Exercise Set 9.7 487
8. (a) The matrix form for the quadric is xT Ax + 150 = 0 where
The eigenvalues of A are λ1 = –25, λ2 = 3, and λ3 = 50. The corresponding eigenspacesare spanned by the orthogonal vectors
Thus we can let x = Px′ where
Note that det(P) = 1. (This is the only reason for actually calculating P in thisproblem.) Using the above transformation, we obtain the quadric
–25(x′)2 + 3(y′)2 + 50(z′)2 + 150 = 0
or
25(x′)2 – 3(y′)2 – 50(z′)2 – 150 = 0
This is a hyperboloid of two sheets.
(c) The matrix form for the quadric is xT Ax + Kx = where
The eigenvalues of A are λ1 = 225, λ2 = 100, and λ3 = 0. The eigenspaces are spannedby the vectors
4
0
3
0
1
0
3
0
4−
A K=−
−
= − −144 0 108
0 100 0
108 0 81
540 0and 7720
P P APT=
−
=−4 5 0 3 5
0 1 0
3 5 0 4 5
25 0 0
0 3 0
0
and
00 50
4
0
3
0
1
0
3
0
4−
A =
2 0 36
0 3 0
36 0 23
488 Exercise Set 9.7
Thus we can let x = Px′ where
Note that det(P) = 1. Hence the quadric becomes
xT (PT AP)x + KPx′ = 0
or
225(x′)2 + 100(y′)2 – 900z′ = 0
This is the elliptic paraboloid
9(x′)2 + 4(y′)2 – 36z′ = 0
9. The matrix form for the quadric is xT Ax + Kx = –9 where
The eigenvalues of A are λ1 = λ2 = –1 and λ3 = 2, and the vectors
span the corresponding eigenspaces. Note that u1 • u3 = u2 • u3 = 0 but that u1 • u2 ≠ 0.Hence, we must apply the Gram-Schmidt process to u1, u2. We must also normalize u3.This gives the orthonormal set
−
−
−
1
2
0
1
2
1
65
61
6
−
1
31
31
3
u
-1
0
0
u and u1 1 1
1
1
0
1
1
1
=
=−
=
A K=
= − − −
0 1 1
1 0 1
1 1 0
6 6 4and
P P APT=
−
=4 5 0 3 5
0 1 0
3 5 0 4 5
225 0 0
0 100and 00
0 0 0
Exercise Set 9.7 489
Thus we can let
Note that det(P) = 1,
Therefore the transformation x = Px′ reduces the quadric to
If we complete the squares, this becomes
Letting and yie′′ = ′ + ′′ = ′ + ′′ = ′ −x x y y z z1
2
1
16
4
3, , llds
This is the hyperb
( ) ( ) ( )′′ + ′′ − ′′ = −x y z2 2 22 1
ooloid of two sheets
( ) ( ) ( )′′−
′′−
′′=
z x y2 2 2
1 2 1 11
( ) ( )
(
′ + ′ +
+ ′ + ′ +
− ′
x x y y
z
2 221
2
2
6
1
6
2 ))2 8
3
16
39
1
2
1
6
32
3− ′ +
= + + −z
− ′ − ′ + ′ − ′ − ′ ′ = −( ) ( ) ( ) –x y z x y z2 2 22 2
2
6
16
39
P AP KPT =
= −− −
-1 0 0
0 -1 0
0 0 2
and 22
6
16
3
P =
−
− −
1
2
1
6
1
3
02
6
1
31
2
1
6
1
3
490 Exercise Set 9.7
11. The matrix form for the quadric is xT Ax + Kx – 31 = where
The eigenvalues of A are λ1 = 1, λ2 = –1, and λ3 = 0, and the corresponding eigenspaces arespanned by the orthogonal vectors
Thus, we let x = Px where
Note that det(P) = 1,
Therefore, the equation of the quadric is reduced to
(x′)2 – (y′)2 + 2√2x′ + 8√2y′ + z′ – 31 = 0
If we complete the squares, this becomes
[(x′)2 + 2√2x′ + 2] – [(y′)2 – 8√2y′ + 32] + z′ = 31 + 2 –32
Letting x′′ = x′ + √2, y′′ = y′ – 4√2, and z′′ = z′ – 1 yields
(x′′)2 – (y′′)2 + z′′ = 0
This is a hyperbolic paraboloid.
P AP KPT = −
=
1 0 0
0 1 0
0 0 0
2 2 8 2 1and
P =
−
1
2
1
20
1
2
1
20
0 0 1
1
1
0
1
1
0
0
0
1
−
A K=
= −
0 1 0
1 0 0
0 0 0
6 10 1and
Exercise Set 9.7 491
13. We know that the equation of a general quadric Q can be put into the standard matrix formxT Ax + Kx + j = where
Since a is a symmetric matrix, then A is orthogonally diagonalizable by Theorem 7.3.1.Thus, by Theorem 7.2.1, A has 3 linearly independent eigenvectors. Now let T be the matrixwhose column vectors are the 3 linearly independent eigenvectors of A. It follows from theproof of Theorem 7.2.1 and the discussion immediately following that theorem, that T–1 AT
will be a diagonal matrix whose diagonal entries are the eigenvalues λ1, λ2, and λ3 of A.Theorem 7.3.2 guarantees that these eigenvalues are real.
As noted immediately after Theorem 7.3.2, we can, if necessary, transform the matrix Tto a matrix S whose column vectors form an orthonormal set. To do this, orthonormalize thebasis of each eigenspace before using its elements as column vectors of S.
Furthermore, by Theorem 6.5.1, we know that S is orthogonal. It follows from Theorem6.5 2 that det(S) = ±1.
In case det(S) = –1, we interchange two columns in S to obtain a matrix P such thatdet(P) = 1. If det(S) = 1, we let P = S. Thus, P represents a rotation. Note that P isorthogonal, so that P–1 = PT, and also, that P orthogonally diagonalizes a. In fact,
Hence, if we let x = Px′, then the equation of the quadric Q becomes
(x′)T (PT AP)x′ + KPx′ + j = 0
or
λ1(x′)2 + λ2(y′)2 + λ3(z′)2 + g′x′ + h′y′ + i′z′ + j = 0
where
[g′ h′ i′] = KP
Thus we have proved Theorem 9.7.1.
P APT =
λλ
λ
1
2
3
0 0
0 0
0 0
A
a d e
d b f
e f c
K g h i=
= and
492 Exercise Set 9.7
EXERCISE SET 9.8
1. If AB = C where A is m × n, B is n × p, and C is m × p, then C has mp entries, each of theform
cij
= ai1b1j
+ ai2b2j
+ … + ain
bnj
Thus we need n multiplications and n – 1 additions to compute each of the numbers cij.
Therefore we need mnp multiplications and m(n – 1)p additions to compute C.
2. If A is an n × n matrix, then for any other n × n matrix, B, it will take (by Exercise 1) n3
multiplications and n2(n – 1) additions to compute AB. To compute Ak, we must carry outk – 1 of these matrix multiplications, one to compute a2, a second to compute a3 = A(A2),and in all, k – 1 to compute Ak = A(Ak – 1). Therefore we must carry out a total of n3(k –1) multiplications and n2(n – 1)(k – 1) additions to compute the product Ak.
5. Following the hint, we have
Sn
= 1 + 2 + 3 + … + n
Sn
= n + (n – 1) + (n – 2) + … + 1
or
2Sn
= (n + 1) + (n + 1) + (n + 1) + … + (n + 1)
Thus
Sn n
n =+( )1
2
493
7. (a) By direct computation,
(k + 1)3 – k3 = k3 + 3k2 + 3k + 1 – k3 = 3k2 + 3k + 1
(b) The sum “telescopes”. That is,
[23 – 13] + [33 – 23] + [43 – 33] + … + [(n + 1)3 – n3]
= 23 – 13 + 33 – 23 + 43 – 33 + … + (n + 1)3 – n3
= (n + 1)3 – 1
(c) By Parts (a) and (b), we have
3(1)2 + 3(1) + 1 + 3(2)2 + 3(2) + 1 + 3(3)2 + 3(3) + 1 + … + 3n2 + 3n + 1
= 3(12 + 22 + 32 + … + n2) + 3(1 + 2 + 3 + … + n) + n
= (n + 1)3 – 1
(d) Thus, by Part (c) and exercise 6, we have
1 2 31
31 1 3
1
22 2 2 2 3+ + + + = + − −
+−
=
n nn n
n( )( )
(nn n n n
n n n n
+− −
+−
=+ − − + −
1
3
1
3
1
2 3
2 1 2 3 1 2
6
3
3
) ( )
( ) ( )
==+ + − −
=+ + + − −
( )[ ( ) ]
( )( )
n n n
n n n n
1 2 1 3 2
6
1 2 4 2 3 2
2
2
66
1 2
61 2 1
6
2
=+ +
=+ +
( )( )
( )( )
n n n
n n n
494 Exercise Set 9.8
9. Since R is a row-echelon form of an invertible n × n matrix, it has ones down the maindiagonal and nothing but zeros below. If, as usual, we let x = [x1 x2
… xn]T and b = [b1 b2
… bn]T, then we have x
n= b
nwith no computations. However, since x
n–1 = bn–1 – cx
nfor
some number c, it will require one multiplication and one addition to find xn–1. In general,
xi
= bi
– some linear combination of xi+1, xi+2, …, x
n
Therefore it will require two multiplications and two additions to find xn–2, three of each to
find xn–3, and finally, n–1 of each to find x1. That is, it will require
multiplications and the same number of additions to solve the system by back substitution.
10. To reduce an invertible n × n matrix to In, we first divide Row 1 by a11 (which we are
assuming is not zero since no row interchanges are required). This requires n – 1multiplications because we can ignore the first row, first column position which must beone. We then subtract a
i1 times Row 1 from Row i for i = 2, 3, … , n to reduce the firstcolumn to that of I
n. This requires (n – 1)2 multiplications and the same number of
additions because we can ignore those entries which we know must be zero. Thus we haveused n – 1 + (n – 1)2 = n(n – 1) multiplications and (n – 1)2 additions.
We then repeat the process on the (n – 1) × (n – 1) submatrix obtained from theoriginal by deleting the first row and the first column. This will require (n – 1)(n – 2)multiplicaitons and (n – 2)2 additions.
We continue in this fashion until we have reduced every diagonal entry to one and everyentry below the diagonal to zero. The number of multiplications required so far is
n n n n n n
n n
( ) ( )( ) ( )( )− + − − + − − + + • + •
= −
1 1 2 2 3 3 2 2 12 ++ − − − + − − − + − + −
= +
( ) ( ) ( ) ( )
(
n n n n
n
1 1 2 2 3 3 2 22 2 2 2
2nn n n
n n
− + + +
− + − + + +
=+
1 3 2 1 3 2
1
2 2 2) ( )
( )(( ) ( )2 1
61
1
21
3
3
n n n
n n
+− −
+−
=−
1 2 3 11
1+ + + + − =
−( )
( )n
n n
Exercise Set 9.8 495
and the number of additions is
Finally, we must reduce the entries above the main diagonal to zero. We start with thenth column. We multiply Row n by a
infor i = 1, 2, … , n – 1 and subtract the result from
Row i. This would appear to involve n – 1 multiplications and the same number of additionssince we can ignore everything but the entries in the last column. However, the result is aforegone conclusion. In fact, once we have reduced the matrix to row-echelon form, thefinal reduction is automatic. Thus the above formulas give all the multiplications andadditions required.
11. To solve a linear system whose coefficient matrix is an invertible n × n matrix, A, we formthe n × (n + 1) matrix [A|b] and reduce A to I
n. Thus we first divide Row 1 by a11, using n
multiplications (ignoring the multiplication whose result must be one and assuming thata11 ≠ 0 since no row interchanges are required). We then subtract a
i1 times Row 1 fromRow i for i = 2, … , n to reduce the first column to that of I
n. This requires n(n – 1)
multiplications and the same number of additions (again ignoring the operations whoseresults we already know). The total number of multiplications so far is n2 and the totalnumber of additions is n(n – 1).
To reduce the second column to that of In, we repeat the procedure, starting with Row
2 and ignoring Column 1. Thus n – 1 multiplications assure us that there is a one on themain diagonal, and (n – 1)2 multiplications and additions will make all n – 1 of theremaining column entries zero. This requires n(n – 1) new multiplications and (n – 1)2
new additions.
In general, to reduce Column i to the ith column of In, we require n + 1 – i
multiplications followed by (n + 1 – i)(n – 1) multiplications and additions, for a total ofn(n + 1 – i) multiplications and (n + 1 – i)(n – 1) additions.
If we add up all these numbers, we find that we need
n n n n n n n n n n
n n
2
2
1 2 2 1 1+ − + − + + + = + −
=
( ) ( ) ( ) ( ) ( ( )
( ++
= +
1
2
2 2
3 2
)
n n
( ) ( )( )( ) )
n nn n n
n n
− + − + + + =− −
= −
1 2 2 11 2 1
6
3
2 2 2 2
3 2
22 6+
n
496 Exercise Set 9.8
multiplications and
additions to compute the reduction.
12.
Since we have
lim li
lim ( ) ,
( )
x
xk
k
x
P x
a x
→+∞
→+∞
=
=
1 0
mm
lim
x
kk
kk
xk
kk
a a x a x
a x
a
a x
a
a x
→+∞
→+∞
+ + +
= +
0 1
0 1kk
k
k
a
a x−−+ +
= + + + +
=
11
0 0 0 1
1
n n n n n n n
n
( ) ( ) ( )( ) ( ) ( )
(
− + − + − − + + − + −
= −
1 1 2 1 2 1 12
11 1 2 1
1 1
2
2 2
3
)( ( ) )
( )( )( )
n n
n n n
n n
+ − + + +
=− +
= −
Exercise Set 9.8 497
EXERCISE SET 9.9
1 The system in matrix form is
This reduces to two matrix equations
and
The second matrix equation yields the system
3y1 = 0
–2y1 + y2 = 1
which has y1 = 0, y2 = 1 as its solution. If we substitute these values into the first matrixequation, we obtain the system
x1 – 2x2 = 0
x2 = 1
This yields the final solution x1 = 2, x2 = 1.
3 0
2 1
0
11
2−
=
y
y
1 2
0 11
2
1
2
−
=
x
x
y
y
3 6
2 5
3 0
2 1
1 2
0 11
2
−− −
=
−
−
x
x
=
x
x
1
2
0
1
499
3. To reduce the matrix of coefficients to a suitable upper triangular matrix, we carry out thefollowing operations:
These operations involve multipliers 1/2, 1, and 1/3. Thus the corresponding lowertriangular matrix is
We therefore have the two matrix equations
and
The second matrix equation yields the system
2y1 = –2
–y1 + 3y2 = –2
which has y1 = –1, y2 = –1 as its solution. If we substitute these values into the first matrixequation, we obtain the system
x1 + 4x2 = –1
x2 = –1
This yields the final solution x1 = 3, x2 = –1.
2 0
1 3
2
21
2−
=
−−
y
y
1 4
0 11
2
1
2
=
x
x
y
y
L =−
2 0
1 3
2 8
1 1
1 4
1 1
1 4
0 3
1 4
0 1− −
→
− −
→
→
= u
500 Exercise Set 9.9
5. To reduce the matrix of coefficients to a suitable upper triangular matrix, we carry out thefollowing operations:
These operations involve multipliers of 1/2, 0 (for the 2 row), 1, –1/2, –4, and 1/5. Thus thecorresponding lower triangular matrix is
We therefore have the matrix equations
and
The second matrix equation yields the system
2y1 = –4
–2y2 = –2
–y1+ 4y2 + 5y3= 6
2 0 0
0 2 0
1 4 5
4
2
6
1
2
3
−−
=−−
y
y
y
1 1 1
0 1 1
0 0 1
1
2
3
1
2
− −−
=x
x
x
y
y
yy3
L = −−
2 0 0
0 2 0
1 4 5
2 2 2
0 2 2
1 5 2
1 1 1
0 2 2
1 5 2
− −−
−
→− −−
−
→− −−
→
− −−
1 1 1
0 2 2
0 4 1
1 1 1
0 1 1
0 4 1
→− −
−
→− −
−1 1 1
0 1 1
0 0 5
1 1 1
0 1 1
0 00 1
= U
Exercise Set 9.9 501
which has y1 = –2, y2 = 1 and y3 = 0 as its solution. If we substitute these values into thefirst matrix equation, we obtain the system
x1 – x2 – x3 = –2
x2 – x3 = 1
x3 = 0
This yields the final solution x1 = –1, x2 = 1, x3 = 0.
11. (a) To reduce A to row-echelon form, we carry out the following operations:
This involves multipliers 1/2, 2, -2, 1 (for the 2 diagonal entry), and –1. Where nomultiplier is needed in the second entry of the last row, we use the multiplier 1, thusobtaining the lower triangular matrix
In fact, if we compute LU, we see that it will equal A no matter what entry we choosefor the lower right-hand corner of L.
If we stop just before we reach row-echelon form, we obtain the matrices
which will also serve.
U L=−
= −
1 1 2 1 2
0 0 1
0 0 1
2 0 0
2 1 0
2 0 1
L = −
2 0 0
2 1 0
2 1 1
2 1 1
2 1 2
2 1 0
1 1 2 1 2
2 1 2
2 1 0
−− −
→−
− −
→−
→−
1 1 2 1 2
0 0 1
2 1 0
1 1 2 1 2
0 0 1
0 0 1
→−
=1 1 2 1 2
0 0 1
0 0 0
U
502 Exercise Set 9.9
(b) We have that A = LU where
If we let
then A = L1DU as desired. (See the matrices at the very end of Section 9.9.)
(c) Let U2 = DU and note that this matrix is upper triangular. Then A = L1U2 is of therequired form.
12. (a) Suppose that
This implies that ad = 0 but that ae = bd = 1. That is, if the matrix has an LU-decomposition, then we must have 4 numbers which satisfy the above equations. Nowif ad = 0, then either a or d must be 0. However, this cannot be true if ae = bd = 1.Thus the matrix has no such decomposition.
(b)
13. (a) If A has such an LU-decomposition, we can write it as
a b
c d w
x y
z
x y
wx yw z
=
=
+
1 0
1 0
A P L U
P L
=
=
=
=
0 1
1 0
0 1
1 0
1 0
0 1
,
,where
,
0 1
1 0
0
0
=
=
+a
b c
d e
f
ad ae
bd be cf
L D1
1 0 0
1 1 0
1 1 1
2 0 0
0 1 0
0 1 1
= −
=
and
L U= −
=−
2 0 0
2 1 0
2 1 1
1 1 2 1 2
0 0 1
0 0 1
Exercise Set 9.9 503
This yields the system of equations
x = a y = b wx = c yw + z = d
Since a ≠ 0, this has the unique solution
x = a y = b w = c/a z = (ad – bc)/a
The uniqueness of the solution guarantees the uniqueness of the LU-decomposition.
13. (b) By the above,
15. We have that L = E1–1 E2
–1 … Ek–1 where each of the matrices E
iis an elementary matrix
which does not involve interchanging rows. By Exercise 27 of Section 2.4, we know that ifE is an invertible lower triangular matrix, then E–1 is also lower triangular. Now the matricesE
iare all lower triangular and invertible by their construction. Therefore for i = 1, … , k we
have that Ei–1 is lower triangular. Hence L, as the product of lower triangular matrices, must
also be lower triangular.
17. Let A be any n × n matrix. We know that a can be reduced to row-echelon form and thatthis may require row interchanges. If we perform these interchanges (if any) first, wereduce A to the matrix
Ek
… E1A = B
where Ei
is the elementary matrix corresponding to the ith such interchange. Now we knowthat B has an LU-decomposition, call it LU where U is a row-echelon form of A. That is,
Ek
… E1A = LU
where each of the matrices Ei
is elementary and hence invertible. (In fact, Ei–1 = E
ifor all
Ei. Why?) If we let
P = (Ek
… E1)–1 = E1
–1 … Ek–1 if k > 0
and P = I if no row interchanges are required, then we have A = PLU as desired.
a b
c d c a
a b
ad bc a
=
−( )
1 0
1 0
504 Exercise Set 9.9
18. First we interchange Row 2 and Row 3 of A to obtain the equation
Next we find an LU-decomposition of B as follows
This gives us
so that B = EA = LU or A = E–1LU. Since
we have A = ELU, or
A =
−1 0 0
0 0 1
0 1 0
3 0 0
0 2 0
3 0 1
1 1 33 0
0 1 1 2
0 0 1
E E− =
=11 0 0
0 0 1
0 1 0
L =
3 0 0
0 2 0
3 0 1
B →−
−
→−
1 1 3 0
0 2 1
3 1 1
1 1 3 0
0 2 1
0 0 1
→−
=1 1 3 0
0 1 1 2
0 0 1
U
B =−
−
=
3 1 0
0 2 1
3 1 1
1 0 0
0 0 1
0 1 0
33 1 0
3 1 1
0 2 1
−−
= EA
Exercise Set 9.9 505
19. Assume A = PLU, where P is a permutation matrix. Then note P–1 = P. To solve AX = B,where A = PLU, set C = P–1B = PB and Y = UX.
First solve LY = C for Y.
Then solve UX = Y for X.
To solve
3 1 0
3 1 1
0 2 1
0
1
0
1
2
3
−−
=
x
x
x
=
=
, ,or AX
set
e
C Pe
2
2
1 0 0
0 0 1
0 1 0
=
0
1
0
0
0
1
Solve LY ==
C
y
y
y
, or =
3 0 0
0 2 0
3 0 1
1
2
3
00
0
1
0
0
1
1
2
3
, to get =
y
y
y
=−
Solve orUX Y ,
1 1 3 0
0 1 1 2
0 0 1
x
x
x
x
x
x
1
2
3
1
2
3
0
0
1
=
so
=−
1 6
1 2
1
If thenA A P L U=−−
=3 1 0
3 1 1
0 2 1
, ,
wwith P L=
=
1 0 0
0 0 1
0 1 0
3 0 0
0 2 0
3 0 1
,
=−
, U
1 1 3 0
0 1 1 2
0 0 1
506 Exercise Set 9.9
EXERCISE SET 10.1
3. (b) Since two complex numbers are equal if and only if both their real and imaginary partsare equal, we have
x + y = 3
and
x – y = 1
Thus x = 2 and y = 1.
4. (a) z1 + z2 = 1 – 2i + 4 + 5i = 5 + 3i
(c) 4z1 = 4(1 – 2i) = 4 – 8i
(e) 3z1 + 4z2 = 3(1 – 2i) + 4(4 + 5i)
= 3 – 6i + 16 + 20i
= 19 + 14i
5. (a) Since complex numbers obey all the usual rules of algebra, we have
z = 3 + 2i – (1 – i) = 2 + 3i
(c) Since (i – z) + (2z – 3i) = –2 + 7i, we have
i + (–z + 2z) – 3i = –2 + 7i
or
z = –2 + 7i – i + 3i = –2 + 9i
507
6. (a) z1 + z2 = 4 + 5i and z1 – z2 = 2 – 3i
7. (b) –2z = 6 + 8i
8. (a) We have k1i + k2(1 + i) = k2 + (k1 + k2)i = 3 – 2i. Hence, k2 = 3 and k1 + k2 = – 2or k1 = –5.
508 Exercise Set 10.1
9. (c)
10. (a) First, z1z2 = (2 – 5i)(–1 – i) = –7 + 3i. Thus,
z1 – z1z2 = (2 – 5i) – (–7 + 3i) = 9 – 8i
(c) Since 1 + z2 = –i, then z1 + (1 + z2) = 2 – 6i. Thus,
[z1 + (1 + z2)]2 = (2 – 6i)2 = 22(1 – 3i)2 = 4(–8 – 6i) = –32 – 24i
11. Since (4 – 6i)2 = 22(2 – 3i)2 = 4(–5 – 12i) = –4(5 + 12i), then
(1 + 2i)(4 – 6i)2 = –4(1 + 2i)(5 + 12i) = –4(–19 + 22i) = 76 – 88i
13. Since (1 – 3i)2 = –8 – 6i = –2(4 + 3i), then
(1 – 3i)3 = –2(1 – 3i)(4 + 3i) = –2(13 – 9i)
15. Since then( ) ,
( )
21
2
3
4
1
42
21
2
+ +
= +
+ +
i i i
i33
4
1
42
63
16
2 2
i i i
= +
= − +
z z i i i i i1 21
62 4 1 5
2
61 2 1 5
1
31 3= + − = + − = − +( )( ) ( )( ) ( 110
11
34
91 2
4
93 4
1
41 5
12 2
22
)
( ) ( )
(
= −
= + = − +
= −
i
z i i
z ii i i) ( )2 1
424 10 6
5
2= − − = − −
Exercise Set 10.1 509
17. Since i2 = –1 and i3 = –i, then 1 + i + i2 + i3 = 0. Thus (1 + i + i2 + i3)100 = 0.
19. (a)
(d)
Hence
20. (a)
(c) We have
Hence
( )CA Bi i
i i
2 6 1
6 5 9=
+− − −
CAi i
i i=
− − −+ − +
=
−6 1
6 5 9
1 0and B2
00 1−
A BC
i
i
i i
i i
i( ) =
+−+ −
+ −3 2 0
2
1 1
5 4 11
3 −− −
=+ − + − −++ −
2 5
13 13 8 12 33 22
1 0
7 9
i
i i i
i i
i 66 6 16 16+ − −
i i
B Ai i
i i
2 2 9 12 2
18 2 13− =
+ +− +
Ai
i
i i
i i
2 2 4
4 10
11 12 6
18 6 23=
−
=
+ +− +
and B2
A iBi
i
i i
i i
i
+ =−
+
− ++
=+
31
3
6 3 6
3 9 12
1 6 −− ++ +
3 7
3 8 3 12
i
i i
510 Exercise Set 10.1
21. (a) Let z = x + iy. Then
Im(iz) = Im[i(x + iy)] = Im(–y + ix) = x
= Re(x + iy) = Re(z)
22. (a) We have
Let z1 = –1 + i and z2 = – 1 – i. Then
z12 + 2z1 + 2 = (–1 + i)2 + 2(–1 + i) + 2
= –2i – 2 + 2i + 2
= 0
Thus, z1 is a root of z2 + 2z + 2. The verification that z2 is also a root is similar.
23. (a) We know that i1 = i, i2 = –1, i3 = –i, and i4 = 1. We also know that im + n = imin
and imn = (im)n where m and n are positive integers. The proof can be broken intofour cases:
1. n = 1, 5, 9, … or n = 4k + 1
2. n = 2, 6, 10, … or n = 4k + 2
3. n = 3, 7, 11, … or n = 4k + 3
4. n = 4, 8, 12, … or n = 4k + 4
where k = 0, 1, 2, . . . . In each case, in = i4k+ for some integer between 1 and 4.Thus
in = i4ki = (i4)k i = 1 ki = i
This completes the proof.
(b) Since 2509 = 4 ⋅ 627 + 1, Case 1 of Part (a) applies, and hence i2509 = i.
z z z i2 2 2 02 4 8
21+ + = = − ± − = − ±⇔
Exercise Set 10.1 511
24. Let z1 = x1 + iy1 and z2 = x2 + iy2 and observe that
z1z2 = 0 ⇔ x1x2 = y1y2 and x1 y2 = –x2y1
If we suppose that x1 ≠ 0, then we have
This implies that
But since – (y1/x1)2 ≤ 0, the above equation can hold if and only if y2 = 0. Recall that
x2 = (y1y2)/x1; then y2 = 0 implies that x2 = 0 and hence that z2 = 0. Thus if x1 ≠ 0,then z2 = 0. A similar argument shows that if y1 ≠ 0, then z2 = 0. Therefore, if z1 ≠ 0,then z2 = 0. Similarly, if z2 ≠ 0, then z1 = 0.
25. Observe that zz1 = zz2 ⇔ zz1 – zz2 = 0 ⇔ z(z1 – z2) = 0. Since z ≠ 0 by hypothesis, itfollows from Exercise 24 that z1 – z2 = 0, i.e., that z1 = z2.
26. (a) Let z1 = x1 + iy1 and z2 = x2 + iy2. Then
z1 + z2 = (x1 + iy1) + (x2 + iy2)
= (x1 + x2) + i(y1 + y2)
= (x2 + x1) + i(y2 + y1)
= (x2 + iy2) + (x1 + iy2)
= z2 + z1
yy
xy x2
1
1
2
2= −
xy y
xy
x y
x21 2
12
2 1
1= =
−and
512 Exercise Set 10.1
27. (a) Let z1 = x1 + iy1 and z2 = x2 + iy2. Then
z1z2= (x1 + iy1)(x2 + iy2)
= (x1x2 – y1y2) + i(x1y2 + x2y1)
= (x2x1 – y2y1) + i(y2x1 + y1x2)
= (x2 + iy2)(x1 + iy1)
= z2z1
30. They lie on a circle of radius one, centered at the origin.
If z1 = a1b1i, and z2 = a2 + b2i, where a12 + b1
2 = 1, a22 + b2
2 = 1, thenz1z2 = (a1a2 – b1b2) + i(a1b2 + a2b1).
Note that (a1a2 – b1b2)2 + (a1b2 + a2b1)
2
= a12a2
2 – 2a1a2b1b2 + b12b2
2 + a12b2
2 + 2a1a2b1b2 + a22b1
2
= a12a2
2 + b12b2
2 + a12b2
2 + a22b1
2
= (a12 + b1
2)(a22 + b2
2) = 1.
Hence z1z2 also lies on the unit circle.
Exercise Set 10.1 513
EXERCISE SET 10.2
2. (a) Since i = 0 + 1i, then
(c)
(e)
3. (a) We have
z–z = (2 – 4i)(2 + 4i) = 20
On the other hand,
z2 = 22 + (–4)2 = 20
(b) We have
z–z = (–3 + 5i)(–3 – 5i) = 34
On the other hand,
z2 = (–3)2 + 52 = 34
4. (a) From Equation (5), we have
z
z
i
i
i i i1
22 2
1 5
3 4
1 5 3 4
3 4
17 19
25= −
+= − −
+= − −( )( )
− = − + = − + =8 8 0 8 0 82 2i ( )
− − = − + − = =3 4 3 4 25 52 2i ( ) ( )
i = + =0 1 12 2
515
ch10.2.qxd 9/30/05 1:29 PM Page 515
4. (c) Again using Equation (5), we have
(e) Since z2 = 5, we have
5. (a) Equation (5) with z1 = 1 and z2 = i yields
(c)
6. (a) Since
then
(c) Since
and
then
ziz
zi i i1
2 1
2
23
5
1
5
3
5
11
5− = − − −
= +
z i21 2=
iz
z
i i
i
i
i
i i
i
1
2
1
1 2
1
1 2
1 1 2
1 2= +
−= − +
−= − + +
−( ) ( )( )
( ))( )1 2
3
5
1
5+= − −
ii
zz
zi i i1
1
2
11
5
3
5
6
5
2
5− = + − − +
= +( )
z
z
i
i
i ii1
2
1
1 2
1 1 2
5
1
5
3
5= +
−= + + = − +( )( )
1 7 7
17
z i
ii=
−= =
( )
1 1
1i
ii=
−= −
( )
z
z
ii
1
2
1 5
5
1
5=
−= −
z
z
i
i
i i i1
22 2
1 5
3 4
1 5 3 4
3 4
23 11
25=
−
−=
− +
+ −=
−( )( )
( )
516 Exercise Set 10.2
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7. Equation (5) with z1 = i and z2 = 1 + i gives
9. Since (3 + 4i)2 = –7 + 24i, we have
11. Since
13. We have
and
(1 – 2i) (1 + 2i) = 5
Thus
i
i i i
ii
( )( )( )1 1 2 1 2 5
1
10
1
10
12
12
− − +=
− += − +
i
i
i ii
1
1
2
1
2
1
2−= + = − +( )
3
3
3
4
2 2 3
4
1
2
3
2
3
1 3
2+−
=+
=+
= +
+− −
i
i
i ii
i
i
( )
( )(
then
ii
i
i
i i
)=
+
−=
+( ) +( )=
−+
+
12
32
12
32
1
1
2
1 3
4
1 3
4 i
1
3 4
7 24
7 24
7 24
62 2 2( ) ( ) ( )+=
− −− + −
=− −
i
i i
225
i
i
i ii
1
1
2
1
2
1
2+= − = +( )
Exercise Set 10.2 517
ch10.2.qxd 9/30/05 1:29 PM Page 517
15. (a) If iz = 2 – i, then
16. (a)
(c) The result follows easily from the fact that This result holdswhenever z ≠ i.
17. (a) The set of points satisfying the
equation z = 2 is the set of all
points representing vectors of length 2.
Thus, it is a circle of radius 2 and center
at the origin.
Analytically, if z = x + iy, then
which is the equation of the above circle.
(c) The values of z which satisfy the equation z – i = z + i are just those z whosedistance from the point i is equal to their distance from the point –i. Geometrically,then, z can be any point on the real axis.
We now show this result analytically. Let z = x + iy. Then
z – i = z + i ⇔ z –i2 = z + i2
⇔ x + i(y – 1)2 = x + i(y + 1)2
⇔ x2 + (y – 1)2 = x2 + (y + 1)2
⇔ – 2y = 2y
⇔ y = 0
z x y
x y
= ⇔ + =
⇔ + =
2 2
4
2 2
2 2
i z i z z i+ = + = − .
z i z i z i+ = + = −5 5 5
zi
i
i ii= − = − − = − −2 2
11 2
( )( )
518 Exercise Set 10.2
ch10.2.qxd 9/30/05 1:29 PM Page 518
18. (a) This inequality represents the set of
all points whose distance from the
point –i is at most 1.
Analytically let
Thus, the inequality represents the set of all points in or on the circle with center at
–i and radius 1.
(c) Since
this inequality represents the set
of all points whose distance from
the point 2i is less than 1/2.
Analytically, if z = x + iy, then
19. (a) Re(iz–) = Re(i
–z–) = Re[(–i)(x – iy)] = Re(–y – ix) = –y
(c) Re(iz–) = Re[i(x – iy)] = Re(y + ix) = y
20. (a) Since 1/i = –i and (–i)n = (–1)n(i)n, this problem is a variation on Exercise 23 ofSection 10.1.
(b) Since 2509 = 4(627) + 1 and (i)4 = 1, then (i)2509 = i and therefore (1/i)2509 =(–1)2509 • i = –i.
2 4 1 2 4 1
2 2 4 1
2
2
2 2
2 2
z i z i
x y
x y
− < ⇔ − <
⇔ + − <
⇔ + − <
( ) ( )
( )11
4
2 4 1 21
2z i z i− < ⇔ − < ,
z x iy
z i z i
x y
= +
+ ≤ ⇔ + ≤
⇔ + + ≤
.
( )
Then
1 1
1 1
2
2 2
Exercise Set 10.2 519
ch10.2.qxd 9/30/05 1:30 PM Page 519
21. (a) Let z = x + iy. Then
22. Let z = x + iy. Then
23. (a) Equation (5) gives
Thus
25.
26. Let z1 = x1 + iy1 and z2 = x2 + iy2.
(a) z z x iy x iy
x x i y y
x
1 2 1 1 2 2
1 2 1 2
− = + − +
= − + −=
( ) ( )
( ) ( )
( 11 2 1 2
1 1 2 2
1 2
− − −= − − −= −
x i y y
x iy x iy
z z
) ( )
( ) ( )
z x y x y z= + = + − =2 2 2 2( )
Rez
z
x x y y
x y
1
2
1 2 1 2
22
22
=
+
+
z
z z
z z
x yx iy x iy
x
1
2 22 1 2
22
22 1 1 2 2
22
1
1
1
=
=+
+ −
=
( ) ( )
+++ + −
yx x y y i x y x y
22 1 2 1 2 2 1 1 2( ) ( )
z z x iy x iy y z= ⇔ + = − ⇔ = ⇔0 is real.
1
2
1
2
1
22( ) ( ) ( ) ( ) Re( )z z x iy x iy x x z+ = + + − = = =
520 Exercise Set 10.2
ch10.2.qxd 9/30/05 1:30 PM Page 520
(c)
27. (a)
(b) We use mathematical induction. In Part (a), we verified that the result holds whenn = 2. Now, assume that (z
–)n = zn—. Then
and the result is proved.
28. Let A denote the matrix of the system, i.e.,
Thus det(A) = i – (–2i) = 3i. Hence
and
xi
i
i ii
ii2
21
3
2
2
1
34
1=−
= +( ) = = −
xi
i
i ii
ii1
21
3
2
1
1
32
1=− −
= − +( ) = − =
Ai i
=−
2 1
( ) ( )z z z
z z
z
n n
n
n
+
+
=
=
=
1
1
z zz z z z2 2= = = ( )
( / )
( )
/ (
z z
z
z z
z
z z
z z
1 2
22 1 2
22 1 2
1 2
1
1
=
=
=
• Part (b)
PPart (d))
Exercise Set 10.2 521
ch10.2.qxd 9/30/05 1:30 PM Page 521
30. Let A denote the matrix of the system, i.e.,
Then det(A) = –4 and hence
32.
Therefore x1 = –(1 + i)t and x2 = t where t is arbitrary.
− − −− + −
→
+− + −
→
+1 1
1 2
1 1
1 2
1 1
0 0
i
i
i
i
i
x i i
x i
1
2
1
4
3 1 1
2 2 1 1
1 1 1
1
2
1
4
1 3 1
1 2 2 1
1
= − + −− −
= +
= − + −−−
=
= +− −
= −
1 1
2
1
4
1 1 3
1 1 2 2
1 1 1
1
23x i i
A = −−
1 1 1
1 1 1
1 1 1
522 Exercise Set 10.2
ch10.2.qxd 9/30/05 1:30 PM Page 522
34.
If we let x3 = t, where t is arbitrary, then x2 = –it and x1 = –(1 – i)t.
35. (a)
It is easy to verify that AA–1 = A–1 A = I.
36. From Exercise 27(b), we have that . Since ak
is real, it follows that
Thus,
Hence, p z p z( ) ( ) .= ⇒ = =0 0 0
p z a a z a z a z
a a z a z a
nn( ) ( ) ( )= + + + +
= + + + +
0 1 22
0 1 22
nnn
nn
z
a a z a z a z
p z
= + + + +
=
0 1 22
( )
. . .
. . .
. . .
a z a zkk
kk( ) .=
( )z zn n=
Ai
i
i
i
i
− =+ −
=
−
1
2
1
2
2
1
2
1
1
1 1 2
2 1 2 3
1
0 1
0 1
i i
i i
i i
i i
i
i
−− −
− + −
→−
− −
→
−
→−1
0 1
0 0 0
1 0 1
0 1
0 0
i i
i
i
i
00
Exercise Set 10.2 523
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39. (a)
Thus
A
i i i
i
i i
− =− − − − +
−
12 2 1
1 2
1
1 1 0 1 0 0
0 1 0 1 0
1 2 2 0 0 1
1 1 0 1 0 0
0
+
− −
+
i
i
i i
i
11 0 1 0
0 2 0 1
1 1 0 1 0 0
0 1 0 1 0
0 0 1 1
i
i i
i
i
i i
−
+
+−
1 1 0 1 0 0
0 1 0 1 2
0 0 1 1
1
i
i
i i
00 0 2 2 1
0 1 0 1 2
0 0 1 1
− − − − +−
i i i
i
i i
524 Exercise Set 10.2
R R iR3 3 1→ +
R R iR3 3 2→ +
R R iR2 2 3→ −
R R i R2 1 21→ − +( )
ch10.2.qxd 9/30/05 1:30 PM Page 524
40. We have
and
Thus, z – 1 = z– – 1.Since z and z– are symmetric about the real axis, they are, in fact, equidistant from any
point on that axis.
41. (a) We have , which is just the distance between the two
numbers z1 and z2 when they are considered as points in the complex plane.
(b) Let z1 = 12, z2 = 6 + 2i, and z3 = 8 + 8i. Then
z1 – z22 = 62 + (–2)2 = 40
z1 – z32 = 42 + (–8)2 = 80
z2 – z32 = (–2)2 + (–6)2 = 40
Since the sum of the squares of the lengths of two sides is equal to the square of thethird side, the three points determine a right triangle.
z z a a b b1 2 1 22
1 22− = − + −( ) ( )
z z z z z z z− = − − = − − = − −1 1 1 1 1 1 12
( )( ) ( )( ) ( )( )
z z z z z− = − − = − −1 1 1 1 12
( )( ) ( )( )
Exercise Set 10.2 525
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ch10.2.qxd 9/30/05 1:30 PM Page 526
EXERCISE SET 10.3
1. (a) If z = 1, then arg z = 2kπ where k = 0, ±1, ±2,. . .. Thus, Arg(1) = 0.
(c) If z = –i, then arg z = + 2kπ where k = 0, ±1, ±2,. . .. Thus, Arg(–i) = –π/2.
(e) If then arg z = + 2kπ where k = 0, ±1, ±2,. . .. Thus, Arg
2. We have
(a) Put k = 0 in the above equation.
(b) Put k = –1 in the above equation.
3. (a) Since 2i = 2 and Arg(2i) = π/2, we have
(c) Since and Arg(5 + 5i) = π/4, we have
5 5 5 24 4
cos sin+ =
+
i i
π π
5 5 50 5 2+ = =i
2 22 2
i icos sin=
+
π π
arg( ) , ,1 35
32 0 2− = + = ±i k k
π π K
( ) .− + =1 32
3i
π
2
3
πz i,= − +1 3
3
2
π
527
ch10.3.qxd 9/30/05 2:25 PM Page 527
3. (e) since we have
4. We have z1 = 2, Arg(z1) = , z2 = 3, and Arg(z2) = .
(a) Here z1z2 = z1z2 = 6 and Arg(z) = Arg(z1) + Arg(z2) = . Hence
(c) Here z2/z1 = z2/z1 = 3/2 and Arg(z2/z1) = Arg(z2) – Arg(z1) = – . Hence
5. We have z1 = 1, Arg(z1) = , z2 = 2, Arg(z2) = – , z3 = 2, and Arg(z3) = . So
and
Therefore
= cos(0) + i sin(0) = 1
6. (a) We have θ = , and n = 12. Thus
(1 + i)12 = 26[cos(3π) + i sin(3π)] = –64
π4
r ,= 2
z z
z
1 2
3
Arg Arg( Arg Arg11 2 3
z z
zz z z2
3
0
= + − =) ( ) ( )
z z
z
z z
z
1 2
3
1 2
3
1= =
π6
π3
π2
z
zi2
1
3
2 12 12cos sin= −
+ −
π π
π12
z z i1 2 65
12
5
12=
+
cos sin
π π
5
12
π
π6
π4
− − = −
+ −
3 3 3 2
3
4
3
4i icos sin
π π
− − = = − − −3 3 18 3 2 3 33
4i iand Arg( ) =
π,
528 Exercise Set 10.3
ch10.3.qxd 9/30/05 2:25 PM Page 528
(c) We have r = 2, θ = , and n = 7. Thus
7. We use Formula (10).
(a) We have r = 1, θ = – , and n = 2. Thus
Thus, the two square roots of –i are:
(c) We have r = 27, θ = π, and n = 3. Thus
( ) cos sin− = +
+ +
27 33
2
3 3
2
31 2 π π π πk
ik
=k 0 1 2, ,
cos sin
cos
−
+ −
= −
π π
π4 4
1
2
1
2
3
4
i i
+
= − +i isin3
4
1
2
1
2
π
( ) cos sin ,− = − +
+ − +
=i k i k k1 2
4 40
π π π π 11
π2
3 27
6
7
6
2
7 7+( ) =
+
=
i icos sinπ π
77 3
2
1
264 3 64− −
= − −i i
π6
Exercise Set 10.3 529
ch10.3.qxd 9/30/05 2:25 PM Page 529
Therefore, the three cube roots of –27 are:
7. (e) Here r = 1, θ = π, and n = 4. Thus
Therefore the four fourth roots of –1 are:
cos sin
cos sin
π π
π π4 4
1
2
1
23
4
3
4
+ = +
+ =
i i
i
cos sin
cos
− +
+ = − −
1
2
1
25
4
5
4
1
2
1
27
i
i iπ π
ππ π4
7
4
1
2
1
2sin+ = −i i
( ) cos sin/− = +
+ +
14 2 4 2
1 4 π π π πki
k
=k 0 1 2 3, , ,
33 3
3
2
3 3cos sin
π π
+
= +i
22
3 3
35
3
i
icos( ) sin( )
cos
π π
π
+ = −
+ ii isin5
3
3
2
3 3
2
π
= −
530 Exercise Set 10.3
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9. We observe that w = 1 is one
sixth root of 1. Since the
remaining 5 must be equally
spaced around the unit circle, any
two roots must be separated
from one another by an angle of
= 60°. We show all six
sixth roots in the diagram.
11. We have z4 = 16 ⇔ z = 161/4. The fourth roots of 16 are 2, 2i, –2, and –2i.
12. All solutions to z4 + 8 = 0 are:
z i i
z
13 4 3 4
2
22
2
2
22
4 4= + +
= +
cos sinπ π
== − +
= +
22
2
2
22
3
4
3
43 4 3 4
i i
z
cos sinπ π
333 4 3 42
2
2
2
22
5
4
5
4= − −
= +
i icos sinπ π
zz i i43 4 3 42
2
2
2
22
7
4
7
4= + −
= +
cos sinπ π
2
6 3
π π=
Exercise Set 10.3 531
ch10.3.qxd 9/30/05 2:25 PM Page 531
Note: z4 = , z3 = .
Then (z – z1)(z – z4) = z2 – (z1 + )z + z1 = z2 – 25/4z + 23/2
and (z – z2)(z – z3) = z2 – (z2 + )z + z2 = z2 + 25/4z + 23/2
Then z4 + 8 = (z2 – 25/4z + 23/2)(z2 + 25/4z + 23/2).
Since z4 + 8 has exactly four zeros, it follows that
z4 + 8 = (z – z1)(z – z2)(z – z3)(z – z4)
But
(z – z1)(z – z4) = z2 + 25/4z + 23/2
and
(z – z2)(z – z3) = z2 + 25/4z + 23/2
Thus,
z4 + 8 =
14. (a) We have θ = , and n = 8. Thus
(1 + i)8 = 24[cos(2π) + i sin(2π)] = 16
15. (a) Since
z = 3eiπ = 3[cos(π) + i sin(π)] = –3
then Re(z) = –3 and Im(z) = 0.
(c) Since
then and hence Re(z) = 0 and .
16. The values of z1/n given in Formula (10) are all of the form
zk
= r1/n [cos θk
+ i sin θk]
Im( )z = − 2z i= − 2
z e i ii= =
+
=2 2
2 222π π π
cos sin
π4
r ,= 2
z z z z2 5 4 3 2 2 3 22 2 2 2− +( ) + +( )5 4
z2z2
z1z1
z2z1
532 Exercise Set 10.3
ch10.3.qxd 9/30/05 2:25 PM Page 532
where
Thus, two roots, z and zm
, can be equal if and only if cos θ + i sin θ = cos θm
+ i sin θm
.This can occur if and only if cos θ = cos θ
mand sin θ = sin θ
m, which, in turn, can occur
if and only if θ and θm
differ by an integral multiple of 2π. But
Thus we have:
(a) If ≠ m and and m are integers between 0 and n – 1, then 0 <
cannot be an integral multiple of 2π and hence the two resulting values
of z1/n are different.
(b) If, say, is not in the range 0 to n – 1, then we can write in the form
= nk + m
for some integer m between 0 and n – 1 and for some integer k ≠ 0. Thus
which guarantees that z = zm
.
17. Case 1. Suppose that n = 0. Then
(cos θ + i sin θ)n = 1 = cos(0) + i sin(0)
So Formula (7) is valid if n = 0.
θ θ π π
− =
−( ) ( ) =m
m
nk2 2( )
m
n2
( - )( )π
− <m
n1 so that
θ θ π π π − = − = −m
n
m
n
m
n
2 22
( )( )
θ θ πk
n
k
n= +
2
Exercise Set 10.3 533
ch10.3.qxd 9/30/05 2:25 PM Page 533
Case 2. In order to verify that Formula (7) holds if n is a negative integer, we first letn = – 1. Then
Thus, Formula (7) is valid if n = –1.
Now suppose that n is a positive integer (and hence that –n is a negative integer).Then
This completes the proof.
18. From
Substitute x = iθ to get
Rearranging, to get
or
eiθ = cos θ + i sin θ
e iiθ θ θ θ θ= − + −
+ −1
2 4 3
2 4 3
! ! !…
! !+ − +
θ θ5 7
5 7…
e ii iiθ θ θ θ θ θ= + − − + + −1
2 3 4 5
2 3 4 5
! ! ! !
θθ θ θ6 7 8
6 7 8! ! !− + +i …
e xx x xx = + + + + +12 3 4
2 3 4
! ! !…
(cos sin ) cos sin,
[cos
θ θ θ θ+ = +( )
= −
− −i i
nn
1
θθ θ
θ θ
( )+ −( )= − + −
i
n i n
nsin ]
cos( ) sin( ) [By Formulla (7)]
(cos sin )cos sin
cos sin
cos(
θ θθ θ
θ θ
+ =+
= −
= −
−i
i
i
1 1
θθ θ) sin( )+ −i
534 Exercise Set 10.3
ch10.3.qxd 9/30/05 2:25 PM Page 534
19. We have and . But (see Exercise 17)
If we replace z2 by 1/z2 in Formula (3), we obtain
which is Formula (5).
20. The first equation, when multiplied out, becomes
cos2 θ + 2i cos θ sin θ – sin2 θ = cos 2θ + i sin 2θ
If we equate real and imaginary parts, we obtain the identities
cos2 θ – sin2 θ = cos 2θ
and
2 cos θ sin θ = sin 2θ
Similarly, the second equation yields
cos3 θ – 3 cos θ sin2 θ = cos 3θ
and
3 cos2 θ sin θ – sin3 θ = sin 3θ
z
zz
z
r
ri
1
21
2
1
21 2 1
1=
= + −( )( )+cos sinθ θ θ ++ −( )( )
= −( )+ −( )
θ
θ θ θ θ
2
1
21 2 1 2
r
ricos sin
1 1 1
2 2 22
2
z r e re
i
i= = −θ
θ
z r ei
2 22= θ
z r ei
1 11= θ
Exercise Set 10.3 535
ch10.3.qxd 9/30/05 2:25 PM Page 535
21. If
eiθ = cos θ + i sin θ
then replacing θ with –θ yields
e–iθ = cos(–θ) + i sin(–θ) = cos θ – i sin θ
If we then compute eiθ + e–iθ and eiθ – e–iθ, the results will follow.
22. If (a + bi)3 = 8, then a + bi = 2. That is, a2 + b2 = 4.
23. Let z = r(cos θ + i sin θ). Formula (5) guarantees that 1/z = r–1 (cos(–θ) + i sin(–θ))since z ≠ 0. Applying Formula (6) for n a positive integer to the above equation yields
which is just Formula (6) for –n a negative integer.
zz
r n i nn
n
n− −=
= −( ) + −( )( )1cos sinθ θ
536 Exercise Set 10.3
ch10.3.qxd 9/30/05 2:25 PM Page 536
EXERCISE SET 10.4
1. (a) u – v = (2i – (–i), 0 – i, –1 – (1 + i), 3 – (–1))
= (3i, –i, –2 – i, 4)
(c) –w + v = (–(1 + i) – i, i + i, –(–1 + 2i) + (1 + i), 0 + (–1))
= (–1 – 2i, 2i, 2 – i, –1)
(e) –iv = (–1, 1, 1 – i, i) and 2iw = (–2 + 2i, 2, –4 – 2i, 0). Thus
–iv + 2iw = (–3 + 2i, 3, –3 – 3i, i)
2. Observe that ix = u – v – w. Thus
x = –i(u – v – w)
3. Consider the equation c1u1 + c2u2 + c3u3 = (–3 + i, 3 + 2i, 3 – 4i). The augmented matrixfor this system of equations is
The row-echelon form for the above matrix is
1 2 0 3
1 2 3 2
0 1 2 3 4
− − ++ +
− −
i i i
i i i i
i i
537
5.
6. (a) Since u + v = (3i, 3 + 4i, –3i), then
(c) Here
It follows that
(e)
8. Since kv = (3ki, 4ki), then
Thus k kvv = ⇔ = ±1 1 5.
k ki ki kvv = + =3 4 52 2
.
1 1
6
2
60
wwww =
+
i i, ,
Since thenww = + + = + =1 2 2 4 62 2
i i ,
− + = +i i iuu uu 10 10
uu uu= + − = + = − = − − =3 9 1 10 3 0 1 12 2
i i i i, ( , , ),and uu 00.
uu vv+ = + + + − = + + =3 3 4 3 9 25 9 432 2 2
i i i
((aa)) vv
((cc)) vv
= + =
= + + + + − =
1 2
2 0 2 1 1 4
2 2
2 2 2 2
i
i i ( ) ++ + + =0 5 1 10
Hence, ,c i c i i3 223
2
1
2
1
2
1
2= − = + − +
= = − −c c i3 10 2, .and
1 1 0 2
0 11
2
1
2
3
2
1
20 0 1 1
− + − −
+ +
−
i i
i i
i
538 Exercise Set 10.4
9. (a) u • v = (–i)(–3i) + (3i)(–2i) = –3 + 6 = 3.
(c) u • v = (1 – i)(4 – 6i) + (1 + i)(5i) + (2i)(–1 – i) + (3)(–i)
= (–2 – 10i) + (–5 + 5i) + (2 – 2i) + (–3i)
= –5 – 10i
11. Let V denote the set and let
We check the axioms listed in the definition of a vector space (see Section 5.1).
(1)
So u + v belongs to V.
(2) Since u + v = v + u and , it follows that u + v = v + u.
(3) Axiom (3) follows by a routine, if tedious, check.
(4) The matrix serves as the zero vector.
(5)
(6) Since ku = , ku will be in V if and only if , which is true if and
only if k is real or u = 0. Thus Axiom (6) fails.
(7)–(9) These axioms all hold by virtue of the properties of matrix addition and scalarmultiplication. However, as seen above, the closure property of scalar multiplicationmay fail, so the vectors need not be in V.
(10) Clearly 1u = u.
Thus, this set is not a vector space because Axiom (6) fails.
ku ku=ku
ku
0
0
Let = Then−−
−
=
−−
uu uu
u
u
u
u
0
0
0
0. ++ ( ) = .− uu 00
0 0
0 0
u v v u+ = +
uu vv+ =+
+
=
+
+
u v
u v
u v
u v
0
0
0
0
uu = a du
u
v
v
0
0
0
0
=
n vv
Exercise Set 10.4 539
12. (a) Since
(z1, 0, 0) + (z2, 0, 0) = (z1 + z2, 0, 0)
and
k(z, 0, 0) = (kz, 0, 0)
this set is closed under both addition and scalar multiplication. Hence it is a subspace.
(b) Since
(z1, i, i) + (z2, i, i) = (z1 + z2, 2i, 2i)
this is not a subspace. Part (b) of Theorem 5.2.1 also is not satisfied.
13. Suppose that T(x) = Ax = 0. It is easy to show that the reduced row echelon form for A is
Hence, x1 = (–(1 + 3i)/2)x3 and x2 = (–(1 + i)/2)x3 where x3 is an arbitrary complexnumber. That is,
spans the kernel of T and hence T has nullity one.
Alternatively, the equation Ax = 0 yields the system
ix1 – ix2 – x3 = 0
x1 – ix2 + (1 + i)x3 = 0
0 + (1 – i)x2 + x3 = 0
The third equation implies that x3 = –(1 – i)x2. If we substitute this expression for x3into the first equation, we obtain x1 = (2 + i)x2. The second equation will then be valid
xx =
− +( )− +( )
1 3 2
1 2
1
i
i
1 0 1 3 2
0 1 1 2
0 0 0
+( )+( )
i
i
540 Exercise Set 10.4
for all such x1 and x3. That is, x2 is arbitrary. Thus the kernel of T is also spanned by thevector
If we multiply this vector by –(1 + i)/2, then this answer agrees with the previous one.
14. (a) While this set is closed under vector addition, it is not closed under scalarmultiplication since the real entries, when multiplied by a non-real scalar, need notremain real. Hence it does not form a subspace.
(b) This set does form a subspace since
and
That is, the set is closed under both operations.
15. (a) Since
(f + g)(1) = f(1) + g(1) = 0 + 0 = 0
and
kf(1) = k(0) = 0
for all functions f and g in the set and for all scalars k, this set forms a subspace.
(c) Since
( )( ) ( ) ( ) ( ) ( )
( ) ( ) (
f g x f x g x f x g x
f x g x
+ − = − + − = +
= + = ff g x+ )( )
kz z
z z
kz kz
kz kz
1 2
3 1
1 2
3 1−
=
−( )
u u
u u
v v
v v
u v u1 2
3 1
1 2
3 1
1 1 2
−
+
−
=
+ +++ − +( )
v
u v u v
2
3 3 1 1
x
x
x
i
i
1
2
3
2
1
1
=+
− +
Exercise Set 10.4 541
the set is closed under vector addition. It is closed under scalar multiplication by a realscalar, but not by a complex scalar. For instance, if f(x) = xi, then f(x) is in the set butif (x) is not.
16. (a) Consider the equation k1u + k2v = (3i, 3i, 3i). Equating components yields
k1i + k22i = 3i
–k1i + k24i = 3i
k13i = 3i
The above system has the solution k1 = k2 = 1. Thus, (3i, 3i, 3i) is a linear combinationof u and v.
(c) Consider the equation k1u + k2v = (i, 5i, 6i). Equating components yields
k1i + k22i = i
–k1i + k24i = 5i
k13i = 6i
The above system is inconsistent. Hence, (i, 5i, 6i) is not a linear combination of u and v.
17. (a) Consider the equation k1u + k2v + k3w = (1, 1, 1). Equating components yields
k1 + (1 + i)k2 = 1
k2 + ik3 = 1
–ik1 + (1 – 2i)k2 + 2k3 = 1
Solving the system yields k1 = –3 – 2i, k2 = 3 – i, and k3 = 1 + 2i.
(c) Let A be the matrix whose first, second and third columns are the components of u, v,and w, respectively. By Part (a), we know that det(A) ≠ 0. Hence, k1 = k2 = k3 = 0.
18. We let A denote the matrix whose first, second and third columns are the components of v1,v2, and v3, respectively.
(a) Since det(A) = 6i ≠ 0, it follows that v1, v2, and v3 span C3.
(d) Since det(A) = 0, it follows that v1, v2, and v3 do not span C3.
542 Exercise Set 10.4
19. (a) Recall that eix = cos x + i sin x and that e–ix = cos(–x) + i sin(–x) = cos x – i sin x.Therefore,
and so cos x lies in the space spanned by f and g.
(b) If af + bg = sin x, then (see Part (a))
(a + b)cos x + (a – b)i sin x = sin x
Thus, since the sine and cosine functions are linearly independent, we have
a + b = 0
and
a – b = –i
This yields a = –i/2, b = i/2, so again the vector lies in the space spanned by f and g.
(c) If af + bg = cos x + 3i sin x, then (see Part (a))
a + b = 1
and
a – b = 3
Hence a = 2 and b = –1 and thus the given vector does lie in the space spanned byf and g.
20. (a) Note that iu1 = u2.
21. Let A denote the matrix whose first, second, and third columns are the components of u1,u2, and u3, respectively.
(a) Since the last row of A consists entirely of zeros, it follows that det(A) = 0 and henceu1, u2, and u3, are linearly dependent.
(c) Since det(A) = i ≠ 0, then u1, u2, and u3 are linearly independent.
cos xe e
ix ix
=+
= +−
2
1
2
1
2ff gg
Exercise Set 10.4 543
22. Solving k1, .
Equating components:
k1a + k2(b – ci) = 0, k1(b + ci) + k2a = 0.
Equating real and imaginary parts:
k1a + k2b = 0
k1b + k2a = 0
k1c = 0
k2c = 0.
The above has non-trivial solutions for k1, k2 when a = ±b, c = 0.
23. Observe that f – 3g – 3h = 0.
24. (a) Since the dimension of C2 is two, any basis for C2 must contain precisely two vectors.
25. (a) Since = –4 ≠ 0, the vectors are linearly independent and hence form a basis
for C2.
(d) Since = 0, the vectors are linearly dependent and hence are not a
basis for C2.
26. Since the number of vectors is the same as the dimension of C3, the vectors will form a basisif and only if they are linearly independent.
(a) Since = –i ≠ 0, the vectors are linearly independent. Hence, they form
a basis.
(c) From Problem 21(c), we know that the vectors are linearly independent. Hence, theyform a basis.
i i i
i i
i
0
0 0
2 3 3 2
1
− +−
i i
i
2 4
0
i i
i−
v k w k a b ci, ,+ = + +2 0 we get 1 kk b Ci a2 0,− =
544 Exercise Set 10.4
Exercise Set 10.4 545
27. The row-echelon form of the matrix of the system is
So x2 is arbitrary and x1 = –(1 + i)x2. Hence, the dimension of the solution space is one
and is a basis for that space.
29. The reduced row-echelon form of the matrix of the system is
So x3 is arbitrary, x2 = (–3i)x3, and x1 = (3 + 6i)x3. Hence, the dimension of the solution
space is one and is a basis for that space.
31. Let u = (u1, u2, . . . , un) and v = (v1, v2, . . . , v
n). From the definition of the Euclidean inner
product in Cn, we have
32. (a) Let u = (u1, u2, . . . , un), v = (v1, v2, . . . , v
n) and w = (w1, w2, . . . , w
n). Then, since
we have u + v = (u1 + v1, u2 + v2, . . . , un
+ vn),
33. Hint: Show that
u + kv2 = u2 + –k(v • u) + k(u • v) + k–
k v2
and apply this result to each term on the right-hand side of the identity.
( ) ( ) ( ) . . . ( )uu vv ww+ = + + + + + +• u v w u v w u v wn n n1 1 1 2 2 2
== + + + + + + +[ . . . ] [ . . . ]u w u w u w v w v w v wn n n n1 1 2 2 1 1 2 2
== • •uu ww ++ vv ww
uu vv• = + + +=
( ) ( ) ( ) . . . ( )
(
k u kv u kv u kv
u kv
n n1 1 2 2
1 1)) ( ) . . . ( )
( ) ( ) . . .
+ + += + +
u kv u kv
k u v u u v
n n2 2
1 1 2 2 2 ++= + + += •
u u v
k u v u v u v
k
n n n
n n
( )
[ . . . ]
( )1 1 2 2
uu vv
3 6
3
1
+−
i
i
1 0 3 6
0 1 3
0 0 0
− −
i
i
− +( )
1
1
i
1 1
0 0
+
i
34. Let f(x) = f1(x) + if2(x) and g(x) = g1(x) + ig2(x). Then
(f + g)(x) = f(x) + g(x)
= [ f1(x) + g1(x)] + i[f2(x) + g2(x)]
and
(kf)(x) = kf(x)
= kf1(x) + ikf2(x)
Thus, if f and g are continuous, then f + g is continuous; similarly, if f is continuous, thenkf is continuous. Hence, the result follows from Theorem 5.2.1.
546 Exercise Set 10.4
EXERCISE SET 10.5
1. Let u = (u1, u2), v = (v1, v2), and w = (w1, w2). We proceed to check the four axioms
(1)
(2) ⟨u + v, w⟩ = 3(u1 + v1) + 2(u2 + v2)
= [3u1 + 2u2 ] + [3v1 + 2v2 ]
= ⟨u, w⟩ + ⟨v, w⟩
(3) ⟨ku, v⟩ = 3(ku1) + 2(ku2)
= k[3u1 + 2u2 ] = k⟨u, v⟩
(4) ⟨u, u⟩ = 3u1 + 2u2
= 3|u1|2 + 2|u2|
2 (Theorem 10.2.1)
≥ 0
Indeed, ⟨u, u⟩ = 0 ⇔ u1 = u2 = 0 ⇔ u = 0.
Hence, this is an inner product on C2.
2. (a)
(c) uu vv, ( ) ( ) ( ) ( )
( )
= + − + − +
= +
3 1 1 2 1 1
3 2
i i i i
i ( )2 2 2− =i i
uu vv, ( ) ( ) ( )( )= − + − = − − = −3 2 2 3 6 6 12i i i i
u2u1
v2v1
v2v1
w2w1w2w1
w2w1
vv uu
uu vv
, = +
= + =
3 2
3 2
1 1 2 2
1 1 2 2
v u v u
u v u v ,
547
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3. Let u = (u1, u2) and v = (v1, v2). We check Axioms 1 and 4, leaving 2 and 3 to you.
(1)
(4) Recall that |Re(z)| ≤ |z| by Problem 29 of Section 10.1. Now
Moreover, ⟨u, u⟩ = 0 if and only if both |u2| and |u1| – |u2| = 0, or u = 0.
4. (a) ⟨u, v⟩ = (2i)i + (1 + i)(2i)(–3i) + (1 – i)(–i)(i) + 3(–i)(–3i)
= –4 + 5i
(c) ⟨u, v⟩ = (1 + i)(1 + i) + (1 + i)(1 + i)(1 – i) + (1 – i)(1 – i)(1 + i) + 3(1 – i)(1 – i)
= 4 – 4i
5. (a) This is not an inner product on C2. Axioms 1–3 are easily checked. Moreover,
⟨u, u⟩ = u1 = |u1| ≥ 0
However, ⟨u, u⟩ = 0 ⇔ u1 = 0 ⇔/ u = 0. For example, ⟨i, i⟩ = 0 although i ≠ 0.Hence, Axiom 4 fails.
u1
2
uu uu, ( ) ( )= + + + − +
=
u u i u u i u u u u1 1 1 2 2 1 2 21 1 3
uu i u u i u u u
u
12
1 2 1 2 22
12
1 1 3
2
( ) ( )
Re
+ + + + +
= + ((( ) )
( )
1 3
2 1 3
1 2 22
12
1 2 22
+ +
≥ − + +
i u u u
u i u u u
== − +
= −( ) +
u u u u
u u u
12
1 2 22
1 1
2
22
2 2 3
2
vv uu, ( ) ( )= + + + − +v u i v u i v u v u1 1 1 2 2 1 2 21 1 3
== + − + + +
=
u v i u v i u v u v1 1 2 1 1 2 2 21 1 3( ) ( )
uu vv,
548 Exercise Set 10.5
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(c) This is not an inner product on C2. Axioms 1 and 4 are easily checked. However, for w = (w1, w2), we have
⟨u + v, w⟩ = |u1 + v1|2|w1|
2 + |u2 + v2|2|w2|
2
≠ (|u1|2 + |v1|
2)|w1|2 + (|u2|
2 + |v2|2)|w2|
2
= ⟨u, w⟩ + ⟨v, w⟩
For instance, ⟨1 + 1, 1⟩ = 4, but ⟨1, 1⟩ + ⟨1, 1⟩ = 2. Moreover, ⟨ku, v⟩ = |k|2⟨u, v⟩, so that⟨ku, v⟩ ≠ k⟨u, v⟩ for most values of k, u, and v. Thus both Axioms 2 and 3 fail.
(e) Axiom 1 holds since
A similar argument serves to verify Axiom 2 and Axiom 3 holds by inspection. Finally,using the result of Problem 29 of Section 10.1, we have
⟨u, u⟩ = 2u1u–
1 + iu1u–
2– iu2u–
1 + 2u2u–
2
= 2|u1|2 + 2Re(iu1u
–2) + 2|u2|
2
≥ 2|u1|2 – 2|iu1u
–2| + 2|u2|
2
= (|u1| – |u2|)2 + |u1|
2 + |u2|2
≥ 0
Moreover, ⟨u, u⟩ = 0 ⇔ u1 = u2 = 0, or u = 0. Thus all four axioms hold.
6. ⟨u, v⟩ = (–i)(3) + (1 + i)(–2 + 3i) + (1 – i)(–4i) + i(1)
= –9 – 5i
vv uu, = + − +
=
2 2
2
1 1 1 2 2 1 2 2
1 1
v u iv u iv u v u
u v −− + +
=
iu v iu v u v2 1 1 2 2 22
uu vv,
Exercise Set 10.5 549
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8. First,
Second, if h = h1 (x) + ih2(x), then
Third, it is easily checked that ⟨kf, g⟩ = k⟨f, g⟩. And finally,
However, ⟨f, f⟩ = 0 ⇔ f1(0) = f2(0) = 0 ⇔/ f = 0. For instance, if f = ix, then ⟨f, f⟩ = 0, butf ≠ 0. Therefore, this is not an inner product.
9. (a) w = [3(–i)(i) + 2(3i)(–3i)]1/2 =
(c) w = [3(0)(0) + 2(2 – i)(2 + i)]1/2 =
10. (a) w = [|–i|2 + |3i|2]1/2 =
(c) w = [|0|2 + |2 –i|2]1/2 =
11. (a) w = [(1)(1) + (1 + i)(1)(i) + (1 – i)(–i)(1) + 3(–i)(i)]1/2 =
(c) w = [(3 – 4i)(3 + 4i)]1/2 = 5
12. (a) A = [(–i)(i) + (7i)(–7i) + (6i)(–6i) + (2i)(–2i)]1/2 = 3 10
2
5
10
10
21
ff,,ff ( ) ( ) ( ) ( )
( )
= +( ) +( )=
f if f if
f
1 2 1 2
12
0 0 0 0
0 ++
≥
( )f22
0
0
ff ++ gg,,hh ( ) ( ) ( ) ( )= + + + ( ) ⋅f g i f g1 1 2 20 0 0 0 (( ( ) ( ))h ih1 20 0+
= +ff,, hh gg,, hh
gg ff, ( ( ) ( )) ( ( ) ( ))
(
= + +
=
g ig f if1 2 1 20 0 0 0
ff if g ig1 2 1 20 0 0 0( ) ( )) ( ( ) ( ))+ +
= ff gg,
550 Exercise Set 10.5
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13. (a) Since u – v = (1 – i, 1 + i), then
d(u, v) = [3(1 – i)(1 + i) + 2(1 + i)(1 – i)]1/2 =
14. (a) Since u – v = (1 – i, 1 + i),
d(u, v) = [(1 – i)(1 + i) + (1 + i)(1 –i)]1/2 = 2
15. (a) Since u – v = (1 – i, 1 + i),
d(u, v)= [(1 – i)(1 + i) + (1 + i)(1 – i)(1 – i)
+ (1 – i)(1 + i)(1 + i) + 3(1 + i)(1 – i)]1/2
=
16. (a)
d(A, B) = [(6i)(–6i) + (5i)(–5i) + (i)(–i) + (6i)(–6i)]1/2 = = 7
17. (a) Since u • v = (2i)(–i) + (i)(–6i) + (3i)(k–), then u • v = 0 ⇔ 8 + 3ik
– = 0 or k = –8i/3.
18. (a) Let B denote the given matrix. Then
⟨A, B⟩ = (2i)(–3) + (i)(1 + i) + (–i)(1 + i) + (3i)(2) = 0
Thus, A and B are orthogonal.
(d) Let B denote the given matrix. Then
⟨A, B⟩ = i + (–i)(3 + i) = 1 – 2i
Thus, A and B are not orthogonal.
298
Since A Bi i
i i− =
6 5
6,
2 3
10
Exercise Set 10.5 551
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19.
Also
and
20. (a) Since , the set of vectors is not orthonormal. (However, the set is
orthogonal.)
(b) Since = 0 and the norm of each vector is 1, the
set is orthonormal.
21. (a) Call the vectors u1, u2, and u3, respectively. Then u1 = u2 = u3 = 1 and
u1 • u2 = u1 • u3 = 0. However, u2 • u3 = . Hence the set is not
orthonormal.
22. If we use the Euclidean inner product, then . Hence, the vectors cannot be
orthonormal with respect to the Euclidean inner product. In fact, neither vector is normal
and the vectors are not orthogonal.
xx = ≠2 5 1
i i2 2
6 6
2
60+
−( )= − ≠
i i i i i i
2 2 2 2 2 2
2 2
, ,•−
= − +
( , )0 1 − =i 2
xx,1
3
1
3
( , , ) , , , ,•0 1 1 0i i e i i i
e
i
i
− = ( ) −( )
=
θ
θ(( )0
0
− +
=
i i
xx,1
31
3
( , , ) , , , ,
(
•1 0 1 1 1 0i e i i
e i
i
i
= ( ) ( )
=
θ
θ −− +
=
i 0
0
)
Since =1
3we have
1
3
x e i
e i
i
i
θ
θ
( , , ),
, ,
1 1
1 1xx = ( )) = + + =1
31 1 1 1 11 2( )( )
552 Exercise Set 10.5
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We now let ⟨u, v⟩ = 3u1v–
1 + 2u2v–
2. Then
and
Hence x and y are orthonormal with respect to this inner product.
24. (a) We have
Because
25. (a) We have
vv11 =
i i i
3 3 3, ,
vv22 =
3
10 10
i i,
12
5
4
5
4 10
5
i i, ,
= then
uu uu vv vv22 22 11 11− = +−
=
( )• ( , ) ,2 24
10 10
3
10
12
i ii i
ii i
5
4
5,
vvuu
uu1111
11
= =−
=−
( , ),
i i i i3
10 10
3
10
xx yy, =
−
+ −
−35
2
302
5
3
30
i i i i
= 0
xx2
35 5
25 5
=
−
+ −
=i i i i
11
32
30
2
302
3
30
32yy =
−
+
−i i i ii
301
=
Exercise Set 10.5 553
and, since it follows that/ ,•uu vv2 1 4 10= −
ch10.5.qxd 9/30/05 7:16 AM Page 553
and since u2 • v1 = 0, then u2 – (u2 • v1)v1 = u2. Thus,
Since the norm of the above vector is , we have
27. Let u1 = (0, i, 1 –i) and u2 = (–i, 0, 1 + i). We shall apply the Gram-Schmidt process
to u1, u2. Since , it follows that
and because the norm of the above vector is , we have
28. Following Theorem 6.3.6, we orthonormalize the set u1, u2. This gives us
and, since uu vv2 1 2 6/ ,• =
vvuu
uu11
1
= = −
i i i
60
6
2
6, , ,
vv2 =− +
3
15
2
15
1
15
i i, ,
15 3/
uu uu vv vv2 2 1 1− = − + − − +
( ) ( , , ) , ,• i i i0 1 02
3
2
3
2
3
= − +
i i, ,2
3
1
3
1
3
Because thenuu vv2 1 2 3/ ,• = i
vv1 =−
03
1
3, ,
i i
u1 = 3
vv3 =−
i i i
6 6
2
6, ,
1 6/
uu uu vv vv uu vv vv3 1 1 3 2− − =−
( ) ( ) , ,• •3 2 6 6
i i i
33
Also, and anduu vv uu vv3 1 3 24 3 1 2/ /• •= = hhence
vv22 = −
i i
2 20, ,
554 Exercise Set 10.5
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so that
Therefore,
and
29. (a) By Axioms (2) and (3) for inner products,
⟨u –kv, u – kv⟩ = ⟨u, u –kv⟩ + ⟨–kv, u – kv⟩
= ⟨u, u – kv⟩ – k⟨v, u – kv⟩
If we use Properties (ii) and (iii) of inner products, then we obtain
⟨u – kv, u – kv⟩ = ⟨u, u⟩ – k–⟨u, v⟩ – k⟨v, u⟩ + kk–⟨v, v⟩
Finally, Axiom (1) yields
⟨u – kv, u – kv⟩ = ⟨u, u⟩ – k–⟨u, v⟩ – k + kk–⟨v, v⟩
and the result is proved.
(b) This follows from Part (a) and Axiom (4) for inner products.
uu vv,
ww ww ww2 1–=
= −
1
4
9
4
19
4
9
4i i i i, , ,
ww ww vv vv ww vv vv
vv vv
1 1 1 2 2
1 2
( ) ( )• •= +
= +−
= −
7
6
1
2 3
5
44
1
4
5
4
9
4i i i i, , ,−
vv22 = −
i i i i
2 3
3
2 3 2 3 2 3, , ,
uu uu vv vv2 0 01
30( ) ( , , , ) ( , ,•− = − −2 1 1 i i i ii i
ii
i
, )
, , ,
2
3
1
3 3= −
Exercise Set 10.5 555
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30. Suppose that v ≠ 0 and hence that ⟨v, v⟩ ≠ 0. Using the hint, we obtain
Since ⟨v, v⟩ is real, this yields the inequality
or
or
as desired.
In case v = 0, Property (i) for inner products guarantees that both sides of the Cauchy-Schwarz inequality are zero, and thus equality holds.
32. If u and v are linearly dependent, then either v = 0 or u is a scalar multiple of v. In eithercase, a routine computation using properties of the inner product shows that equality holds.
Conversely, if equality holds, then we shall show that u and v must be linearlydependent. For suppose that they are not. Then u – kv ≠ 0 for every scalar k, so it followsthat ⟨u – kv, u – kv⟩ is greater than zero for all k. Thus, strict inequality must hold inProblem 29(b). Now if we choose k as in Problem 30 and rederive the Cauchy-Schwarzinequality, we find that strict inequality must hold there also. That is, if u and v are linearlyindependent, then the inequality must hold. Hence, if the equality holds, then we are forcedto conclude that u and v are linearly dependent.
33. Hint: Let v be any nonzero vector, and consider the quantity ⟨v, v⟩ + ⟨0, v⟩.
34. We have
u, v + w , +
+ ,
=
=
uu vv ww
vv ww uu ( )Axiom 1
, ,= +vv uu ww uu ( )Axiom 2
, ,
, ,
= +
= +
vv uu ww uu
uu vv uu ww ( )Axiom 1
uu vv uu uu vv vv, , ,2
≤
uu vv
vv vvuu uu
,
,,
2
≤
0
2 2 2
≤ − − +uu uuuu vv
vv vv
uu vv
vv vv
uu vv
vv vv,
,
,
,
,
,
,
0 ≤ −
−
uu uu
uu vv
vv vvuu vv
uu vv
vv vv,
,
,,
,
,uu vv
uu vv
vv vvvv vv,
,
,,+
2
556 Exercise Set 10.5
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35. (d) Observe that u + v2 = ⟨u + v, u + v⟩. As in Exercise 29,
⟨u + v, u + v⟩ = ⟨u, u⟩ + 2 Re(⟨u, v⟩) + ⟨v, v⟩
Since (see Exercise 29 of Section 10.1)
|Re(⟨u, v⟩)| ≤ |⟨u, v⟩|
this yields
⟨u + v, u + v⟩ ≤ ⟨u, u⟩ + 2|⟨u, v⟩| + ⟨v, v⟩
By the Cauchy-Schwarz inequality and the definition of norm, this becomes
u + v2 ≤ u2 + 2u v + v2 = (u + v)2
which yields the desired result.
(h) Replace u by u – w and v by w – v in Theorem 6.2.2, Part (d).
37. Observe that for any complex number k,
u + kv2= ⟨u + kv, u + kv⟩
= ⟨u, u⟩ + k⟨v, u⟩ + k–⟨u, v⟩ + kk–⟨v, v⟩
= ⟨u, u⟩ + 2 Re(k⟨v, u⟩) + |k|2⟨v, v⟩
Therefore,
u + v2 – u – v2 + iu + iv2 – iu – iv2
= (1 – 1 + i – i)⟨u, u⟩ + 2 Re(⟨v, u⟩) – 2 Re(–⟨v, u⟩) + 2i Re(i⟨v, u⟩)
–2i Re(–i⟨v, u⟩) + (1 – 1 + i – i)⟨v, v⟩
= 4 Re(⟨v, u⟩) – 4i Im(⟨v, u⟩)
=
= 4 ⟨u, v⟩
4 vv uu,
Exercise Set 10.5 557
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38. By the hint, we have
The last step uses the fact that the basis vectors vj
form an orthonormal set.
39. We check Axioms 2 and 4. For Axiom 2, we have
For Axiom 4, we have
Since |f |2 ≥ 0 and a < b, then ⟨f, f⟩ ≥ 0. Also, since f is continuous, |f |2 dx > 0 unless
f = 0 on [a, b]. [That is, the integral of a nonnegative, real-valued, continuous function (which
represents the area under that curve and above the x-axis from a to b) is positive unless the
function is identically zero.]
a
b
∫
ff ff, = =
= ( )( ) + ( )( )
∫ ∫ff ff dxa
b
a
bf dx
f x f x
2
12
22
∫ dxa
b
ff gg hh+ ,
,
= +
= +
=
∫
∫ ∫
( )ff gg hh
ff hh gg hh
ff
dx
dx dx
a
b
a
b
a
b
hhhh gg hh+ ,
uu ww uu vv ww vv vv vv
uu vv ww vv
, ,
,
=
=
=
=
∑ j k
j k
n
j k
j k
j
n
, ,
,
, 1
1
∑∑
uu uu vv vv ww ww vv= == =∑ ∑, and , v Thereforj j
j
n
k k
k
n
1 1
. ee,
558 Exercise Set 10.5
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40. (a) Since we are working in C[0, 1],
Hence .
(b) Since we are again working in C[0, 1],
41. Let vm
= e2πimx = cos(2πmx) + i sin(2πmx). Then if m ≠ n, we have
Thus the vectors are orthogonal.
vv vvm n mx i mx nx, cos sin cos= ( ) + ( ) ( )∫ 2 2 20
1π π π −− ( )
= ( ) ( ) +
i nx dx
mx nx
sin
cos cos sin
2
2 2 2
π
π π π mmx nx dx
i mx
( ) ( )
+
∫ sin
sin
2
2
0
1π
π(( ) ( ) − ( ) ( )
=
∫ cos cos sin
co
2 2 20
1π π πnx mx nx dx
ss sin2 2
1
0
1
0
1π πm n x dx i m n x dx−( ) + −( )
=
∫∫
222
22
0
1
ππ
ππ
m nm n x
i
m nm n
−( ) −( ) −−( ) −sin[ ] cos[ (( )
= −−( ) +
−( )=
x
i
m n
i
m n
]0
1
2 2
0
π π
ff gg, = +( )= −( )
= −
∫
∫
x ix dx
x ix dx
i
1
1
1
2
1
3
0
1
0
1
gg /= 2 3
gg gg gg gggg2 2
0
1
0
11
4
3= = = +( ) =∫∫, dx x dx
Exercise Set 10.5 559
ch10.5.qxd 9/30/05 7:16 AM Page 559
ch10.5.qxd 9/30/05 7:16 AM Page 560
EXERCISE SET 10.6
4. (b) The row vectors of the matrix are
Since r1 = r2 = 1 and r1 • r2 = 0, it follows that the matrix is unitary.
(c) The row vectors of the matrix are
r1 = (1 + i, 1 + i) and r2 = (1 – i, –1 + i)
Although r1 • r2 = (1 + i)(1 + i) + (1 + i)(–1 – i) = 0 (from which it follows thatr1 and r2 are orthogonal), they are not orthonormal because r1 = r2 = 2. Hence, thematrix is not unitary.
5. (b) The row vectors of the matrix are
Since r1 = r2 = 1 and
the matrix is unitary by Theorem 10.6.2. Hence,
A A A
i
i
T− = = =
− +
−
1
1
2
1
2
1
2
1
2
*
rr rr1 21
2
1
2
1
2
1
20• =
− +
+−
=i i
rr rr1 21
2
1
2
1
2
1
2=
= − + +
, ,and
i i
rr rr1 22
1
2 2
1
2=
= −
i i, ,and
561
5. (d) The row vectors of the matrix are
and
We have r1 = r2 = r3 = 1,
and
Hence, by Theorem 10.6.2, the matrix is unitary and thus
6. Using Formula (11) of Section 10.3, we see that and hence that Let
rr rr rr rr1 1 2 21
2
1
21• •= = + =
rr rr1 22 2 2 2
=
=
−
− −e e ie ie
i i i iθ θ θ θ, ,and . Then
e ei i− =θ θ .e e
i iθ θ= −
A A A
i i i
i
i i
T− = = =
− − −
−−
−
1
1
2 3
3
2 151
2
1
3
4 3
2 151
2 3
5
2
*
115
rr rr1 31
2
3
2 15
1
2
4 3
2• •=
+
−
−−i i i
115
1
2
5
2 150+
−=•
i
rr rr1 21
2 3
1
2
1
3
1
2• • •=
+
−
− +i i i
330
1
3
3
2 15
1
3
4 32 3
=
=
−
+−
rr rr• •i ii i i
2 15 3
5
2 150−
−=•
rr33
2 15
4 3
2 15
5
2 15= + +
i i i, ,
rr
rr
1
2
1
2
1
2
1
2
3
1
3 3
=+
−
=−
i
i i
, ,
, ,
562 Exercise Set 10.6
and
Hence, by Theorem 10.6.2, the row vectors form an orthonormal set and so the matrix isunitary.
7. The characteristic polynomial of A is
Therefore, the eigenvalues are λ = 3 and λ = 6. To find the eigenvectors of A correspondingto λ = 3, we let
This yields x1 = –(1 – i)s and x2 = s where s is arbitrary. If we put s = 1, we see that
is a basis vector for the eigenspace corresponding to λ = 3. We normalize this
vector to obtain
To find the eigenvectors corresponding to λ = 6, we let
This yields and x2 = s where s is arbitrary. If we put s = 1, we have that
is a basis vector for the eigenspace corresponding to λ = 6. We normalize this
vector to obtain
1 2
1
−( )
i
xi
s11
2=
−
2 1
1 1
0
01
2
− +− −
=
i
i
x
x
PP1
1
31
3
=
− +
i
− +
1
1
i
−− −
=
1
1
0
01
2i
x
x
det ( )( ) ( )(λ
λλ λ λ
− − +− − −
= − − − = −
4 1
1 54 5 2 6
i
iλλ − 3)
rr rr1 21
2
1
2• = −( ) + ( )
=e ie e ie
i i i iθ θ θ θ −− + =i i 0
Exercise Set 10.6 563
Thus
diagonalizes A and
9. The characteristic polynomial of A is
Therefore the eigenvalues are λ = 2 and λ = 8. To find the eigenvectors of A correspondingto λ = 2, we let
This yields and x2 = s where s is arbitrary. If we put s = 1, we have that
is a basis vector for the eigenspace corresponding to λ = 2. We normalize this
vector to obtain
− +( )
1 2
1
i
xi
s11
2= −
+
− − −− + −
=
4 2 2
2 2 2
0
01
2
i
i
x
x
det ( )( )λ
λλ λ
− − −− + −
= − −
6 2 2
2 2 46 4
i
i−− = − −( )( )8 8 2λ λ
P AP
i
i
i
− =
− −
+
−1
1
3
1
3
1
6
2
6
4 1
11 5
1
3
1
6
1
3
2
6+
− + −
i
i i
=
3 0
0 6
P
i i
=
− + −
1
3
1
6
1
3
2
6
PP2
1
6
2
6
=
−
i
564 Exercise Set 10.6
To find the eigenvectors corresponding to λ = 8, we let
This yields x1 = (1 + i)s and x2 = s where s is arbitrary. If we set s = 1, we have that
is a basis vector for the eigenspace corresponding to λ = 8. We normalize this
vector to obtain
Thus
diagonalizes A and
P AP
i
i
i
− =
− +
−
+1
1
6
2
61
3
1
3
6 2 2
22 2 4
1
6
1
32
6
1
3−
−+ +
i
i i
=
2 0
0 8
P
i i
=−
+ +
1
6
1
62
6
1
3
PP2 =
+
1
31
3
i
1
1
+
i
2 2 2
2 2 4
0
01
2
− −− +
=
i
i
x
x
PP1
1
62
6
=−
+
i
Exercise Set 10.6 565
11. The characteristic polynomial of A is
Therefore, the eigenvalues are λ = 1, λ = 5, and λ = –2. To find the eigenvectors of Acorresponding to λ = 1, we let
This yields x1 = 0, x2 = – and x3 = s where s is arbitrary. If we set s = 1, we have
that is a basis vector for the eigenspace corresponding to λ = 1. We normalize
this vector to obtain
To find the eigenvectors corresponding to λ = 5, we let
This yields x1 = s and x2 = x3 = 0 where s is arbitrary. If we let s = 1, we have that
is a basis vector for the eigenspace corresponding to λ = 5. Since this vector is already
normal, we let
1
0
0
0 0 0
0 6 1
0 1 5
1
2
3
−+
i
i
x
x
x
=
0
0
0
PP1
2
6
= −−
0
1 i
6
0
1 2
1
− −( )
i
1
2
− is,
−−
+
=4 0 0
0 2 1
0 1 1
0
01
2
3
i
i
x
x
x 00
det ( )( )(
λλ
λλ λ λ
−+ −+
= − −5 0 0
0 1 1
0 1
1 5i
i
++ 2)
566 Exercise Set 10.6
To find the eigenvectors corresponding to λ = – 2, we let
This yields x1 = 0, x2 = (1 – i)s, and x3 = s where s is arbitrary. If we let s = 1, we have that
is a basis vector for the eigenspace corresponding to λ = –2. We normalize this
vector to obtain
Thus
diagonalizes A and
Pi i
= −− −
0 1 0
1
60
1
32
60
1
3
PP3
0
1
31
3
=−
i
0
1
1
−
i
−− −+ −
7 0 0
0 1 1
0 1 2
1
2
3
i
i
x
x
x
=
0
0
0
PP2
1
0
0
=
Exercise Set 10.6 567
13. The eigenvalues of A are the roots of the equation
The roots of this equation, which are λ = are not real. This shows that the
eigenvalues of a symmetric matrix with nonreal entries need not be real. Theorem 10.6.6
applies only to matrices with real entries.
14. (b) Let A be hermetian and unitary. Then A = AH = A–1.
15. We know that det(A) is the sum of all the signed elementary products ,
where aij
is the entry from the ith row and jth column of A. Since the ijth element of
, then det is the sum of the signed elementary products or
. That is, det is the sum of the conjugates of the terms in det(A). But
since the sum of the conjugates is the conjugate of the sum, we have
det
16. (a) We have
det( *) det( )
det( ) ( det(
A A
A
T=
= =since det( )B BB
A
T ))
det( ) ( )= by Exercise 15
A A( ) = det( ).
( )A±a a aj j njn1 21 2…
±a a aj j njn1 21 2…( )AA aijis
±a a aj j njn1 21 2…
4 16 4 19
2
± − ( ),
detλ
λλ λ
− −− −
= − + =
1 4
4 34 19 02i
i
P AP
i
i
− =
−+
+
1
01
6
2
61 0 0
01
3
1
3
− − +− −
−−
5 0 0
0 1 1
0 1 0
0 1 0
1i
i
i
660
1
32
60
1
3
1 0 0
0 5 0
0 0 2
−
=−
i
568 Exercise Set 10.6
(b) If A is Hermitian, then A = A*. Therefore,
det(A) = det(A*)
= (by Part (a))
But z = if and only if z is real.
(c) If A is unitary, then A–1 = A*. Therefore,
det(A) = 1/det(A–1)
= 1/det(A*)
= 1/ (by Part (a))
Thus det(A)2 = 1. But since |z| ≥ 0 for all z, it follows that |det(A)| = 1.
18. (b) We have
(d) We have
19. If A is invertible, then
A*(A–1)* = (A–1 A)* (by Exercise 18(d))
Thus we have (A–1)* = (A*)–1.
I I I IT T* = = =
( )* ( )
( )
* *
AB AB
AB
B A
B A
T
T
T T
=
=
=
=
( )* ( )
( )
( )
* *
A B A B
A B
A B
A B
T
T
T T
+ = +
= +
= += +
det( )A
z
det( )A
Exercise Set 10.6 569
20. Hint: use Exercise 18(a) and Exercise 19.
21. Let ri
denote the ith row of A and let cj
denote the jth column of A*. Then, since
we have cj
= for j = 1,…,n. Finally, let
Then A is unitary ⇔ A–1 = A*. But
A–1 = A* ⇔ AA* = I
⇔ ri
cj = δij for all i, j
⇔ ri rj = δij for all i, j
⇔ r1,…,rn is an orthonormal set
22. We know that if A is unitary, then so is A*. But since (A*)* = A, then if A* is unitary, so isA. Moreover, the columns of A are the conjugates of the rows of A*. Thus we have
A is unitary ⇔ A* is unitary
⇔ the rows of A* form an orthonormal set
⇔ the conjugates of the columns of A form an orthonormalset
⇔ the columns of A form an orthonormal set
23. (a) We know that A = A*, that Ax = λIx, and that Ay = µIy. Therefore
x* Ay = x*(µIy) = µ(x* Iy) = µx*y
and
x* Ay = [(x* Ay)*]* = [y* A*x]*
= [y* Ax]* = [y* (λIx)]*
= [λy*x]* = λx*y
The last step follows because λ, being the eigenvalue of an Hermitian matrix, is real.
δij
i j
i j=
≠=
0
1
if
if
rrjA A AT T* ( ),= =
570 Exercise Set 10.6
(b) Subtracting the equations in Part (a) yields
(λ – µ)(x* y) = 0
Since λ ≠ µ, the above equation implies that x*y is the 1 × 1 zero matrix. Letx = (x1,…, x
n) and y = (y1,…, y
n). Then we have just shown that
so that
and hence x and y are orthogonal.
x y x yn n1 1 0 0+ + = =
x y x yn n1 1 0+ + =
Exercise Set 10.6 571
SUPPLEMENTARY EXERCISES 10
2. If a and b are not both zero, then | a |2 + | b |2 ≠ 0 and the inverse is
This inverse is computed exactly as if the entries were real.
3. The system of equations has solution x1 = –is + t, x2 = s, x3 = t. Thus
where s and t are arbitrary. Hence
form a basis for the solution space.
−
i
1
0
1
0
1
and
x
x
x
i
s
1
2
3
1
0
1
0
1
=−
+
t
a
a b
b
a b
b
a b
a
a b
2 2 2 2
2 2 2 2
+ +
−
+ +
573
5. The eigenvalues are the solutions, λ, of the equation
or
But so that . Thus we have
λ3 – 1 = 0
or
(λ – 1)(λ2 + λ + 1) = 0
Hence λ = 1, ω, or . Note that = ω 2.
6. (a) The eigenvalues of are
z = a + bi and = a – bi
(b) det = a2 + b2 = z = | z |2
which corresponds to the complex number (ac – bd) + i (ad + bc)
((cc)) Ζ Ζ Ζ− =+ −
=12 2
1 1
a b
a b
b a
T
det
whilee zz
z
a bi c di ac bd i
− =
+ + = − +
12
((dd)) ( )( ) ( ) (aad bc
a b
b a
c d
d c
+
−
−
)
while ==− − −+ −
ac bd ad bc
ad bc ac bc
zΖ
z
Ζ
ωω
ωω
ω+ + = + =11
2 1 0Re( )1
ωω= ,
λ ωω
λ ωω
λ3 211
11
1 0+ + +
− + +
− =
det
λ
λ ωω
λ ωω
0 1
1 11
0 1 11
−
− − − −
− + + +
= 00
574 Supplementary Exercises 10
7. (c) Following the hint, we let z = cos θ + i sin θ = eiθ in Part (a). This yields
If we expand and equate real parts, we obtain
But
Re Re1
1
1 111
−−
=
−( ) −+( ) +( )e
e
e en i
i
n iθ
θ
θ −−
−
+( )
( )−( ) −( )
= −
i
i i
n i
e e
e
θ
θ θ1 1
11
Reθθ θ θ
θ− +
−
=− +( )
−e e
e
n
i ni
i2 2
1
2
1 1
Re( )
cos θθ θ θ
θ
θ
− −( )+
−
= −
cos cos
cos
cos
n
1
1
2
1
11
1
1
1
21
−+
− +( )( )−
=
cos
cos cos
cosθθ θ
θn n
++− −
cos cos cos sin sin
sin
n n nθ θ θ θ θθ
22
2
= +−( )+1
21
1 22 2
cos cos sin sin cosn nθ θ θ θ θ
222
1
21
22
2
2
2
sin
cos sin sin
θ
θ θ
= ++n nθθ θ θ
θ
sin cos
sin
2 2
22
2
1 21
1
1
+ + + + =−
−
+cos cos cos Re
( )
θ θ θθ
θ ne
e
n i
i
11
12
1
+ + + + =−
−
+e e e
e
e
i i nin i
i
θ θ θθ
θ( )
Supplementary Exercises 10 575
Observe that because 0 < θ < 2π, we have not divided by zero.
8. Observe that ω 3 = 1. Hence by Exercise 6(b), we have 1 + ω + ω2 = 0. Therefore
Thus, the vectors form an orthonormal set in C3.
9. Call the diagonal matrix D. Then
( )* ( ) *UD UD D U DU DU
T T T= = = = −1
vv vv
vv vv
1 •
•
22 2
1 22
1
31
1
31 0
1
31
= + +( ) = + +( ) =
= + +
ω ω ω ω
ω ω 44 2
1 22 2 4
1
31 0
1
31
1
31
( ) = + +( ) =
= + +( )= +
ω ω
ωω ω ω
ω
vv vv•
55 10 2
2
1 2
1
31
0
1
31 1
+( ) =
= + +( )=
= +
ω ω ω
ω ω
( )
•
since
vv vv ++( ) =
= + +( ) = =
1 1
1
31 1 11 2
2 2
1
vv vv
vv vv
•
•
( )ωω ω ω ωωsince
222 2 41
31 1= + +( ) =ω ω ω ω4 as above( )
= ++
1
21 2 2
2
cos sin sin cos
sin
n nθ θ θ θ
θ
== ++
1
21
12
2
sin
sin
n θ
θ
576 Supplementary Exercises 10
Supplementary Exercises 10 577
We need only check that (UD)* = (UD)–1. But
and
Hence (UD)* = (UD)–1 and so UD is unitary.
10. (a) Since A* = –A, we have
Hence iA is Hermitian.
(b) By Part (a), iA is Hermitian and hence unitarily diagonalizable with real eigenvalues.That is, there is a unitary matrix P and a diagonal matrix D with real entries suchthat
P–1(iA)P = D
Thus
i(P–1 AP) = D
or
P–1 AP = –iD
That is, A is unitarily diagonalizable with eigenvalues which are –i times the eigenvaluesof iA, and hence pure imaginary.
( ) ( )* *iA iA i A iA= = − − =
DD
z
z
z
I
n
=
=
12
22
2
0 0
0 0
0 0
…
…
…
( )*( ) ( )( )UD UD DU UD DD= =−1
...
...
...
578 Supplementary Exercises 10
11. Show the eigenvalues of a unitary matrix have modulus one.
Proof: Let A be unitary. Then A–1 = AH.
Let λ be an eigenvalue of A, with corresponding eigenvector
Then Ax2 = (Ax)H(Ax) = (xHAH)(Ax) = xH(A–1A)x = xHx =
but also
Ax 2 = λx 2 = (λx)H(λx) = ( λ)(xHx)
since λ and are scalars = | λ |2 x 2. So, | λ |2 = 1, and hence the eigenvalues of A havemodulus one.
λ
λ
x1
2
x
EXERCISE SET 11.1
1. (a) Substituting the coordinates of the points into Eq. (4) yields
which, upon cofactor expansion along the first row, yields –3x + y + 4 = 0; that is,y = 3x – 4.
(b) As in (a),
yields 2x + y – 1 = 0 or y = –2x + 1.
2. (a) Using Eq. (9) yields
which, upon first-row cofactor expansion, yields 18(x2 + y2) – 72x – 108y + 72 = 0 or,dividing by 18, x2 + y2 – 4x – 6y + 4 = 0. Completing the squares in x and y yields thestandard form (x – 2)2 + (y – 3)2 = 9.
x y x y2 2 1
40 2 6 1
4 2 0 1
34 5 3 1
0
+
=
x y 1
0 1 1
1 1 1
0
−=
x y 1
1 1 1
2 2 1
0− =
579
2. (b) As in (a),
yields 50(x2 + y2) + 100x – 200y – 1000 = 0; that is, x2 + y2 + 2x – 4y – 20 = 0. Instandard form this is (x + 1)2 + (y – 2)2 = 25.
3. Using Eq. (10) we obtain
which is the same as
by expansion along the second row (taking advantage of the zeros there). Add column fiveto column three and take advantage of another row of all but one zero to get
.
Now expand along the first row and get 160x2 + 320xy + 160(y2 + y) – 320x = 0; that is,x2 + 2xy + y2 – 2x + y = 0, which is the equation of a parabola (since the discriminant c2
2 – 4c1c3 is zero).
x xy y y x2 2
4 0 0 2
4 10 20 2
16 4 0 4
0
+
−−
= .
x xy y x y2 2
0 0 1 0 1
4 0 0 2 0
4 10 25 2 5
16 4 1 4 1
0−
− −− −
=
x xy y x y2 2 1
0 0 0 0 0 1
0 0 1 0 1 1
4 0 0 2 0 1
4 10 25 2 5 1
16 4 1
−
− −− 44 1 1
0
−
=
x y x y2 2 1
8 2 2 1
34 3 5 1
52 4 6 1
0
+−
−
=
580 Exercise Set 11.1
4. (a) From Eq. (11), the equation of the plane is
Expansion along the first row yields –2x – 4y – 2z = 0; that is, x + 2y + z = 0.
(b) As in (a),
yields –2x + 2y – 4z + 2 = 0; that is, –x + y – 2z + 1 = 0 for the equation of the plane.
5. (a) Using Eq. (12), the equation of the sphere is
Expanding by cofactors along the first row yields 16(x2 + y2 + z2) – 32x – 64y – 32z +32 = 0; that is, (x2 + y2 + z2) – 2x – 4y – 2x + 2 = 0. Completing the squares in eachvariable yields the standard form (x – 1)2 + (y – 2)2 + (z – 1)2 = 4.
NOTE: When evaluating the cofactors, it is useful to take advantage of the column ofones and elementary row operations; for example, the cofactor of x2 + y2 + z2
above can be evaluated as follows:
by cofactor expansion of the latter determinant along the last column.
1 2 3 1
1 2 1 1
1 0 1 1
1 2 1 1
1 2 3 1
2 0 2 0
0 2 2 0
0 0 4 0
−
−
=− −
− −−
==16
x y z x y z2 2 2 1
14 1 2 3 1
6 1 2 1 1
2 1 0 1 1
6 1 2 1 1
0
+ +
−
−
= .
x y z 1
2 3 1 1
2 1 1 1
1 2 1 1
0− −
=
x y z 1
1 1 3 1
1 1 1 1
0 1 2 1
0−
−−
= .
Exercise Set 11.1 581
5. (b) As in (a),
yields –24(x2 + y2 + z2) + 48x + 48y + 72 = 0; that is, x2 + y2 + z2 – 2x – 2y – 3 = 0or in standard form, (x – 1)2 + (y – 1)2 + z2 = 5.
6. Substituting each of the points (x1, y1), (x2, y2), (x3, y3), (x4, y4), and (x5, y5) into theequation
c1x2 + c2xy + c3y
2 + c4x + c5y + c6 = 0
yields
These together with the original equation form a homogeneous linear system with a non-trivial solution for c1, c2, … , c6. Thus the determinant of the coefficient matrix is zero,which is exactly Eq. (10).
7. As in the previous problem, substitute the coordinates (xi, y
i, z
i) of each of the three
points into the equation c1x + c2y + c3z + c4 = 0 to obtain a homogeneous system withnontrivial solution for c1, … , c4. Thus the determinant of the coefficient matrix is zero,which is exactly Eq. (11).
8. Substituting the coordinates (xi, y
i, z
i) of the four points into the equation c1(x2 + y2 + z2)
+ c2x + c3y + c4z + c5 = 0 of the sphere yields four equations, which together with theabove sphere equation form a homogeneous linear system for c1, … , c5 with a nontrivialsolution. Thus the determinant of this system is zero, which is Eq. (12).
c x c x y c y c x c y c
c x
1 12
2 1 1 3 12
4 1 5 1 6
1 52
0+ + + + + =
+
.
cc x y c y c x c y c2 5 5 3 52
4 5 5 5 6 0+ + + + = .
x y z x y z2 2 2 1
5 0 1 2 1
11 1 3 1 1
5 2 1 0 1
11 3 1 1 1
0
+ +−
−−
=
582 Exercise Set 11.1
9. By substituting of the coordinates of the three points (x1, y1), (x2, y2) and (x3, y3) weobtain the equations:
c1y + c2x2 + c3x + c4 = 0.
c1y1 + c2x12 + c3x1 + c4 = 0.
c1y2 + c2x22 + c3x2 + c4 = 0.
c1y3 + c2x32 + c3x3 + c4 = 0.
This is a homogeneous system with a nontrivial solution for c1, c2, c3, c4, so the determinantof the coefficient matrix is zero; that is,
y x x
y x x
y x x
y x x
2
1 12
1
2 22
2
3 32
3
1
1
1
1
0= .
Exercise Set 11.1 583
EXERCISE SET 11.2
1.
Applying Kirchhoff’s current law to node A in the figure yields
I1 = I2 + I3.
Applying Kirchhoff’s voltage law and Ohm’s law to Loops 1 and 2 yields
5I1 + 13I2 = 8
and
9I3 – 13I2 + 5I3 = 3.
In matrix form these three equations are
with solution I I I1 2 3255
317
97
317
158
317= = =, , .
1 1 1
5 13 0
0 13 14
1
2
3
− −
−
I
I
I
=
0
8
3
loop 1
loop 2
node A
5 Ω
5 Ω
13
9 Ω
3 V
8 V–
–
+
+
I1
I2
I3
585
2.
Node A gives I1 + I2 = I3.
Loop 1 gives –2I1 + 2I2 = –6.
Loop 2 gives –2I2 – 4I3 = –8.
In matrix form we have
with solution
3.
loop 1 loop 2
node A
4 Ω
6 Ω
2 Ω
2 V
1 V
–
–
+
+
I1
I2
I3
I I I1 2 313
5
2
5
11
5= = − =, , .
1 1 1
2 2 0
0 2 4
1
2
3
−−
− −
I
I
I
== −−
0
6
8
loop 1 loop 2
node A
4 Ω2 Ω2 Ω
3 V
8 V
6 V–
–
+
+
I1 I2 I3
586 Exercise Set 11.2
Node A gives I1 + I3 = I2.
Loop 1 gives –4I1 – 6I2 = –1.
Loop 2 gives 4I1 – 2I3 = –2.
In system form we have
with solution
4.
Node A gives I1 = I2 + I4.
Node B gives I4 = I3 + I5.
Node C gives I6 = I3 + I5.
Loop 1 gives 20I1 + 20I2 = 10.
Loop 2 gives –20I2 + 20I3 = 0.
Loop 3 gives –20I3 + 20I5 = 10.
loop 1 loop 2 loop 3
node A node B
node C
20 Ω20 Ω 10 V10 V—+ —
+
I1
I2
I4 I5
I6
I3
I I I1 2 35
22
7
22
6
11= − = =, , .
1 1 1
4 6 0
4 0 2
1
2
3
−− −
−
I
I
I
== −−
0
1
2
Exercise Set 11.2 587
In matrix form we have
with solution
5.
After setting I5 = 0 we have that:
node A gives I1 = I3
node B gives I2 = I4
loop 1 gives I1R1 = I2R2
loop 2 gives I3R3 = I4R4.
From these four equations it easily follows that R4 = R3R2/R1.
loop 1
loop 2
node AE node B
I1 I2
R2
R5
R1
R3 R4
I3 I4
I5
I0
–+
I I I I I I1 4 5 6 31
20= = = = = =and 2 .
1 1 0 1 0 0
0 0 1 1 1 0
0 0 1 0 1 1
20 20 0 0 0 0
0 20 20 0 0 0
0 0 2
− −−
−
−− 00 0 20 0
1
2
3
4
5
6
I
I
I
I
I
I
=
0
0
0
10
0
10
588 Exercise Set 11.2
6.
Node A in the first circuit gives I = I1 + I2.
Loop 1 in the first circuit gives I1R1 = E.
Loop 2 in the first circuit gives I1R1 = I2R2.
Loop 3 in the first circuit gives IR = E.
From these four equations we obtain
RE
I
E
I I
E
E
R
E
R R R
.= =+
=+
=+1 2
1 2 1 2
11 1
loop 1 loop 2
node A
E E–+
I1R1 R2
I
I2
loop 3
–+
R
I
Exercise Set 11.2 589
EXERCISE SET 11.3
1.
In the figure, the feasible region is shown and the extreme points are labelled. The valuesof the objective function are shown in the following table:
Thus the maximum, 7-1/3, is attained when x1 = 2 and x2 = 2/3.
(3/2, 1)
(2, 2/3)
(2, 0)(0, 0)
x1 = 2x2
x1
x2 = 1
2x1 + 3x2 = 6
(1/2, 1)
591
Extreme point Value of(x1, x2) z = 3x1 + 2x2
(0, 0) 0
(1/2, 1) 3-1/2
(3/2, 1) 6-1/2
(2, 2/3) 7-1/3
(2, 0) 6
2.
The intersection of the five half-planes defined by the constraints is empty. Thus thisproblem has no feasible solutions.
3.
The feasible region for this problem, shown in the figure, is unbounded. The value ofz = –3x1 + 2x2 cannot be minimized in this region since it becomes arbitrarily negativeas we travel outward along the line –x1 + x2 = 1; i.e., the value of z is –3x1 + 2x2 = –3x1 + 2(x1 + 1) = –x1 + 2 and x1 can be arbitrarily large.
3x1 – x2 = ñ5
2x1 + 4x2 = 12
–x1 + x2 = 1
x1
x2
2x1 − x2 = −2
4x1 − x2 = 0x2 = 3
x1
x2
592 Exercise Set 11.3
4.
The feasible region and vertices for this problem are shown in the figure. The maximumvalue of the objective function z = .1x1 + .07x2 is attained when x1 = 6,000 and x2 = 4,000,then z = 880. In other words, she should invest $6,000 in bond A and $4,000 in bond Bfor an annual yield of $880.
5.
The feasible region and its extreme points are shown in the figure. Though the region isunbounded, x1 and x2 are always positive, so the objective function z = 7.5x1 + 5.0x2is also. Thus, it has a minimum, which is attained at the point where x1 = 14/9 andx2 = 25/18. The value of z there is 335/18. In the problem’s terms, we use 7/9 cups ofmilk and 25/18 ounces of corn flakes, a minimum cost of 18.6¢ is realized.
x2
x1
x1 = 3/2 3x1 – 2x1 = 0
x1 – 2x2 = 0
4x1 + 2x2 = 9
(3/2, 9/4)
(14/9, 25/18)
(40/21, 20/21)
x1 + x2 =18
110
13
(3/2, 3/2)
x2
x2 = 2000
x1 – x2 = 0 x1 + x2 = 10,000x1 = 6000
(5000, 5000)
(6000, 4000)
(6000, 2000)
x1
(2000, 2000)
Exercise Set 11.3 593
6. Letting x1 be the number of Company A’s containers shipped and x2 be the number ofCompany B’s, the problem is
Maximize z = 2.20x1 + 3.00x2
subject to
40x1 + 50x2 ≤ 37,000
2x1 + 3x2 ≤ 2,000
x1 ≥ 0
x2 ≥ 0.
The feasible region is shown in the figure. The vertex at which the maximum is attainedis when x1 = 550 and x2 = 300, then z = 2110.
7. We must now maximize z = 2.50x1 + 3.00x2. The feasible region is as in 11.6, but now themaximum happens when x1 = 925 and x2 = 0, then z = 2312.50.
8. Let x1 be the number of pounds of ingredient A used, and x2 be the number of pounds ofingredient B. Then the problem is
Minimize z = 8x1 + 9x2
subject to
2x1 + 5x2 ≥ 10
2x1 + 3x2 ≥ 8
6x1 + 4x2 ≥ 12
x1 ≥ 0
x2 ≥ 0
40x1 + 50x2 = 37,000
(550, 300)
(0, 0) (925, 0)
2x1 + 3x2 = 2000
x2
x1
(0, 666 )23
594 Exercise Set 11.3
Though the feasible region shown in the figure is unbounded, the objective function isalways positive there and hence must have a minimum. This minimum occurs at the vertexwhere x1 = 2/5 and x2 = 12/5. The minimum value of z is 124/5 or 24.8¢.
9. It is sufficient to show that if a linear functions has the same value at two points, then ithas that value along the line connecting the two points. Let ax1 + bx2 be the function, andax′1 + bx2′ = ax′1′ + bx2′′ = c. Then (tx′1 + (1 – t)x′1′, tx2′ + (1 – t)x2′′) is an arbitrary point on theline connecting (x1′, x2′) to (x1′′, x2′′) and a(tx1′ + (1 – t)x1′′) + b(tx2′ + (1 – t) x2′′)= t(ax1′ + bx2′) + (1 – t)(ax1′′ + bx2′′) = c.
10.
The level curves of the objective function –5x1 + x2 are shown in the figure, and it is readilyseen that the objective function remains bounded in the region.
z decreasing
x2
x1
z increasing
(0,3)
x2
x1
(2/5, 12/5)
(5/2, 1)
6x1 + 4x2 = 12 2x1 + 3x2 = 8 2x1 + 5x2 = 10
(5, 0)
Exercise Set 11.3 595
EXERCISE SET 11.4
1. (a) Step 1 yields
and Step 2 yields
We see that two lines are needed to cover the zeros as shown. Thus the test foroptimality fails. Applying Step 5 yields
Now we cannot cover the zeros with less than three lines, so the matrix contains anoptimal assignment. Actually there are two such assignments:
from the original matrix, we see the cost for each is 30.
2 0 2
0 1 0
0 0 0
2 0 2
0 1 0
0 0 0
and
;
2 0 2
0 1 0
0 0 0
.
0 0 0
3 0 3
1 0 1
.
13 0 6
10 0 3
11 0 4
597
1. (b) Step 1 yields and Step 2 yields
The zeros can be covered by three lines (as shown), so we proceed. Step 5 gives
Now we cannot cover the zeros with less than four lines, so there is an optimalassignment. There are two in fact:
each (from the original matrix) with a cost of –2.
(c) Step 1 yields
5 2 0 0 3
7 3 0 5 6
4 1 0 7 7
0 3 7 1 4
0 5 4 1 5
3 0 0 0
8 8 0 6
0 7 3 0
5 0 0 0
3 0 0 0
8 8 0 6
0 7
and33 0
5 0 0 0
,
3 0 0 0
8 8 0 6
0 7 3 0
5 0 0 0
.
5 0 2 3
8 6 0 7
0 5 3 1
7 0 2 3
5 0 2 2
8 6 0 6
00 5 3 0
7 0 2 2
.
598 Exercise Set 11.4
and Step 2 yields
The zeros can be covered by four lines, so, Step 5 yields
Now the zeros cannot be covered with fewer than five lines, so we have an optimalassignment. There are three such (one is shown), each having a cost of 39.
2. From the matrix (1), Step 1 yields
Step 2 yields
and this is not optimal. Step 5 yields
10 13 0
0 0 0
18 10 0
16 59 0
6 46 0
24 56 0
6 1 1 0 0
7 1 0 4 2
5 0 1 7 4
0 1 7 0 0
0 3 4 0 1
..
5 1 0 0 0
7 2 0 5 3
4 0 0 7 4
0 2 7 1 1
0 4 4 1 2
.
Exercise Set 11.4 599
clearly optimal, and we obtain the solutions given in the example.
3. Another such covering is
Then Step 5 yields
in which there is still a covering by three lines. Step 5 again yields
Since this matrix is the same as (8), we must get the same optimal assignments.
.
40 0 5 0
0 25 0 0
55 0 0 5
0 40 30 40
−
−
20 0 5 0
0 45 20 20
35 0 0 5
0 60 50 60
−
15 0 0 0
0 50 20 25
35 5 0 10
0 65 50 65
.
0 3 0
0 0 10
8 0 0
600 Exercise Set 11.4
4. Another such assignment is
giving the matches
Sue - Don (rank = 10)
Ann - Tom (rank = 9)
Bea - Joe (rank = 6)
Fay - Bob (rank = 6)
Hal - unmatched,
a total score of 31.
5. (a) A dummy coin adds a column of zeros, then we negate to obtain a starting matrix:
Step 1 yields
60 145 0 75 210
55 155 0 75 230
65 115 0 60 200
50 120 0 60 1900
30 150 0 40 200
− − − −− − − −− − −
150 65 210 135 0
175 75 230 155 0
135 85 200 −−− − − −− − − −
140 0
140 70 190 130 0
170 50 200 160 0
.
0 3 0 4 0
4 0 6 1 5
3 1 0 4 0
0 1 6 2 1
0 0 0 0 3
Exercise Set 11.4 601
and Step 2 yields
which is not optimal. Step 5 yields
which is not optimal. Step 5 yields
This is optimal, containing only the optimal assignment shown. Thus, bidder 1 getscoin 3 ($210), bidder 2 gets coin 1 ($175), bidder 3 gets coin 2 ($85), bidder 5 getscoin 4 ($160), and bidder 4 gets nothing; a total of $630.
(b) We use bidder 2’s row twice (so he has two chances at getting accepted), and addtwo dummy coin columns. Then negating yields the starting matrix
5 5 0 10 0
0 15 0 10 20
35 0 25 20 15
15 0 20 15 0
0 35 25 0 15
.
10 10 0 15 0
5 20 0 15 20
35 0 20 20 10
20 5 20 20 0
0 35 20 0 10
30 30 0 35 20
25 40 0 35 40
35 0 0 20 10
20 5 0 20 0
0 35 0 0 10
602 Exercise Set 11.4
After Steps 1 and 2 we get
not optimal. Step 5 yields
still not optimal. Step 5 again
10 10 0 15 0 0
5 20 0 15 20 20
5 20 0 15 20 20
35 0 20 20 10 10
20 5 220 20 0 0
0 35 20 0 10 10
,
30 30 0 35 20 20
25 40 0 25 40 40
25 40 0 35 40 40
35 0 0 20 10 10
220 5 0 20 0 0
0 35 0 0 10 10
,
− − − −− − − −− − −
150 65 210 135 0 0
175 75 230 155 0 0
175 75 2330 155 0 0
135 85 200 140 0 0
140 70 190 130 0 0
−− − − −− − − −−− − − −
170 50 200 160 0 0
.
Exercise Set 11.4 603
which is optimal. The matrix contains several optimal assignments, but all result in:bidder 2 gets coins 1 and 3 ($405), bidder 3 gets coin 2 ($85), bidder 5 gets coin 4($160), and bidders 1 and 4 get nothing; this is a total of $650.
6. Negating the original matrix, and applying Steps 1 and 2 yield
This is not optimal, and Step 5 gives
4 6 13 5 5 0 0 20 7
4 1 1 0 3 6 7 8 4
0 0 2 3 0 6 6 0 1
6 2 8 5 2 13 12 0 7
7 3 88 5 3 13 11 0 7
4 2 3 4 2 13 10 0 7
7 2 1 3 5 7 8 0 0
4 0 0 2 2 4 5 0 0
4 0 0 2 22 4 5 0 0
.
5 5 0 10 0 0
0 15 0 10 20 20
0 15 0 10 20 20
35 0 25 20 15 15
15 0 20 15 0 0
0 35 25 0 15 15
,
604 Exercise Set 11.4
This is not optimal. Step 5 again yields
This is still not optimal. Step 5 again gives
6 10 17 17 7 0 0 24 11
4 3 3 0 3 4 5 10 6
0 2 4 3 0 4 4 2 3
0 2 4 3 0 9 8 0 7
5 33 8 3 1 9 7 0 7
2 2 3 2 0 9 6 0 7
5 2 1 1 3 3 4 0 0
2 0 0 0 0 0 1 0 0
2 0 0 0 0 0 1 00 0
.
4 8 15 15 5 0 0 22 9
4 3 3 0 3 6 7 10 6
0 2 4 3 0 6 6 2 3
4 2 8 3 0 11 10 0 7
5 33 8 3 1 11 9 0 7
2 2 3 2 0 11 8 0 7
5 2 1 1 3 5 6 0 0
2 0 0 0 0 2 3 0 0
2 0 0 0 0 22 3 0 0
.
Exercise Set 11.4 605
This is optimal, yielding the following lineup(s):
P - Pete 2B - Jill or Ann LF - Herb
C - Liz 3B - Alex CF - John or Lois
1B - Sam SS - Ann or Jill RF - Lois or John
All possibilities have a “score” of 125.
7. Proof of part (ii) of the test for optimality: Since each zero entry of an optimal assignmentis in a different row, a horizontal line can cover at most one assigned zero; similarly, avertical line can cover at most one assigned zero. Thus, as many lines as there are zeros isneeded, that is, n lines.
Proof of part (i) of the test for optimality: Suppose we are given an n × n matrix that doesnot contain an optimal assignment of zeros. Then we cannot find a set of n zero entries, notwo of which come from the same row or column. But we can always find a set of fewerthan n zero entries, say k zero entries where k < n, no two of which come from the samerow or column (possibly k is zero). A set of k such zero entries for which k is as large aspossible is called a maximal assignment of zeros, and the zero entries in the set are calledthe assigned zeros of the maximal assignment. We shall prove the following theorem:
THEOREM All of the zero entries in matrix that contains a maximal
assignment of zeros with k assigned zeros can be covered with k vertical
and horizontal lines.
8 10 17 19 9 0 0 26 11
4 1 1 0 3 2 3 10 4
0 0 2 3 0 2 2 2 1
4 0 6 3 0 7 6 0 5
5 11 6 3 1 7 5 0 5
2 0 1 2 0 7 4 0 5
7 2 1 3 5 3 4 2 0
4 0 0 2 2 0 1 2 0
4 0 0 2 2 0 1 22 0
606 Exercise Set 11.4
PROOF. The following algorithm shows how the k lines in the statement of the theoremcan be found:
Step 1. Search the matrix for unassigned zeros that do not contain any assigned zeros intheir rows. Cover each of the columns in which these unassigned zeros lie with a verticalline. Label one such unassigned zero in each covered column as 01. Each column coveredin this way must also contain an assigned zero, which we label as 01
*. (If not, then we couldadd the unassigned zero 01 in that column to the maximal assignment of zeros — acontradiction to the definition of a maximal assignment of zeros.)
Step 2. Search the matrix for uncovered, unassigned zeros that contain an assigned zero ofthe form 01
* in their rows. Cover each of the columns in which these uncovered, unassignedzeros lie with a vertical line. Label one such uncovered, unassigned zero in each newly-covered column as 02. Each column covered in this way must also contain an assigned zero,which we label as 02
*. (If not, then we could add the unassigned zero 02 in that column to themaximal assignment of zeros after we reassign the assigned zero of the form 01
* in its row tothe corresponding unassigned zero 01.)
Step 3. Continue Step 2 iteratively, so that at the m-th step we search for uncovered,unassigned zeros that contain an assigned zero of the form 0
m–1* in its row. Cover their
columns with vertical lines and label one such uncovered, unassigned zero from each suchcovered column as 0
m. Each newly covered column must also contain an assigned zero,
which we label as 0m* (otherwise we could increase the number of assigned zeros by one by
replacing m – 1 assigned zeros of the form 01*, 02
*, …, 0*m–1 by m unassigned zeros of the
form 01, 02, …, 0m–1, 0
m). We continue iteratively until we can find no more unassigned
zeros which contain covered, assigned zeros in their rows.
Step 4. After Step 3, all uncovered, unassigned zeros must lie in rows that containuncovered, assigned zeros. Passing horizontal lines through these remaining uncovered,assigned zeros completes the algorithm. All zeros in the matrix are covered and each of thek assigned zeros has either a vertical or a horizontal line through it. There are thus exactlyk covering lines, which is what we wanted to prove.
As an example of this algorithm, the zero entries of the following 7 × 7 matrix containing amaximal assignment of zeros with six assigned zeros (labeled with asterisks):
Exercise Set 11.4 607
can be covered with six lines as follows:
Remark: The above theorem was first proved in 1931 by two Hungarian mathematicians, D. König and E. Egerváry. It is after them that the Hungarian Method is named.
8. (a) Let q be the number of intersections of covering lines. Then there are nm – q entriescovered altogether (since each line covers n entries, but q are covered twice). Thereare thus n2 – (nm – q) uncovered entries. The total amount subtracted from C is thusa(n2 – nm + q), and the total amount added to C is aq. So, the sum of the entries ofC′ is equal to the sum of the entries of C plus aq – a(n2 – nm + q), i.e.,
(b) We assume the original matrix contains integers only. Since the difference Σ (cij
– c′ij)
is positive if a > 0 and n > m, the sum of the entries will decrease each time. If theprocess does not terminate, the sequence of sums (being a decreasing sequence ofnonnegative integers) will reach zero. But then the matrix, non-negative, must beidentically zero, so we must get an optimal assignment of zeros either way.
c c an n mij ij∑ ∑− ′ = −( ).
1 0 0 0 1 1 1
1 1 0 0 1 1 0
0 1 1 0 1 0 0
1 0 1 0 1 1 1
1 1 0 1 0 0 1
1 1 1 0 0 1 1
1 0 1 1 0
1 2
2
1
1
*
*
*
*
*
*
11 1 1
.
1 0 1 0 1 1 1
1 1 0 0 1 1 0
0 1 1 0 1 0 0
1 0 1 0 1 1 1
1 1 0 1 0 0 1
1 1 1
*
*
*
*
*
00 0 1 1
1 0 1 1 0 1 1
*
,
608 Exercise Set 11.4
EXERCISE SET 11.5
1. We have 2b1 = M1, 2b2 = M2, … , 2bn–1 = M
n–1 from (6).
Inserting in (13) yields
6a1h + M1 = M2
6a2h + M2 = M3
6an–2h + M
n–2 = Mn–1,
from which we obtain
Now S′′ (xn) = M
n, or from (6), 6a
n–1h + 2bn–1 = M
n.
Also, 2bn–1 = M
n–1 from (6) and so
6an–1h + M
n–1 = Mn
or
aM M
h
aM M
h
aM M
hn
n n
12 1
23 2
21 2
6
6
6
=−
=−
=−
−− −
.
609
Thus we have
for i = 1, 2, … , n – 1. From (9) and (11) we have
aih3 + b
ih2 + c
ih + d
i= yi+1, i = 1, 2, … , n – 1.
Substituting the expressions for ai, b
i, and d
ifrom (14) yields
cih + y
i= y
i+1, i = 1, 2, …, n – 1. Solving for ci
gives
for i = 1, 2, … , n – 1.
2. (a) Set h = .2 and
x1 = 0, y1 = .00000.
x2 = 2, y2 = .19867.
x3 = 4, y3 = .38942.
x4 = .6, y4 = .56464.
x5 = .8, y5 = .71736.
x6 = 1.0, y6 = .84147.
Then
6(y1 – 2y2 + y3)/h2 = –1.1880
6(y2 – 2y3 + y4)/h2 = –2.3295
6(y3 – 2y4 + y5)/h2 = –3.3750
6(y4 – 2y5 + y6)/h2 = –4.2915
cy y
h
M Mhi
i i i i=−
−−
+ +1 1 2
6
M M
hh
Mhi i i+ −
+ +1 3 2
6 2
aM M
hi
i=−+1
6
aM M
hn
n n−
−=−
11
6.
610 Exercise Set 11.5
and the linear system (21) for the parabolic runout spline becomes
Solving this system yields
M2 = –.15676
M3 = –.40421
M4 = –.55592
M5 = –.74712.
From (19) and (20) we have
M1 = M2 = –0.15676
M6 = M5 = –0.74712.
The specified interval .4 ≤ x ≤ .6 is the third interval. Using (14) to solve for a3, b3,c3 and d3 gives
a3 = (M4 – M3)/6h = –.12643
b3 = M3/2 = –.20211
c3 = (y4 – y3)/h – (M4 + 2M3)h/6 = .92158
d3 = y3 = .38942.
The interpolating parabolic runout spline for .4 ≤ x ≤ .6 is thus
S(x) = –.12643(x – .4)3 – .20211(x –.4)2
+.92158(x – .4) + .38942.
5 1 0 0
1 4 1 0
0 1 4 1
0 0 1 5
2
3
4
5
M
M
M
M
=
−−−−
1 1880
2 3295
3 3750
4 2915
.
.
.
.
.
Exercise Set 11.5 611
2. (b)
S(.5) = –.12643(.1)3 – .20211(.1)2 + .92158(.1) + .38942
= .47943.
Since sin(.5) = S(.5) = .47943 to five decimal places, the percentage error is zero.
3. (a) Given that the points lie on a single cubic curve, the cubic runout spline will agreeexactly with the single cubic curve.
(b) Set h = 1 and
x1 = 0 , y1 = 1
x2 = 1 , y2 = 7
x3 = 2 , y3 = 27
x4 = 3 , y4 = 79
x5 = 4 , y5 = 181.
Then
6(y1 – 2y2 + y3)/h2 = 84
6(y2 – 2y3 + y4)/h2 = 192
6(y3 – 2y4 + y5)/h2 = 300
and the linear system (24) for the cubic runout spline becomes
Solving this system yields
M2 = 14.
M3 = 32.
M4 = 50.
6 0 0
1 4 1
0 0 6
84
1922
3
4
=M
M
M 3300
.
612 Exercise Set 11.5
From (22) and (23) we have
M1 = 2M2 – M3 = –4.
M5 = 2M4 – M3 = 68.
Using (14) to solve for the ai’s, b
i’s, c
i’s, and d
i’s we have
a1 = (M2 – M1)/6h = 3
a2 = (M3 – M2)/6h = 3
a3 = (M4 – M3)/6h = 3
a4 = (M5 – M4)/6h = 3
b1 = M1/2 = –21
b2 = M2/2 = 171
b3 = M3/2 = 161
b4 = M4/2 = 251
c1 = (y2 – y1)/h – (M2 + 2M1)h/6 = 55
c2 = (y3 – y2)/h – (M3 + 2M2)h/6 = 10
c3 = (y4 – y3)/h – (M4 + 2M3)h/6 = 33
c4 = (y5 – y4)/h – (M5 + 2M4)h/6 = 74
d1 = y1 = 1.
d2 = y2 = 7.
d3 = y3 = 2.
d4 = y4 = 79.
Exercise Set 11.5 613
For 0 ≤ x ≤ 1 we have
S(x) = S1(x) = 3x3 – 2x2 + 5x + 1.
For 1 ≤ x ≤ 2 we have
S(x) = S2(x) = 3(x – 1)3 + 7(x – 1)2 + 10(x – 1) + 7
= 3x3 – 2x2 + 5x + 1.
For 2 ≤ x ≤ 3 we have
S(x) = S3(x) = 3(x – 2)3 + 16(x – 2)2 + 33(x – 2) + 27
= 3x3 – 2x2 + 5x + 1.
For 3 ≤ x ≤ 4 we have
S(x) = S4(x) = 3(x – 3)3 + 25(x – 3)2 + 74(x – 3) + 79
= 3x3 – 2x2 + 5x + 1.
Thus S1(x) = S2(x) = S3(x) = S4(x),
or S(x) = 3x3 – 2x2 + 5x + 1 for 0 ≤ x ≤ 4.
4. The linear system (16) for the natural spline becomes
Solving this system yields
4 1 0
1 4 1
0 1 4
0002
3
4
=−.M
M
M
11116
0000816
0000636
−−
.
.
614 Exercise Set 11.5
M2 = –.0000252.
M3 = –.0000108.
M4 = –.0000132.
From (17) and (18) we have
M1 = 0.
M5 = 0.
Solving for the ai’s, b
i’s, c
i’s, and d
i’s from Eqs. (14) we have
a1 = (M2 – M1)/6h = –.00000042
a2 = (M3 – M2)/6h = –.00000024
a3 = (M4 – M3)/6h = –.00000004
a4 = (M5 – M4)/6h = –.00000022
b1 = M1/2 = 0
b2 = M2/2 = –.0000126
b3 = M3/2 = –.0000054
b4 = M4/2 = –.0000066
b5 = M5/2 = 0
c1 = (y2 – y1)/h – (M2 + 2M1)h/6 = –.000214
c2 = (y3 – y2)/h – (M3 + 2M2)h/6 = –.000088
c3 = (y4 – y3)/h – (M4 + 2 M3)h/6 = –.000092
c4 = (y5 – y4)/h – (M5 + 2M4)h/6 = –.000212
Exercise Set 11.5 615
d1 = y1 = .99815.
d2 = y2 = .99987.
d3 = y3 = .99973.
d4 = y4 = .99823.
The resulting natural spline is
Assuming the maximum is attained in the interval [0, 10] we set S′(x) equal to zero in thisinterval:
S′(x) = .00000072x2 –.0000252x + .000088 = 0.
To three significant digits the root of this quadratic equation in the interval [0, 10] is x =3.93, and
S(3.93) = 1.00004.
5. The linear system (24) for the cubic runout spline becomes
Solving this system yields
M2 = –.0000186.
M3 = –.0000131.
M4 = –.0000106.
6 0 0
1 4 1
0 0 6
0002
3
4
=−.M
M
M
11116
0000816
0000636
−−
.
.
.
S x
x x
( )
. ( ) . ( ) .
=
− + + + +00000042 10 000214 10 998153 ,,
. ( ) . ( ) .
− ≤ ≤
− +
10 0
00000024 0000126 0000883 2
x
x x (( ) . ,
. ( ) .
x x
x
+ ≤ ≤
− − −
99987 0 10
0000004 10 0000053 44 10 000092 10 99973 10 20
000
2( ) . ( ) . ,
.
x x x− − − + ≤ ≤
− 000022 20 0000066 20 000212 203 2( ) . ( ) . ( )x x x− − − − − ++ ≤ ≤
. , .99823 20 30x
616 Exercise Set 11.5
From (22) and (23) we have
M1 = 2M2 – M3 = –.0000241.
M5 = 2M4 – M3 = –.0000081.
Solving for the ai’s, b
i’s, c
i’s and d
i’s from Eqs. (14) we have
a1 = (M2 – M1)/6h = .00000009
a2 = (M3 – M2)/6h = .00000009
a3 = (M4 – M3)/6h = .00000004
a4 = (M5 – M4)/6h = .00000004
b1 = M1/2 = –.0000121
b2 = M2/2 = –.0000093
b3 = M3/2 = –.0000066
b4 = M4/2 = –.0000053
c1 = (y2 – y1)/h – (M2 + 2 M1)h/6 = .000282
c2 = (y3 – y2)/h – (M3 + 2M2)h/6 = .000070
c3 = (y4 – y3)/h – (M4 + 2M3)h/6 = .000087
c4 = (y5 – y4)/h – (M5 + 2M4)h/6 = .000207
d1 = y1 = .99815.
d2 = y2 = .99987.
d3 = y3 = .99973.
d4 = y4 = .99823.
Exercise Set 11.5 617
The resulting cubic runout spline is
Assuming the maximum is attained in the interval [0, 10], we set S′(x) equal to zero in thisinterval:
S′(x) = .00000027x2 – .0000186x + .000070.
To three significant digits the root of this quadratic in the interval [0, 10] is 4.00 and
S(4.00) = 1.00001.
6. (a) Since S(x1) = y1 and S(xn) = y
n, then from S(x1) = S(x
n) we have y1 = yn.
By definition S′′(x1) = M1 and S′′(xn) = M
n, and so from S′′(x1) = S′′(x
n) we have
M1 = Mn.
From (5) we have
S′(x1) = c1
S′(xn) = 3a
n–1h2 + 2b
n–1h + cn–1.
Substituting for c1, an–1, bn–1, cn–1 from Eqs. (14) yields
S′(x1) = (y2 – y1)/h – (M2 + 2M1)h/6
S′(xn) = (M
n– M
n–1)h/2 + Mn–1h
+ (yn
– yn–1)/h – (M
n+ 2M
n–1)h/6.
S x
x x
( )
. ( ) . ( ) .
=
+ − + +00000009 10 0000121 10 03 2 000282 10 99815 10 0
00000009 3
( ) . ,
. ( )
x x
x
+ + − ≤ ≤
− .. ( ) . ( ) . ,
.
0000093 000070 99987 0 10
0
2x x x+ + ≤ ≤
00000004 10 0000066 10 0000873 2( ) . ( ) . (x x x− − − + −− + ≤ ≤
− −
10 99973 10 20
00000004 20 003
) . ,
. ( ) .
x
x 000053 20 000207 20 99823 20 302( ) . ( ) . ,x x x− + − + ≤ ≤ ..
618 Exercise Set 11.5
Using Mn
= M1 and yn
= y1, the last equation becomes
S′(xn) = M1h/3 + M
n–1h/6 + (y1 – yn–1)/h.
From S′(x1) = S′(xn) we obtain
(y2 – y1)/h – (y1 – yn–1)/h = M1h/3 + M
n–1h/6 + (M2 + 2M1)h/6
or
4M1 + M2 + Mn–1 = 6(y
n–1 – 2y1 + y2)/h2.
(b) Eqs. (15) together with the three equations in part (a) of the exercise statement give
4M1 + 4M2 + Mn–1 = 6(y
n–1 – 2y1 + y2)/h2
4M1 + 4M2 + M3 = 6(y1 – 2y2 + y3)/h2
4M2 + 4M3 + M4 = 6(y2 – 2y3 + y4)/h2
Mn–3 + 4M
n–2 + Mn–1
= 6(yn–3 – 2y
n–2 + yn–1)/h2
M1 + 4Mn–2 + 4M
n–1 = 6(yn–2 – 2y
n–1 + y1)/h2.
This linear system for M1, M2, … , Mn–1 in matrix form is
4 1 0 0 0 0 0 1
1 4 1 0 0 0 0 0
0 1 4 1 0 0 0 0
0 0
. . .
. . .
. . .
00 0 0 1 4 1
1 0 0 0 0 0 1 4
1
2
. . .
. . .
M
M
MM
M
M
h
y y
n
n
n
3
2
1
2
1
6
2
−
−
−
=
− 11 2
1 2 3
2 3 4
3 2 1
2
2
2
2
+− +− +
− +− − −
−
y
y y y
y y y
y y y
y
n n n
n
−− +
−2 1 1y yn
Exercise Set 11.5 619
7. (a) Since S′(x1) = y1′ and S′(xn) = y
n′, then from (5) we have
S′(x1) = c1
S′(xn) = 3a
n–1h2 + 2b
n–1h + cn–1.
Substituting for c1, an–1, bn–1, c
n–1 from Eqs. (14) yields
y1′ = (y2 – y1)/h – (M2 + 2M1)h/6
yn′ = (M
n– M
n–1)h/2 + Mn–1h
+ (yn
– yn–1)/h – (M
n+ 2M
n–1)h/6.
From the first equation we obtain
2M1 + M2 = 6(y2 – y1 – hy1′)/h2.
From the second equation we obtain
2Mn
+ Mn–1 = 6(y
n–1 – yn
+ hyn′)/h2.
(b) Eqs. (15) together with the two equations in part (a) give
2M1 + 2M2 + M3 = 6(y2 – y1 – hy1′)/h2
2M1 + 4M2 + M3 = 6(y1 – 2y2 + y3)/h2
2M2 + 4M3 + M4 = 6(y2 – 2y3 + y4)/h2
Mn–2 + 4M
n–1 + Mn
= 6(yn–2 – 2y
n–1 + yn)/h2
Mn–1 + 2M
n= 6(y
n–1 – yn
+ hyn′)/h2.
620 Exercise Set 11.5
This linear system for M1, M2, … , Mn
in matrix form is
2 1 0 0 0 0 0 0
1 4 1 0 0 0 0 0
0 1 4 1 0 0 0 0
0 0 1 4 0 0 0
. . .
. . .
. . .
. . . 00
0 0 0 0 0 1 4 0
0 0 0 0 0 1 4 1
0 0 0 0 0 0 1 2
. . .
. . .
. . .
−
−
M
M
M
M
M
M
n
n
1
2
3
2
1
nn
h
hy y y
y
=
− ′ − +−
6
2
2
1 1 2
1 yy y
y y y
y y y
y y hy
n n n
n n n
2 3
2 3 4
2 1
1
2
2
+− +
− +− + ′
− −
−
Exercise Set 11.5 621
EXERCISE SET 11.6
1. (a)
Continuing in this manner yields
(b) P is regular because all of the entries of P are positive. Its steady-state vector q
solves (P – I)q = 0; that is,
This yields one independent equation, .6q1 – .5q2 = 0, or q1 = q2 Solutions are thus
of the form q = s
2. (a)
and xx( ).
.
.
.3273
396
331
=
xx xx xx( ) ( ) ( ).
.
.
;1 0 27
2
1
= =
=P likewise
..
.
.
23
52
25
5 6
1
156
1
6
11
5 1
=
+= =. Set to obtains qq
11
6 11
.
5
6
−−
=
. .
. ..
6 5
6 5
0
01
2
q
q
xx xx( ) ( ).
.
.
.4 54546
5454
45454
54546=
=
and
.
xx(3) =
.
.,
454
546
xx xx xx xx( ) ( ) ( ) ( ).
.,
.
.1 0 2 14
6
46
54= =
= =
P P
.
623
(b) P is regular because all of its entries are positive.To solve (P – I)q = 0, i.e.,
reduce the coefficient matrix to row-echelon form:
This yields solutions (setting q23 = s) of the form
To obtain a probability vector, take
yielding
3. (a) Solve (P – I)q = 0, i.e.,
The only independent equation is yielding Setting
yields qq =
9 17
8 17
s = 8
17qq =
9 8
1.s
2
3q q1 2
3
4= ,
−−
=
2 3 3 4
2 3 3 4
0
01
2.
q
q
qq =
22
72
29
72
21
72, , .
t
s =+ +
=1
2221
2921
1
21
72,
22 21
29 21
1
.s
−−
−
⇒−
−
8 1 7
6 6 2
2 5 9
2 5 9
0 21 29
0 0 0
⇒
−
−
1 022
21
0 129
210 0 0
.
−−
−
. . .
. . .
. . .
8 1 7
6 6 2
2 5 9
1
2
3
q
q
q
=
,
0
0
0
624 Exercise Set 11.6
(b) As in (a), solve
i.e., .19q1 = .26q2. Solutions have the form
Set to get
(c) Again, solve
by reducing the coefficient matrix to row-echelon form:
yielding solutions of the form
Set to get
4. (a) Prove by induction that p12(n) = 0: Already true for n = 1. If true for n – 1, we
have Pn = Pn–1P, so p12(n) = p12 + p22. But p12 = = 0 so
p12(n) = 0 + 0 = 0. Thus, no power of P can have all positive entries, so P is not
regular.
pn
121( )−
pn
121( )−
pn
111( )−
qq =
3 19
4 19
12 19
.s =12
19
qq =
1 4
1 3
1
s.
1 0 1 4
0 1 1 3
0 0 0
−−
−−
−
2 3 1 2 0
1 3 1 0
1 3 1 2 1 4
1
2
3
q
q
q
=
0
0
0
qq =
26 45
19 45.s =
19
45
qq =
26 19
1s.
−−
=
. .
. .
19 26
19 26
0
01
2
q
q
Exercise Set 11.6 625
4. (b)
etc. We use induction to show Pnx =
Already true for n = 1, 2. If true for n – 1, then
Since we get if x is a state vector.
(c) Theorem 3 says that the entries of the steady state vector should be positive; they
are not for
5. Then since the row sums of
sums of P are 1. Thus (Pq)i
= qi for all i.
P p qk
pk
pki ij j
j
k
ij
j
k
ij
j
k
qq( ) = = = == = =
∑ ∑ ∑1 1 1
1 1 1,Let qq =
1 1 1
k k k
t
.
0
1
.
limn→∞
=+
=
P
x x
nxx0 0
11 2lim
n→∞
=1
20
n
,
P P P P
xn n
n
n= =
−
−−
( )11
1
1
2
11
2
xx
xx x
x
n
1 2
1
11
2
1
2
+
=
−
111
2
1
2
1
1 2−
+
+
−n n
x x
=
−
+
1
2
11
2
1
1 2
n
n
x
x x
.
1
2
11
2
1
1 2
−
+
n
n
x
x x
.
If xx xx=
=
+
x
xP
x
x x
1
2
1
1 2
1
21
2
,
=+ +
, P
x
x x x
21
1 1 2
1
41
4
1
2
xx
626 Exercise Set 11.6
6. Since P has zero entries, consider P2 = , so P is regular.
Note that all the rows of P sum to 1. Since P is 3 × 3, Exercise 5 implies q =
7. Let x = [x1 x2]t be the state vector, with x1 = probability that John is happy and
x2 = probability that John is sad. The transition matrix P will be
since the columns must sum to one. We find the steady state vector for P by solving
i.e., Let and get so 10/13 is the
probability that John will be happy on a given day.
8. The state vector x = [x1 x2 x3]t will represent the proportion of the population living in
regions 1, 2 and 3, respectively. In the transition matrix, pij
will represent the proportionof the people in region who move to region i, yielding
P =
. . .
. . .
. . .
.
90 15 10
05 75 05
05 10 85
qq =
10 13
3 13,s =
3
131
5
2
3
10 3
11 2q q s= =
, .so qq
−−
=
1 5 2 3
1 5 2 3
0
01
2,
q
q
P =
4 5 2 3
1 5 1 3
1
3
1
3
1
3
t
.
1
2
1
4
1
41
4
1
2
1
41
4
1
4
1
2
Exercise Set 11.6 627
To find the steady state vector, solve
First reduce to row echelon form , yielding
Set and get , i.e., in the long run 13/24 (or 54 %) of the people
reside in region 1, 4/24 (or 16 %) in region 2 and 7/24 (or 29 %) in region 3.1
6
2
3
1
6q =
13 24
4 24
7 24
s =7
24
qq =
13 7
4 7
1
.s
1 0 13 7
0 1 4 7
0 0 0
−−
−−
−
. . .
. . .
. . .
10 15 10
05 25 05
05 10 15
q11
2
3
0
0
0
q
q
=
.
628 Exercise Set 11.6
EXERCISE SET 11.7
1. Note that the matrix has the same number of rows and columns as the graph has vertices,and that ones in the matrix correspond to arrows in the graph. We obtain
2. See the remark in problem 1; we obtain
P3
P3
P4P5P6
P5
P3P4
P1 P1 P1P2
P2
P2P4
(c)(b)(a)
((aa)) ((bb))
0 0 0 1
1 0 1 1
1 1 0 1
0 0 0 0
0 1 1 0 0
0 0 0 0
11
1 0 0 1 0
0 0 1 0 0
0 0 1 0 0
0 1 0 1 0 0
1
((cc))
00 0 0 0 0
0 1 0 1 1 1
0 0 0 0 0 1
0 0 0 0 0 1
0 0 1 0 1 0
629
3. (a) As in problem 2, we obtain
(b) m12 = 1, so there is one 1-step connection from P1 to P2.
So m12(2) = 2 and m12
(3) = 3 meaning there are two 2-step and three 3-step connectionsfrom P1 to P2 by Theorem 1. These are:
1-step: P1 → P2
2-step: P1 → P4 → P2 and P1 → P3 → P2
3-step: P1 → P2 → P1 → P2, P1 → P3 → P4 → P2,
and P1 → P4 → P3 → P2.
(c) Since m14 = 1, m14(2) = 1 and m14
(3) = 2, there are one 1-step, one 2-step and two 3-stepconnections from P1 to P4. These are:
1-step: P1 → P4
2-step: P1 → P3 → P4
3-step: P1 → P2 → P1 → P4 and P1 → P4 → P3 → P4.
4. (a) Note that to be contained in a clique, a vertex must have “two-way” connections withat least two other vertices. Thus, P4 could not be in a clique, so P1, P2, P3 is the onlypossible clique. Inspection shows that this is indeed a clique.
(b) Not only must a clique vertex have two-way connections to at least two other vertices,but the vertices to which it is connected must share a two-way connection. Thisconsideration eliminates P1 and P2, leaving P3, P4, P5 as the only possible clique.Inspection shows that it is indeed a clique.
M M2 3
1 2 1 1
0 1 1 1
1 1 1 0
1 1 0 1
2 3 2 2
1 2=
=and11 1
1 2 1 2
1 2 2 1
.
P2
P4
P1
P3
630 Exercise Set 11.7
(c) The above considerations eliminate P1, P3 and P7 from being in a clique. Inspectionshows that each of the sets P2, P4, P6 , P4, P6, P8 , P2, P6, P8 , P2, P4, P8 and P4, P5, P6 satisfy conditions(i) and (ii) in the definition of a clique. But note that P8 can be added to the firstset and we still satisfy the conditions. P5 may not be added, so P2, P4, P6, P8 is aclique, containing all the other possibilities except P4, P5, P6 , which is also a clique.
5. (a) With the given M we get, as in Example 5, that
Since sii(3) = 0 for all i, there are no cliques in the graph represented by M.
(b) As in (a),
The elements along the main diagonal tell us that only P3, P4 and P6 are members ofa clique. Since a clique contains at least three vertices, we must have P3, P4, P6 asthe only clique.
6. As in problem 1,
M =
0 0 1 1
1 0 0 0
0 1 0 1
0 1 0 0
.
S =
0 1 0 1 0 0
1 0 1 0 1 0
0 1 0 1 0 1
1 0 1 0 1 1
0 1 0 1 0 0
0 0 1 1 0 0
=, S3
0 6 1 7 0 2
6 0 7 1 6 3
1 7 2 8 1 4
7 1 8 2 7 55
0 6 1 7 0 2
2 3 4 5 2 2
.
S =
0 1 0 1 0
1 0 1 0 0
0 1 0 0 1
1 0 0 0 1
0 0 1 1 0
,, S3
0 3 1 3 1
3 0 3 1 1
1 3 0 1 3
3 1 1 0 3
1 1 3 3 0
=
.
Exercise Set 11.7 631
Then
By summing the rows of M + M2, we get that the power of P1 is 2 + 1 + 2 = 5, the powerof P2 is 3, of P3 is 4, and of P4 is 2.
7. Associating vertex P1 with team A, P2 with B, … , P5 with E, the game results yield thefollowing dominance-directed graph:
which has vertex matrix
Now,
M2
0 1 1 1 2
1 0 0 2 1
1 1 0 1 0
0 0 1 0 1
0 2 1 1 0
=
+ =
, M M2
0 2 2 2 2
1 0 1 2 2
1 1 0 2 1
0 1 1 0 1
1 2 1 2 0
.
M =
0 1 1 1 0
0 0 1 0 1
0 0 0 1 1
0 1 0 0 0
1 0 0 1 0
.
P2
P4
P1
P3
P5
M M M2 2
0 2 0 1
0 0 1 1
1 1 0 0
1 0 0 0
0 2 1 2
=
+ =and11 0 1 1
1 2 0 1
1 1 0 0
.
632 Exercise Set 11.7
Summing the rows, we get that the power of A is 8, of B is 6, of C is 5, of D is 3, and of E is 6. Thus ranking in decreasing order we get A in first place, B and E tie for secondplace, C in fourth place, and D last.
Exercise Set 11.7 633
EXERCISE SET 11.8
1. (a) From Eq. (2), the expected payoff of the game is
(b) If player R uses strategy [p1 p2 p3] against player C’s strategy
his payoff will be pAq = (–1/4)p1 + (9/4)p2 – p3. Since p1, p2 and p3 are nonnegativeand add up to 1, this is a weighted average of the numbers –1/4, 9/4 and –1. Clearlythis is the largest if p1 = p3 = 0 and p2 = 1; that is, p = [0 1 0].
(c) As in (b), if player C uses [q1 q2 q3 q4]t against , we get pAq = –6q1 +
3q2 + q3 – 12q4. Clearly this is minimized over all strategies by setting q1 = 1 and
q2 = q3 = q4 = 0. That is q = [1 0 0 0]t.
2. As per the hint, we will construct a 3 × 3 matrix with two saddle points, say a11 = a33 = 1.Such a matrix is
A .=
1 2 1
0 7 0
1 2 1
1
20
1
2
1
4
1
4
1
4
1
4
t
,
p qA =
− −−
− −
1
20
1
2
4 6 4 1
5 7 3 8
8 0 6 2
= − .
1
41
41
41
4
5
8
635
(The reader is invited to prove that if the matrix A has two saddle points ars
and atu
, thenwe must necessarily have a
rs= a
tu).
3. (a) Calling the matrix A, we see a22 is a saddle point, so the optimal strategies are pure,namely: p = [0 1], q = [0 1]t; the value of the game is a22 = 3.
(b) As in (a), a21 is a saddle point, so optimal strategies are p = [0 1 0], q = [1 0]t; the value of the game is a21 = 2.
(c) Here, a32 is a saddle point, so optimal strategies are p = [0 0 1], q = [0 1 0]t
and v = a32 = 2.
(d) Here, a21 is a saddle point, so p = [0 1 0 0], q = [1 0 0]t and v = a21 = –2.
4. (a) Calling the matrix A, the formulae of Theorem 2 yield p = [5/8 3/8], q = [1/8 7/8]t,v = 27/8 (A has no saddle points).
(b) As in (a),
(c) For this matrix, a11 is a saddle point, so p = [1 0], q = [1 0]t and v = a11 = 3.
(d) This matrix has no saddle points, so, as in (a),
(e) Again, A has no saddle points, so as in (a),
qq =
=−1
13
12
13
29
13
t
vand .
pp ,=
3
13
10
13
pp qq,=−−
−−
=
=−−
−3
5
2
5
3
5
2
5
3
5
2
−−
=
=−−
=
5
3
5
2
5
19
5
19
5
t t
v .
and
pp qq,=
=
=40
60
20
60
2
3
1
3
10
60
50
600
1
6
5
6
1400
60
70
3
=
= =
,
(
t
v Agaain, A has no saddle points).
636 Exercise Set 11.8
5. Let a11 = payoff to R if the black ace and black two are played = 3.
a12 = payoff to R if the black ace and red three are played = –4.
a21 = payoff to R if the red four and black two are played = –6.
a22 = payoff to R if the red four and red three are played = 7.
So, the payoff matrix for the game is
A has no saddle points, so from Theorem 2,
that is, player R Should play the black ace 65 percent of the time, and
player C should play the black two 55 percent of the time. The value of the game is ,that is, player C can expect to collect on the average 15 cents per game.
6. If p* = [0 0 … 1 … 0] (one in r-th place) and
q* = [0 0 … 1 … 0]t (one in s-th place) and
A is an m × n matrix (m = length of p, n = length of q), then
(since other entries of p*, q* are zero), verifying Eq. (6). If ars
is a saddle point then a
rs≤ a
rjfor all j, since a
rsis the smallest in its row. So if q = [q1
… qn]t is any
column strategy, we have
pp** qq pp** qqAA AA= ( ) =
= ≥
= =∑ ∑p a q a qi
i
m
ij j rj j
j
n*
1 1jj
n
rs j
j
n
rs j rs
j
n
rs
a q
a q a a
= =
=
∑ ∑
∑= = =
1 1
1
1( ) ,
pp** qq** == pp** qq **AA AA( ) =
=
= =∑ ∑* *
p a qi
i
m
ij j
j
n
1 1
*p a ai
i
m
is rs
=∑ =
1
−3
20
qq ;=
11
20
9
20
t
pp ,=
13
20
7
20
A .=−
−
3 4
6 7
Exercise Set 11.8 637
demonstrating Eq. (7). For Eq. (8), note that ars is a saddle point implies ars
≥ ais
for all i, since a
rsis the largest in its column. So, if p = [p1
… pn] is any row strategy, we
have
7. Take . Suppose a22 – a21 > 0. Since the game is not strictly
determined, a21 is not a saddle point, so a21 < a11. Similarly, a11 is not a saddle point, so
a11 > a12. Then . All the parenthesized expressions being
positive yields If a22 – a21 < 0, get a22 < a12, then a12 > a11, so all the parenthesized
expressions are negative; again
If a11 > a12, then a12 < a22, yielding a22 > a21; i.e., all parenthesized expressions are positive,
so If a11 < a12, then a11 < a21, so a21 > a22, all parenthesized expressions are
negative and
Expressing as respectively,
and using the non-strictly determined property of A as above, we get and as well.0 12
*< <q
0 11*< <q
a a
a a a a
a a
a
22 12
22 12 11 21
11 21−( )−( ) + −( )
−( )and
111 21 22 12−( ) + −( )a a aq q1 2
* *and
0 12 .*< <p
0 12 .*< <p
pa a
a a a a
a a
a a2
11 12
11 22 12 21
11 12
11 12
* =−
+ − −=
−( )−(( ) + −( )a a22 21
.0 11 .*< <p
0 11 .*< <p
pa a
a a a a1
22 21
22 21 11 12
* =−( )
+( ) + −( )
pa a
a a a a1
22 21
11 22 12 21
* =−
+ − −
pp qq** pp qq **AA AA= ( ) =
= ≤
= =∑ ∑p a q p ai ij j
j
n
i is
i
m
*
1 1ii
m
rs i
i
n
rs rs
a p
a a
= =∑ ∑
= =1 1
1( ) .
638 Exercise Set 11.8
EXERCISE SET 11.9
1. (a) Calling the given matrix E, we need to solve
This yields , that is, p = s[1 3/2]t. Set s = 2 and get p = [2 3]t.
(b) As in (a), solve
In row-echelon form, this reduces to
Solutions of this system have the form p = s[1 5/6 1]t. Set s = 6 and get p = [6 5 6]t.
(c) As in (a), solve
( )
. . .
. . .
. . .
I E− =− −
− −− −
pp
65 50 30
25 80 30
40 30 60
=
p
p
p
1
2
3
0
0
0
,,
1 0 1
0 1 5 6
0 0 0
1
2
3
−−
=p
p
p
00
0
0
.
( )
/ /
/ /
/
I E− =−
− −− −
pp
1 2 0 1 2
1 3 1 1 2
1 6 1 1
.
p
p
p
1
2
3
0
0
0
=
1
2
1
31 2p p=
( )I Ep
p− =
−−
=
pp
1 2 1 3
1 2 1 3
0
01
2
.
639
which reduces to
Solutions are of the form p = s[78/79 54/79 1]t. Let s = 79 to obtain p = [78 54 79]t.
2. (a) By Corollary 1, this matrix is productive, since each of its row sums is .9.
(b) By Corollary 2, this matrix is productive, since each of its column sums is less thanone.
(c) Try x = [2 1 1]t. Then Cx = [1.9 .9 .9]t, i.e., x > Cx, so this matrix is productive by Theorem 3.
3. Theorem 2 says there will be one linearly independent price vector for the matrix E if somepositive power of E is positive. Since E is not positive, try E2:
4. The exchange matrix for this arrangement (using A, B, and C in that order) is
For equilibrium, we must solve (I – E)p = 0. That is
1 2 1 3 1 4
1 3 2 3 1 4
1 6 1 3 1 2
1
2
3
− −− −− −
p
p
p
=
.
0
0
0
1 2 1 3 1 4
1 3 1 3 1 4
1 6 1 3 1 2
/
.
E2
2 34 1
2 54 6
6 12 3
0=
>. . .
. . .
. . .
.
1 3 4 3 2
0 1 54 79
0 0 0
1
2
3
−−
p
p
p =
.
0
0
0
640 Exercise Set 11.9
Row reduction yields solutions of the form p = [18/16 15/16 1]ts.
Set s = 1600/15 and obtain p = [120 100 106.67]t; i.e., the price of tomatoes was $120,corn was $100 and lettuce was $106.67.
5. Taking the CE, EE, and ME in that order, we form the consumption matrix C, where c
ij= the amount (per consulting dollar) of the i-th engineer’s services purchased by the
j-th engineer. Thus,
We want to solve (I – C)x = d, where d is the demand vector, i.e.
In row-echelon form this reduces to
Back-substitution yields the solution x = [1256.48 1448.12 1556.19]t.
6. The i-th column sum of E is eji, and the elements of the i-th column of I – E are
the negatives of the elements of E, except for the ii-th, which is 1 – eii. So, the i-th
column sum of I – E is 1 – eji
= 1 – 1 = 0. Now, (I – E)t has zero row sums, so
the vector x = [1 1 … 1]t solves (I – E)tx = 0. This implies det(I – E)t = 0. But det(I – E)t = det(I – E), so (I – E)p = 0 must have nontrivial (i.e., nonzero) solutions.
j
n
=∑
1
j
n
=∑
1
1 2 3
0 1 43877
0 0 1
1
2
3
− −−
. .
.
x
x
x
=
500
785 31
1556 19
.
.
.
1 2 3
1 1 4
3 4 1
1
2
3
− −− −− −
. .
. .
. .
x
x
x
=
500
700
600
.
C =
0 2 3
1 0 4
3 4 0
. .
. .
. .
.
Exercise Set 11.9 641
7. Let C be a consumption matrix whose column sums are less than one; then the row sumsof Ct are less than one. By Corollary 1, Ct is productive so (I – Ct)–1 ≥ 0. But
(I – C)–1 = (((I – C)t))–1)t = ((I – Ct)–1)t ≥ 0.
Thus, C is productive.
8. (I) Let y be a strictly positive vector, and x = (I – C)–1y. Since C is productive (I – C)–1 ≥ 0, so x = (I – C)–1y ≥ 0. But then (I – C)x = y > 0, i.e., x – Cx > 0, i.e., x > Cx.
(II) Step 1: Since both x* and C are ≥ 0, so is Cx*. Thus x* > Cx* ≥ 0.
Step 2: Since x* > Cx*, x* – Cx* > 0. Let ε be the smallest element in x* – Cx*, andM the largest element in x*. Then x* – Cx* > 2M
ε—–x* > 0, i.e., x* – 2Mε—–x* > Cx*.
Setting λ = 1 – 2Mε—– < 1, we get Cx* < λx*.
Step 3: First, we show that if x > y, then Cx > Cy. But this is clear since (Cx)i
=
Now we prove Step 3 by induction on n, the case n = 1
having been done in Step 2. Assuming the result for n – 1, then Cn–1x* < λn–1x*. But then Cnx* = C(Cn–1x*) < C(λn–1x*) = λn–1(Cx*) < λn–1(λx*) = λnx*, proving Step 3.
Step 4: Clearly, Cnx* ≥ 0 for all n. So we have
Denote the elements of Then we have for all i. But
–c
ij≥ 0 and x* < 0 imply –c
ij= 0 for all i and j, proving Step 4.
Step 5: By induction on n, the case n = 1 is trivial. Assume the result true for n – 1.Then
(I – C)(I + C + C2 + … + Cn–1) = (I – C)
(I + C + … + Cn–2) + (I – C)Cn–1 = (I – Cn–1) + (I – C)Cn–1
= I – Cn–1 + Cn–1 – Cn,
= I – Cn,
proving Step 5.
01
==∑ c xij j
j
n*lim .
n
nijC c
→∞by
0 0≤ ≤ = =→∞ →∞ →∞
lim * lim * , lim *n
n
n
n
n
nC Cx x i.e., xλ 00.
c x c y Cyij j
j
n
ij j
j
n
i
= =∑ ∑> =
1 1
( ) .
642 Exercise Set 11.9
Step 6: First we show (I – C)–1 exists. If not, then there would be a nonzero vector z such that Cz = z. But then Cnz = z for all n, so z = lim
n → ∞ Cnz = 0, acontradiction, thus I – C is invertible. Thus, I + C + … + Cn–1 = (I – C)–1(I – Cn), so S = lim
n → ∞ (I – C)–1(I – Cn) = (I – C)–1 (I – limn → ∞ Cn) = (I – C)–1, proving
Step 6.
Step 7: Since S is the (infiite) sum of nonnegative matrices, S itself must be non-negative.
Step 8: We have shown in Steps 6 and 7 that (I – C)–1 exists and is nonnegative,thus C is productive.
Exercise Set 11.9 643
EXERCISE SET 11.10
1. Using Eq. (18), we calculate
So all the trees in the second class should be harvested for an optimal yield (since s = 1000) of $15,000.
2. From the solution to Example 1, we see that for the fifth class to be harvested in theoptimal case we must have
p5s/ (.28–1 + .31–1 + .25–1 + .23–1) > 14.7s,
yielding p5 > $222.63.
3. Assume p2 = 1, then Yld2 = = .28s. Thus, for all the yields to be the same we
must have
s
.281( )−
Ylds
s
Ylds s
2
3
30
215
50
232
100
7.
= =
=+
=
645
p3s/(.28–1 + .31–1) = .28s
p4s/(.28–1 + .31–1 + .25–1) = .28s
p5s/(.28–1 + .31–1 + .25–1 + .23–1) = .28s
p6s/(.28–1 + .31–1 + .25–1 + .23–1 + .27–1) = 28s
Solving these sucessively yields p3 = 1.90, p4 = 3.02, p5 = 4.24 and p6 = 5.00. Thus the ratio
p2 p3 p4 p5 p6 = 1 1.90 3.02 4.24 5.00 .
4. From Eq. (17) we have
Then Eq. (16) yields
for 1 ≤ i ≤ k – 1. Then (*) and Eq. (13) yield Eq. (19).
5. Since y is the harvest vector, N = yi
is the number of trees removed from the forest.
Then Eq. (7) and the first of Eqs. (8) yield N = g1x1, and from Eq. (17) we obtain
Ng s
g
g
g
g
s
g gk k
.=+ + ⋅ ⋅ ⋅ +
=+ ⋅ ⋅ ⋅ +
− −
1
1
2
1
1 1 11
1 1
i
n
=∑
1
(*)xg x
g
g
g
s
gg g
ii i
k
= =+ ⋅ ⋅ ⋅ +
−
1 1 1
11 1
1 1
=+ ⋅ ⋅ ⋅ +
−
1
1 1
1 1
g
s
g g
i
k
xs
g
g
g
g
g
g
s
gg gk
11
2
1
3
1
11
1 2
1 1 1=
+ + + ⋅ ⋅ ⋅ +=
+ + ⋅−
⋅⋅ ⋅ +
−
1
1gk
.
646 Exercise Set 11.10
6. Set g1 = … = gn–1 = g, and p2 = 1. Then from Eq. (18), Yld2 = = gs. Since we want all
of the Yldk’s to be the same, we need to solve Yld
k= = gs for p
kfor 3 ≤ k ≤ n.
Thus pk
= k – 1. So the ratio
p2 p3 p4
… pn
= 1 2 3 … (n – 1).
p s
kg
k
( )−11
p s
g
2
1
1
Exercise Set 11.10 647
EXERCISE SET 11.11
1. (a) Using the coordinates of the points as the columns of a matrix we obtain
(b) The scaling is accomplished by multiplication of the coordinate matrix on the left by
resulting in the matrix
0 3
2
3
2
0
0 0 1
2
1
2
0 0 0 0
,
3
20 0
01
20
0 0 1
,
0 1 1 0
0 0 1 1
0 0 0 0
.
649
which represents the vertices (0, 0, 0), as shown below.
1. (c) Adding the matrix
to the original matrix yields
which represents the vertices (–2, –1, 3), (–1, –1, 3), (–1, 0, 3), and (–2, 0, 3) asshown below.
(d) Multiplying by the matrix
cos( ) sin( )
sin( ) cos( )
− − −
− −
30 30 0
30 30 0
0 0 1
,
− − − −− −
2 1 1 2
1 1 0 0
3 3 3 3
,
− − − −− − − −
2 2 2 2
1 1 1 1
3 3 3 3
3
20 0
3
2
1
20 0, , , , , ,
and 0,1
2
650 Exercise Set 11.11
we obtain
The vertices are then (0, 0, 0), (.866, – .500, 0), (1.366, .366, 0), and (.500, .866, 0)as shown:
2. (a) Simply perform the matrix multiplication
(b) We multiply
yielding the vertices (0, 0, 0), (1, 0, 0), 3
21 0
1
21 0, , , , , .
and
11
20
0 1 0
0 0 1
0 1 1 0
0 0 1 1
0 0 0 0
=
0 13
2
1
20 0 1 1
0 0 0 0
11
20
0 1 0
0 0 1
x
y
z
i
i
i
=
+
.
x y
y
z
i i
i
i
1
2
0 30 30 30 30
0
cos( ) cos( ) sin( ) sin( )
sin
− − − − − −
(( ) cos( ) sin( ) cos( )− − + − −
30 30 30 30
0 0 0 0
=
− 0 866 1 366 500
0 500 366 866
0 0 0 0
. . .
. . .
.
Exercise Set 11.11 651
2. (c) Obtain the vertices via
yielding (0, 0, 0), (1, .6, 0), (1, 1.6, 0) and (0, 1, 0), as shown:
3. (a) This transformation looks like scaling by the factors 1, –1, 1, respectively and indeedits matrix is
(b) For this reflection we want to transform (xi, y
i, z
i) to (–x
i, y
i, z
i) with the matrix
Negating the x-coordinates of the 12 points in view 1 yields the 12 points (–1.000,–.800, .000), (–.500, –.800, –.866), etc., as shown:
−
1 0 0
0 1 0
0 0 1
.
1 0 0
0 1 0
0 0 1
−
.
1 0 0
6 1 0
0 0 1
0 1 1 0
0 0 1 1
0 0 0 0
.
=
. . ,
0 1 1 0
0 6 1 6 1
0 0 0 0
652 Exercise Set 11.11
(c) Here we want to negate the z-coordinates, with the matrix
This does not change View 1.
4. (a) The formulas for scaling, translation and rotation yield the matrices
(b) Clearly, P′ = M5M4M3(M1P + M2).
M M1 2
1 2 0 0
0 2 0
0 0 1 3
1
2
1
2
1
2
1
2
=
=
⋅⋅⋅/
/
, 00 0 0 0
0 0 0 0
1 1 0
0 23
⋅⋅⋅
⋅⋅⋅
=
,
M cos 00 20
0 20 20
4
−
=
sin
sin cos
,
cos(
M
−− −
− − −
45 0 45
0 1 0
45 0 45
) sin( )
sin( ) cos( )
=
−
,
cos sin
sin cos
and
M5
90 90 0
90 90 0
00 0 1
.
1 0 0
0 1 0
0 0 1−
.
Exercise Set 11.11 653
5. (a) As in 4(a),
(b) As in 4(b), P′ = M7(M6 + M5M4(M3 + M2 M1P)).
6. Using the hint given, we have
R R1 2
0
0 1 0
0
=−
=cos sin
sin cos
,
cosβ β
β β
α −−
=
sin
sin cos ,
cos sin
αα α
θ θ
0
0
0 0 1
0
03R 11 0
0
0
4
−
=− − −
sin cos
,
cos( ) sin( )
θ θ
α αR ssin( ) cos( ) ,− −
α α 0
0 0 1
M M1 2
3 0 0
0 5 0
0 0 1
1 0 0
0 45 4=
= −.
. , cos sin 55
0 45 45
1 1 1 1
0 0 03
sin cos
,
=⋅⋅⋅⋅⋅⋅M 00
0 0 0 0
35 0 35
0 1 04
⋅⋅⋅
=
−
,
cos sin
sin
M
335 0 45
45 45
5
cos
,
cos( ) sin(
=
− − −
M
))
sin( ) cos( ) ,
0
45 45 0
0 0 1
0 0
6
− −
=
M
00 0
0 0 0 0
1 1 1 1
⋅⋅⋅⋅⋅⋅⋅⋅⋅
=, and
2 0 0
0 1 07M
00 0 1
.
654 Exercise Set 11.11
Exercise Set 11.11 655
7. (a) We rewrite the formula for vi′ as
So
(b) We want to translate xi
by –5, yi
by +9, zi
by –3, so x0 = –5, y0 = +9, z0 = –3. The matrix is
8. This can be done most easily performing the multiplication RRt and showing that this is I.For example, for the rotation matrix about the x-axis we obtain
RRt = −
1 0 0
0
0
1 0 0
0cos sin
sin cos
cθ θθ θ
oos sin
sin cos
θ θθ θ0
1 0 0
0 1 0
0 0 1−
=
.
1 0 0 5
0 1 0 9
0 0 1 3
0 0 0 1
−
−
.
v
x
y
z
x
y
zi
i
i
i
′ =
1 0 0
0 1 0
0 0 1
0 0 0 1
0
0
0
11
.
v
x x
y y
z zi
i
i
i
′ =
⋅ + ⋅⋅ + ⋅⋅ + ⋅
⋅
1 1
1 1
1 1
1 1
0
0
0
.
and R5
0
0 1 0
0
=− −
− − −
cos( ) sin( )
sin( ) cos( )
β β
β β
.
EXERCISE SET 11.12
1. (a) The discrete mean value property yields the four equations
Translated into matrix notation, this becomes
t
t
t
t
1
2
3
4
01
4
1
40
1
40 0
=
11
41
40 0
1
4
01
4
1
40
1
2
t
t
tt
t
3
4
0
1
2
0
1
2
+
.
t t t
t t t
t t t
t
1 2 3
2 1 4
3 1 4
4
1
4
1
41 1
1
4
= +
= + + +
= +
=
( )
( )
( )
11
41 12 3( ).t t+ + +
657
1. (b) To solve the system in part (a), we solve (I – M)t = b for t:
In row-echelon form, this is
Back substitution yields the result t = [1/4 3/4 1/4 3/4]t.
(c)
t(2) = Mt(1) + b = [1/8 5/8 1/8 5/8]t
t(3) = Mt(2) + b = [3/16 11/16 3/16 11/16]t
tt tt bb( )1
01
4
1
40
1
40 0
1
41
40 0
1
4
01
4
1
40
= + =
M(0)
0
0
0
0
+
=
0
1
2
0
1
2
0
1
2
0
11
2
11
4
1
40
0 1 15 4
0 0 11
2
0 0 0 1
− −
−
−
=
t
t
t
t
1
2
3
4
00
0
1 8
3 4
−
.
01
4
1
40
1
41 0
1
41
40 1
1
4
01
4
1
41
− −
− −
− −
− −
t
t
t
t
1
2
3
4
=
.
0
1
2
0
1
2
658 Exercise Set 11.12
t(4) = Mt(3) + b = [7/32 23/32 7/32 23/32]t
t(5) = Mt(4) + b = [15/64 47/64 15/64 47/64]t
from Eq. (10).
(d) Using percentage error we have that the
percentage error for t1 and t3 was and for t2 and t4 was
2. The average value of the temperature on the circle is
where r is the radius of the circle and f(θ) is the temperature at the point of the circumfer-ence where the radius to that point makes the angle θ with the horizontal. Clearly f(θ) = 1for –π/2 < θ < π/2 and is zero otherwise. Consequently, the value of the integral above(which equals the temperature at the center of the circle) is 1/2.
3. As in 1(c), but using M and b as in the problem statement, we obtain
t(1) = Mt(0) + b = [3/4 5/4 1/2 5/4 1 1/2 5/4 1 3/4]t
t(2) = Mt(1) + b = [13/16 9/8 9/16 11/8 13/16 7/16 21/16 15/8]t.
4. Using the telephone number 555-1212, we start in row 1, column 2. The ten boundaryvalues we obtain are 0, 0, 2, 2, 0, 0, 0, 2, 1, and 1. The last arrow used was in row 4,column 9, and the average value obtained was .8.
1
2πθ θ
π
π
rf rd( ) ,
−∫
−× = −
.
.% . %.
0129
7629100 1 7
.
.% . %,
0129
2371100 5 4× =
=−
×computed value actual value
actual value1100%
Exercise Set 11.12 659
EXERCISE SET 11.13
1. (a) Using the lines in Eq. (5) we have
a1ta1 = 2 a2
ta2 = 5 a3ta3 = 10
b1 = 2 b2 = –2 b3 = 3
Setting x(p)k
= we have
Writing xx xxaa xx
aa aaaak
pkp k k
tkp
kt
k
k
b( ) ( )( )
= +−
−
−1
1
aa xx
aa xx
1 0 01 02
2 1 11 122
t p p p
t p p
x x
x x
( ) ( ) ( )
( ) ( ) (
= +
= − pp
t p p p
t p
t
x x
b
)
( ) ( ) ( )
( )
aa xx
aa xx
aa aa
3 2 21 22
1 1 0
1 1
3= −
−=
22
2
2
01 02
2 2 1
2 2
11
− −
−=
− −
x x
b x
p p
t p
t
p
( ) ( )
( ) ( )aa xx
aa aa
−−
−=
− −
x
b
a a
x x
p
t p
t
p p
12
3 3 2
3 3
21 22
5
3 3
1
( )
( ) ( ) ( )aa xx
00.
x
x
kp
kp
1
2
( )
( )
aa aa aa1 2 31
1
1
2
3
1=
=
−
=
−
661
in the form:
we have
Substituting for the expressions
for k = 2:
for k = 3:
where , which means
x x x xp p p p
011
021
31 32( ) ( ) ( ) ( ), , .+ +( ) = ( )
x xp p
01
3( ) ( )+ =
x xx xp p
p p
31 2121 223 3
103
1
109( ) ( )
( ) ( )
= +− +
= +• xx x
x xx
p p
p pp
21 22
32 2221
3
3 3
( ) ( )
( ) ( )( )
+
= +− + xx
x x
pp p22
21 22101
1
103 3 9
( )( ) ( )( )• − = − + +
x xx x
xp p
p pp
12 0201 02
012
21
1
22( ) ( )
( ) ( )(= +
− −= +•
)) ( )
( ) ( )( ) (
−
= +− − +
x
x xx x
p
p pp p
02
21 1111 122 2 ))
( ) ( )
( ) (
51
1
52 4 211 12
22 12
• = − + +
=
x x
x x
p p
p p))( ) ( )
( )( )+− − +
− = + +•2 2
52
1
54 2 211 12
11 1x x
x x
p pp
22( )p
x xx x
xp p
p p
11 0101 02
012
21
1
22( ) ( )
( ) ( )
= +− −
= +•(( ) ( )p p
x−
02
bkk k
tkp
kt
k
−
=−aa xx
aa aa1 1
( )
gives for :
x xb
akp
kp k k
tkp
kt
k
k1 1 11( )
,( )
( )
= +−
−
−aa xx
aa aa11
2 1 21x x
bk
pk
p k kt
kp
kt
k
( ),
( )( )
= +−
−
−aa xx
aa aaaak2.
x
x
x
x
kp
kp
kp
kp
1
2
1 1
1 2
( )
( )
,( )
,( )
=
−
−
+−
−b
a
a
k kt
kp
kt
k
k
k
aa xx
aa aa1
1
2
( )
662 Exercise Set 11.13
1. (b) From part (a) we have
Subsituting the expressions for from part (a) gives
Subsituting the expressions for from part (a) gives
From part (a) we have
Subsituting for from part (a) gives
Subsituting for from part (a) gives
x x x xp p p p
32 01 02 013
101
2
22
1
22( ) ( ) ( ) ( )= + + −( ) + − ++( )
= + −
x
x x
p
p p
02
01 021
2024 3 3
( )
( ) ( ) .
x xp p
11 12( ) ( )and
x x x xp p p
32 11 121
103
3
52 4 2
9
54 2( ) ( ) ( )= − + − + +( ) + + 111 12
11 123
101 2
( ) ( )
( ) ( )
p p
p p
x
x x
+( )
= + +
.
x xp p
21 22( ) ( )and
x x xp p p
32 21 221
103 3 9( ) ( ) ( ) .= − + +
x x x xp p p
31 01 02 011
5055
10
22
5
22( ) ( ) ( ) (= + + −( ) + − pp p
p p
x
x x
) ( )
( ) ( )
+( )
= + −
02
01 021
2028 .
x xp p
11 12( ) ( )and
x x x xp p p
31 11 12 11
109
1
52 4 2
3
54 2( ) ( ) ( )= + − + +( ) + + 11 12
11 121
5055 10 5
( ) ( )
( ) (
p p
p p
x
x x
+( )
= + + )) .
x xp p
21 22( ) ( )and
x x xp p p
31 21 221
109 3( ) ( ) ( ) .= + +
Exercise Set 11.13 663
Now
and so
1. (c) The linear system
can be rewritten as
which has the solution
2. (a) Setting and using part (b) of Exercise 1,
we have
x
x
311
321
1
2028 1 40000
1
2024 1
( )
( )
.
.
= =
= = 220000
1
2028 1 4 1 2 1 4100031
2
322
x
x
( )
(
. . .= + − =
)) ( . ) ( . ) .= + − =1
2024 3 1 4 3 1 2 1 23000
x01
011
021
310
320 0( ) ( ) ( ) ( ) ( ), , ( ,= ( ) = ( ) =x x x x ),0
x
x
31
32
31 22
27 22
*
* .
=
=
19 28
3 23 24
31 32
31 32
x x
x x
* *
* * ,
+ =
− + =
x x x
x x x
31 31 32
32 31
1
2028
1
2024 3 3
* = + −
= + −
* *
* *332*
x x x
x
p p p
p
31 311
321
32
1
2028
1
2
( ) ( ) ( )
( )
= + −
=
− −
0024 3 331
102
1+ −
− −x x
p p( ) ( ) .
orx x x x xp p p p p
0 31
01 02 311( ) ( ) ( ) ( ) (,= ( ) =− − )) ( ), ,x
p32
1−( )
664 Exercise Set 11.13
(b)
Since x3(1) in this part is the same as x3
(1) in part (a), we will get x3(2) as in part (a)
and therefore x3(3), … , x3
(6) will also be the same as in part (a).
(c)
x01
310
320
311
148 15
1
202
( ) ( ) ( )
( )
( , ) ,= − = ( )=
x x
x 88 148 15 9 55000
1
2024 3 14832
1
+ − − =
= +
( ) .
( )( )x −− − =3 15 25 65000( ) .
x01
310
320
311
1 1
1
2028 1 1
( ) ( ) ( )
( )
( , ) ,= = ( )= + −
x x
x =
= + − =
1 4
1
2024 3 1 3 1 1 232
2
.
( ) ( ) .( )x
x
x
313
323
1
2028 1 41 1 23 1 40900
1
2
( )
( )
. . .= + − =
=00
24 3 1 41 3 1 23 1 22700
1
20231
4
+ − =
=
( . ) ( . ) .
( )x 88 1 409 1 227 1 40910
1
2024 3 132
4
+ − =
= +
. . .
( .( )x 4409 3 1 227 1 22730
1
2028 1 4031
5
) ( . ) .
.( )
− =
= +x 991 1 2273 1 40909
1
2024 3 1 409132
5
− =
= +
. .
( .( )x )) ( . ) .
.( )
− =
= +
3 1 2273 1 22727
1
2028 1 409031
6x 99 1 22727 1 40909
1
2024 3 1 409032
6
− =
= +
. .
( .( )x 99 3 1 22727 1 22727) ( . ) . .− =
Exercise Set 11.13 665
666 Exercise Set 11.13
3. (a) The three lines are
L1 x1 + x2 = 2
L2 x1 – 2x2 = –2
L3 3x1 – x2 = 3
x L
x
1 1
2
12
11
10
11
22
112*
*
lies on since
lie
+ = =
ss on sinceL
x
2
3
46
552
78
55
110
552−
=−
= −
* lies on sinceL3 331
22
27
22
66
223
− = = .
= + −1
2028 9 55 25 631
2 . .( )x 55 0 59500
1
2024 3 9 55 3 25 6532
2
=
= + −
.
( . ) ( . )( )x = −
= + + =
1 21500
1
2028 0 595 1 21531
3
.
. .( )x 11 49050
1
2024 3 0 595 3 1 21532
3
.
( . ) ( . )( )x = + + = 11 47150
1
2028 1 4905 1 4715 1 4031
4
.
. . .( )x = + − = 0095
1
2024 3 1 4905 3 1 4715 132
4x
( ) ( . ) ( . ) .= + − = 220285
1
2028 1 40095 1 20285 1 4031
5x
( ) . . .= + − = 9991
1
2024 3 1 40095 3 1 2028532
5x
( ) ( . ) ( . )= + − = 11 22972
1
2028 1 40991 1 22972 131
6
.
. . .( )x = + − = 440901
1
2024 3 1 40991 3 1 2297232
6x
( ) ( . ) ( . )= + − = 1 22703.
Exercise Set 11.13 667
The slope of line L1 is –1.
The slope of the line
Since (–1) • 1 = –1, it follows——x1*x2* is perpendicular to L1.
The slope of line L2 is 12.
The slope of the line .
Since 12 • (–2) = –1, it follows
——x1*x2* is perpendicular to L2.
The slope of line L3 is 3.
The slope of the line
Since , it follows is perpendicular to L3.
(b) Substituting in the equations for x1(1)from part (a) of Exercise 1
gives
Then gives
x211 1
52 4
12
112
10
11( ) = − +
+
• • =
= +
+
=•
46
55
1
54 2
12
11
10
11
7221
x( ) 88
55
xx11 12
11
10
11( ) ,=
x
x
111
12
1
22
31
22
27
22
48
44
12
11( )
(
= + −
= =
11 1
22
31
22
27
22
40
44
10
11)
.
= − +
= =
xx01 31
22
27
22( ) ,=
x x2 3* *3
1
31• −
= −
x x2 327 22 78 55
31 22 46 55
21
63
1
3* * is
−−
=−
=−
..
isx x1 278 55 10 11
46 55 12 11
28
142* * −
−=
−= −
x x1 327 22 10 11
31 22 12 11
7
71* * .is
−−
= =
668 Exercise Set 11.13
and gives
or
4. Referring to the figure below and starting with x0(1) = (0, 0)
x0(1) is projected to x1
(1) on L1,
x1(1) is projected to x2
(1) on L2,
x2(1) is projected to x3
(1) on L3, and so on.
As seen from the graph the points of the limit cycle are
x 1* = A, x2* = B, x3* = A.
Since x1* is the point of intersection of L1 and L3 it follows on solving the system
x12* = 1
x11* – x12* = 0
L1 x1 x1
x2
x0
x3
A
B
x2
x2 = 1
x1 – x2 = 0 x1 – x2 = 2
x1
L3
L2(1)
(1)
(1)
(1)
(2)
xx31 31
22
27
22( ) , .=
x311 1
109
46
553
78
55
775
550( ) = + + ⋅
= =
331
22
1
103 3
46
559
78
55321
x( ) = − + ⋅
+ ⋅
= =
675
550
27
22
xx21 46
55
78
55( ) ,=
Exercise Set 11.13 669
that x1* = (1, 1). Since x2* = (x21*, x22*) is on L2, it follows that x21* – x22* = 2. Now is perpendicular to L2; therefore
so we have x22* – 1 = 1 – x21* or x21* + x22* = 2.
Solving the system
x21* – x22* = 2
x21* + x22* = 2
gives x21* = 2 and x22* = 0. Thus the points on the limit cycle are
x1* = (1, 1), x2* = (2, 0), x3* = (1, 1).
5. (a) From Theorem 1:
and so
= atx* + b – atx* = b.
Therefore xp
lies on the line atx = b.
(b) Let z be a point on the line L, so that atz = b. We will prove that xp
– x* isperpendicular to x
p– z by showing that (x
p– z)t (x
p– x* = 0 (i.e., their dot product
is 0). Now
xxp p p
b
b
−( ) − = −−
=
zz xx xx xx zzaa xx
aa aaaa
t t)( ) (**t
t
−−
−
aa xx
aa aaxx zz aa
tt)
*
t( p
aa xx aa xxaa xx
aa aaaa aat
pt
t
t
tb= +
−
**
xx xxaa xx
aa aaaap
t
t
b= +
−
**
x
x
22
21
1
11 1
*
*( )
−
−
= −
x x1 2* *
670 Exercise Set 11.13
Since xp
and z are on L, we have
atxp
= b
atz = b ,
and so at (xp
– z) = 0 or (xp
– z)t a = 0.
Thus (xp
– z)t (xp
– x*) = 0, and so xp
– x* is perpendicular to L.
6. Suppose that the center of some pixel, say the K-th pixel, is not crossed by any of theM beams in the scan. Then we have a
iK= 0 for i = 1, 2, … , M. That is, the K-th component
of each of the N-dimensional vectors a1, a2, … , aM
is zero. But then they cannot spanN since, in particular, no linear combination of them could equal e
K, where e
Kis the
N-dimensional vector whose K-th component is one and whose other components are allzero.
7. Let us choose units so that each pixel is one unit wide. Then
aij
= length of the center line of the i-th beam that lies in the j-th pixel
If the i-th beam crosses the j-th pixel squarely, it follows that aij
= 1. From Fig. 11 in thetext, it is then clear that
a17 = a18 = a19 = 1
a24 = a25 = a26 = 1
a31 = a32 = a33 = 1
a73 = a76 = a79 = 1
a82 = a85 = a88 = 1
a91 = a94 = a97 = 1
since beams 1, 2, 3, 7, 8, and 9 cross the pixels squarely.
Next, the centerlines of beams 5 and 11 lie along the diagonals of pixels 3, 5, 7 and 1, 5, 9,respectively. Since these diagonals have length , we have
a53 = a55 = a57 = = 1.41421
a11,1 = a11,5 = a11,9 = = 1.41421.2
2
2
Exercise Set 11.13 671
In the following diagram, the hypotenuse of triangle A is the portion of the centerline of the 10th beam that lies in the 2nd pixel. The length of this hypotenuse is twice the height of triangle A, which in turn is – 1. Thus,
a10,2 = 2( – 1) = .82843.
By symmetry we also have
a10,2 = a10,6 = a12,4 = a12,8
= a62 = a64 = a46 = a48 = .82843.
Also from the diagram, we see that the hypotenuse of triangle B is the portion of thecenterline of the 10th beam that lies in the 3rd pixel. Thus,
a10,3 = 2 – = .58579.
By symmetry we have
a10,3 = a12,7 = a61 = a49 = .58579.
1− 1−2
1−2
1
1
B
A
1
2
22
2
2 − 1
2( )2 − 1
2− 2
2
2
2
672 Exercise Set 11.13
The remaining aij’s are all zero, and so the 12 beam equations (4) are
x7 + x8 + x9 = 13.00
x4 + x5 + x6 = 15.00
x1 + x2 + x3 = 8.00
.82843 (x6 + x8) + .58579x9 = 14.79
1.41421 (x3 + x5 + x7) = 14.31
.82843 (x2 + x4) + .58579 x1 = 3.81
x3 + x6 + x9 = 18.00
x2 + x5 + x8 = 12.00
x1 + x4 + x7 = 6.00
.82843 (x2 + x6) + .58579x3 = 10.51
1.41421 (x1 + x5 + x9) = 16.13
.82843 (x4 + x8) + .58579x7 = 7.04
8. Let us choose units so that each pixel is one unit wide. Then
aij
= area of the i-th beam that lies in the j-th pixel
Since the width of each beam is also one unit it follows that aij
= 1 if the i-th beam crossesthe j-th pixel squarely. From Fig. 11 in the text, it is then clear that
a17 = a18 = a19 = 1
a24 = a25 = a26 = 1
a31 = a32 = a33 = 1
a73 = a76 = a79 = 1
a82 = a85 = a88 = 1
a91 = a94 = a97 = 1
since beams 1, 2, 3, 7, 8, and 9 cross the pixels squarely.
Exercise Set 11.13 673
For the remaining aij’s, first observe
from the figure on the right that anisosceles right triangle of height h, asindicated, has area h2. From the dia-gram of the nine pixels, we then have
Area of triangle B ( ) ( )= −
= − =1
22 1
1
43 2 2
2
..
.
04289
1
2
1
4250
2
Area of triangle C =
= = 000
3
22 1
9
43 2
2
Area of triangle F ( ) (= −
= − 22 38604) .=
1−2
3−2 3−
2
1−2
3−2
1
1
A
B
B
D
C E
F
2
2
( )2 − 1
( )2 − 1
Area = h2 h
674 Exercise Set 11.13
We also have
Referring back to Fig. 11, we see that
a11,1 = Area of polygon A = .91421
a10,1 = Area of triangle B = .04289
a11,2 = Area of triangle C = .25000
a10,2 = Area of polygon D = .75000
a10,3 = Area of polygon E = .61396
By symmetry we then have
a11,1 = a11,5 = a11,9 = a53 = a55 = a57 = .91421
a10,1 = a10,5 = a10,9 = a12,1 = a12,5 = a12,9
= a63 = a65 = a67 = a43 = a45 = a47 = .04289
a11,2 = a11,4 = a11,6 = a11,8
= a52 = a54 = a56 = a58 = .25000
a10,2 = a10,6 = a12,4 = a12,8
= a62 = a64 = a46 = a48 = .75000
a10,3 = a12,7 = a61 = a49 = .61396.
The remaining aij’s are all zero, and so the 12 beam equations (4) are
x7 + x8 + x9 = 13.00
x4 + x5 + x6 = 15.00
x1 + x2 + x3 = 18.00
0.04289 (x3 + x5 + x7) + 0.75 (x6 + x8) + 0.61396x9 = 14.79
Area of polygon Area of triangle(A = − ×1 2
Area of polygon Area o
B
D
)
.
(
= − =
= −
21
291421
1 ff triangle C
Area of polygon
)
.= − = =11
4
3
47500
( )
.
E = −
= −( ) =
1
1
418 2 23 6139
Area of triangle F
66
0.91421 (x3 + x5 + x7) + 0.25 (x2 + x4 + x6 + x8) = 14.31
0.04289 (x3 + x5 + x7) + 0.75 (x2 + x4) + 0.61396x1 = 3.81
x3 + x6 + x9 = 18.00
x2 + x5 + x8 = 12.00
x1 + x4 + x7 = 6.00
0.04289 (x1 + x5 + x7) + 0.75 (x2 + x6) + 0.61396x3 = 10.51
0.91421 (x1 + x5 + x7) + 0.25 (x2 + x4 + x6 + x8) = 16.13
0.04289 (x1 + x5 + x7) + 0.75 (x4 + x8) + 0.61396x7 = 7.04
Exercise Set 11.13 675
EXERCISE SET 11.14
1. Each of the subsets S1, S2, S3, S4 in the figure is congruent to the entire set scaled by afactor of 12/25. Also, the rotation angles for the four subsets are all 0°. The displacementdistances can be determined from the figure to find the four similitudes that map the entireset onto the four subsets S1, S2, S3, S4. These are, respectively,
where the four values of
Because s = 12/25 and k = 4 in the definition of a self-similar set, the Hausdorff dimensionof the set is d
H(S) = ln(k)/ln(1/s) = ln(4)/ln(25/12) = 1.889 … . The set is a fractal because
its Hausdorff dimension is not an integer.
e
f
i
i
are
0
0
13 25
0
0, ,
113 25
13 25
13 25
, and .
Tx
y
x
yi
=
12
25
1 0
0 1
+
=,
e
fi
i
i
1, 2, 3, 4,
y
x
S3
S1 S2
S4
1
1
1225 1
25
677
2. The rough measurements indicated in the figure give an approximate scale factor of s ≈(15/16)/(2) = .47 to two decimal places. Since k = 4, the Hausdorff dimension of the set isapproximately d
H(S) = ln(k)/ln(1/s) ≈ ln(4)/ln(1/.47) = 1.8 to two significant digits.
Examination of the figure reveals rotation angles of 180°, 180°, 0°, and –90° for the sets S1,S2, S3, and S4, respectively.
3. (a) The figure shows the original self-similar set and a decomposition of the set into sevennonoverlapping congruent subsets, each of which is congruent to the original setscaled by a factor s = 1/3. By inspection, the rotations angles are 0° for all sevensubsets. The Hausdorff dimension of the set is d
H(S) = ln(k)/ln(1/s) = ln(7)/ln(3) =
1.771 … . Because its Hausdorff dimension is not an integer, the set is a fractal.
2"
15 "16
s3
s1 s2
s4
678 Exercise Set 11.14
(b) The figure shows the original self-similar set and a decomposition of the set into threenonoverlapping congruent subsets, each of which is congruent to the original setscaled by a factor s = 1/2. By inspection, the rotation angles are 180° for all threesubsets. The Hausdorff dimension of the set is d
H(S) = ln(k)/ln(1/s) = ln(3)/ln(2) =
1.584 … . Because its Hausdorff dimension is not an integer, the set is a fractal.
3. (c) The figure shows the original self-similar set and a decomposition of the set into threenonoverlapping congruent subsets, each of which is congruent to the original setscaled by a factor s = 1/2. By inspection, the rotation angles are 180°, 180°, and –90° for S1, S2, and S3, respectively. The Hausdorff dimension of the set is d
H(s) = ln(k)/ln(1/s) = ln(3)/ln(2) = 1.584 … . Because its Hausdorff dimension is not
an integer, the set is a fractal.
(d) The figure shows the original self-similar set and a decomposition of the set into threenonoverlapping congruent subsets, each of which is congruent to the original setscaled by a factor s = 1/2. By inspection, the rotation angles are 180°, 180°, and –90°for S1, S2, and S3, respectively. The Hausdorff dimension of the set is d
H(S) =
ln(k)/ln(1/s) = ln(3)/ln(2) = 1.584 … . Because its Hausdorff dimension is not aninteger, the set is a fractal.
S3
S1 S2
Exercise Set 11.14 679
4. The matrix of the affine transformation in question is The matrix
portion of a similitude is of the form Consequently, we must have
. Solving this pair of equations gives
5. Letting be the vector to the tip of the fern and using the hint, we have
Solving this matrix equation gives
rounded to three decimal places.
6. As the figure indicates, the unit square can be expressed as the union of 16 nonoverlappingcongruent squares, each of side length 1/4. Consequently, the Hausdorff dimension of theunit square as given by Equation (2) of the text is d
H(S) = ln(k)/ln(1/s) = ln(16)/ln(4) = 2.
x
y
=
−
−.15 .
. .
.
.
04
04 15
075
180
1
=
.
.
766
996
Tx
y
x
y
=or2
.
. .
.
.
.85 04
04 85
075
18−
+
x
y 00
.
x
y
=x
y
= −−. ... tan ( /8509 041and θ .. ) . ... .85 2 69= − °
s = ( ) + −( ) =. .85 042 2
s sandcos . sin .θ θ= = −85 04
ssin
sin cos.
cosθ θθ θ
−
.85 .
. ..
04
04 85−
S3S2
S1
680 Exercise Set 11.14
7. The similitude T1 maps the unit square (whose vertices are (0, 0), (1, 0), (1, 1), and (0, 1))onto the square whose vertices are (0, 0), (3/4, 0), (3/4, 3/4), and (0, 3/4). The similitudeT2 maps the unit square onto the square whose vertices are (1/4, 0), (1, 0), (1, 3/4), and(1/4, 3/4). The similitude T3 maps the unit square onto the square whose vertices are (0, 1/4), (3/4, 1/4), (3/4, 1), and (0, 1). Finally, the similitude T4 maps the unit square ontothe square whose vertices are (1/4, 1/4), (1, 1/4), (1, 1), and (1/4, 1). The union of thesefour smaller squares is the unit square, but the four smaller squares overlap. Each of thefour smaller squares has side length of 3/4, so that the common scale factor of thesimilitudes is s = 3/4. The right-hand side of Equation (2) of the text gives ln(k)/ln(1/s) =ln(4)/ln(4/3) = 4.818 …. This is not the correct Hausdorff dimension of the square (whichis 2) because the four smaller squares overlap.
8. Because s = 1/2 and k = 8, Equation (2) of the text gives dH(S)= ln(k)/ln(1/s) =
ln(8)/ln(2) = 3 for the Hausdorff dimension of a unit cube. Because the Hausdorffdimension of the cube is the same as its topological dimension, the cube is not a fractal.
9. A careful examination of Figure 27 of the text shows that the Menger sponge can beexpressed as the union of 20 smaller nonoverlapping congruent Menger sponges each ofside length 1/3. Consequently, k = 20 and s = 1/3, and so the Hausdorff dimension of theMenger sponge is d
H(S)= ln(k)/ln(1/s) = ln(20)/ln(3) = 2.726 … . Because its Hausdorff
dimension is not an integer, the Menger sponge is a fractal.
10. The figure shows the first four iterates as determined by Algorithm 1 and starting with theunit square as the initial set. Because k = 2 and s = 1/3, the Hausdorff dimension of theCantor set is d
H(S)= ln(k)/ln(1/s) = ln(2)/ln(3) = 0.6309 … . Notice that the Cantor set is
a subset of the unit interval along the x-axis and that its topological dimension must be 0(since the topological dimension of any set is a nonnegative integer less than or equal to itsHausdorff dimension).
141
Exercise Set 11.14 681
11. The area of the unit square S0 is, of course, 1. Each of the eight similitudes T1, T2, …, T8given in Equation 8 of the text has scale factor s = 1/3, and so each maps the unit squareonto a smaller square of area 1/9. Because these eight smaller squares are nonoverlapping,their total area is 8/9, which is then the area of the set S1. By a similar argument, the areaof the set S2 is 8/9-th the area of the set S1. Continuing the argument further, we find thatthe areas of S0, S1, S2, S3, S4, …, form the geometric sequence 1, 8/9, (8/9)2, (8/9)3,(8/9)4, …. (Notice that this implies that the area of the Sierpinski carpet is 0, since thelimit of (8/9)n as n tends to infinity is 0.)
Initial set
First Iterate
Second Iterate
Third Iterate
Fourth Iterate
682 Exercise Set 11.14
EXERCISE SET 11.15
1. Because 250 = 2 • 53 it follows from (i) that
Because 25 = 52 it follows from (ii) that
Because 125 = 53 it follows from (ii) that
Because 30 = 6 • 5 it follows from (ii) that
Because 10 = 2 • 5 it follows from (i) that
Because 50 = 2 • 52 it follows from (i) that
Because 3750 = 6 • 54 it follows from (ii) that
Because 6 = 6 • 50 it follows from (ii) that
Because 5 = 51 it follows from (ii) that
2. The point (0, 0) is obviously a 1-cycle. We now choose another of the 36 points of the form(m/6, n/6), say (0, 1/6). Its iterates produce the 12-cycle
So far we have accounted for 13 of the 36 points. Taking one of the remaining points,
say (0, 2/6), we arrive at the 4-cycle We
continue in this way, each time starting with some point of the form (m/6, n/6) that has not
0
2 6
2 6
4 6
0
4 6
→
→
→44 6
2 6
.
0
1 6
1 6
2 6
3 6
5 6
→
→
→2 6
1 6
3 6
4 6
1 6
5 6
→
→
→
→
→
0
5 6
5 6
4 6
3 6
1 6
→
→
→
4 6
5 6
3 6
6 6
5 6
1 6
.
( ) .5 2 5 10= =•∏
( ) .6 2 6 12= =•∏
( ) .3750 2 3750 7500= =•∏
( ) .50 3 50 150= =•∏
( ) .10 3 10 30= =•∏
( ) .30 2 30 60= =•∏
( ) .125 2 125 250= =•∏
( ) .25 2 25 50= =∏ •
( ) .250 3 250 750= =∏ •
683
yet appeared in a cycle, until we exhaust all such points. This yields a 3-cycle:
The possible periods of points for the form (m/6, n/6) are thus 1, 3, 4 and 12. The leastcommon multiple of these four numbers is 12, and so
3. (a) We are given that x0 = 3 and x1 = 7. With p = 15 we have
x3 = x2 + x1 mod 15 = 7 + 3 mod 15 = 10 mod 15 = 10,
x4 = x3 + x2 mod 15 = 10 + 7 mod 15 = 17 mod 15 = 2,
x5 = x4 + x3 mod 15 = 2 + 10 mod 15 = 12 mod 15 = 12,
x6 = x5 + x4 mod 15 = 12 + 2 mod 15 = 14 mod 15 = 14,
x7 = x6 + x5 mod 15 = 14 + 12 mod 15 = 26 mod 15 = 11,
x8 = x7 + x6 mod 15 = 11 + 14 mod 15 = 25 mod 15 = 10,
x9 = x8 + x7 mod 15 = 10 + 11 mod 15 = 21 mod 15 = 6,
x10 = x9 + x8 mod 15 = 6 + 10 mod 15 = 16 mod 15 = 1,
x11 = x10 + x9 mod 15 = 1 + 6 mod 15 = 7 mod 15 = 7,
x12 = x11 + x10 mod 15 = 7 + 1 mod 15 = 8 mod 15 = 8,
x13 = x12 + x11 mod 15 = 8 + 7 mod 15 = 15 mod 15 = 0,
x14 = x13 + x12 mod 15 = 0 + 8 mod 15 = 8 mod 15 = 8,
x15 = x14 + x13 mod 15 = 8 + 0 mod 15 = 8 mod 15 = 8,
x16 = x15 + x14 mod 15 = 8 + 8 mod 15 = 16 mod 15 = 1,
x17 = x16 + x15 mod 15 = 1 + 8 mod 15 = 9 mod 15 = 9,
( ) .6 12=∏
3 6
0
3 6
3 6
0
3 6
→
→
→
; another 4-cycle:
4 6
0
4 6
4 6
→
→
;
2 6
0
2 6
2 6
and anoother 12-cycle:1 6
0
1 6
1 6
→
→
→
2 6
3 6
5 6
2 6
→
→
→
1 6
3 6
4 6
1 6
5 6
0
5 6
5 6
4 6
3 6
→
→
→
→
1 6
4 6
55 6
3 6
→
.2 6
5 6
684 Exercise Set 11.15
x18 = x17 + x16 mod 15 = 9 + 1 mod 15 = 10 mod 15 = 10,
x19 = x18 + x17 mod 15 = 10 + 9 mod 15 = 19 mod 15 = 4,
x20 = x19 + x18 mod 15 = 4 + 10 mod 15 = 14 mod 15 = 14,
x21 = x20 + x19 mod 15 = 14 + 4 mod 15 = 18 mod 15 = 3,
x22 = x21 + x20 mod 15 = 3 + 14 mod 15 = 17 mod 15 = 2,
x23 = x22 + x21 mod 15 = 2 + 3 mod 15 = 5 mod 15 = 5,
x24 = x23 + x22 mod 15 = 5 + 2 mod 15 = 7 mod 15 = 7,
x25 = x24 + x23 mod 15 = 7 + 5 mod 15 = 12 mod 15 = 12,
x26 = x25 + x24 mod 15 = 12 + 7 mod 15 = 19 mod 15 = 4,
x27 = x26 + x25 mod 15 = 4 + 12 mod 15 = 16 mod 15 = 1,
x28 = x27 + x26 mod 15 = 1 + 4 mod 15 = 5 mod 15 = 5,
x29 = x28 + x27 mod 15 = 5 + 1 mod 15 = 6 mod 15 = 6,
x30 = x29 + x28 mod 15 = 6 + 5 mod 15 = 11 mod 15 = 11,
x31 = x30 + x29 mod 15 = 11 + 6 mod 15 = 17 mod 15 = 2,
x32 = x31 + x30 mod 15 = 2 + 11 mod 15 = 13 mod 15 = 13,
x33 = x32 + x31 mod 15 = 13 + 2 mod 15 = 15 mod 15 = 0,
x34 = x33 + x32 mod 15 = 0 + 13 mod 15 = 13 mod 15 = 13,
x35 = x34 + x33 mod 15 = 13 + 0 mod 15 = 13 mod 15 = 13,
x36 = x35 + x34 mod 15 = 13 + 13 mod 15 = 26 mod 15 = 11,
x37 = x36 + x35 mod 15 = 11 + 13 mod 15 = 24 mod 15 = 9,
x38 = x37 + x36 mod 15 = 9 + 11 mod 15 = 20 mod 15 = 5,
x39 = x38 + x37 mod 15 = 5 + 9 mod 15 = 14 mod 15 = 14,
x40 = x39 + x38 mod 15 = 14 + 5 mod 15 = 19 mod 15 = 4,
x41 = x40 + x39 mod 15 = 4 + 14 mod 15 = 18 mod 15 = 3,
x42 = x41 + x40 mod 15 = 3 + 4 mod 15 = 7 mod 15 = 7,
and finally: x41 = x0 and x42 = x1. Thus this sequence is periodic with period 41.
(b) Step (ii) of the algorithm is
xn+1 = x
n+ x
n–1 mod p. (A)
Exercise Set 11.15 685
Replacing n in this formula by n + 1 gives
xn + 2 = xn + 1
+xn
mod p = (xn+ xn–1) + x
nmod p = 2x
n+ x
n–1 mod p. (B)
Equations (A) and (B) can be written as
xn+1 = x
n–1 + xn
mod p
xn+2 = x
n–1 + 2xn
mod p
which in matrix form are
3. (c) Beginning with we obtain
x
x
2
3
1 1
1 2
5
5
=
modd mod2110
1521
10
15
4
5
=
=
x
x
=
mod1 1
1 2
10
15221
25
4021
4
19
6
7
=
mod =
x
x
=
=mod1 1
1 2
4
1921
233
4221
2
0
8
9
mod =
x
x==
=
mod
1 1
1 2
2
021
2
2
=
mod 212
2
1 1
1 2
10
11
=
x
x
=
mod mo2
221
4
6dd 21
4
6=
x
x
0
1
5
5
=
,
x
x
x
x
n
n
n
n
+
+
−
=
1
2
11 1
1 2
mod .p
686 Exercise Set 11.15
and we see that
4. (a) We have that
Γ25101
101
7 778m
n
=
, ,, , , , ,
, , , ,
742 049 12 586 269 025
12 586 269 025 20 365,, ,011 074
101
101
m
n
=
+
mod
, , , , , ,
1
7 778 742 049
101
12 586 269 025
101m nn
m n12 586 269 025
101
20 365 011 074
101
, , , , , ,+
=
+
+
+ +
mod1
1
101
1
101
r m sn
sm t
n
m n r swhere , , , ,mod (1 and are integers)t
rm snm
sm tn
=
+( ) +
+( ) +
101
nn
m
n
101
1
101
101
=
mod
+ +(because and are integersrm sn sm tn )).
x
x
x
x
16
17
0
1
=
.
1 1
1 2
12
13
=
x
x
=
mod mod4
621
10
1621 ==
10
16
1 1
1 2
14
15
=
x
xmod mod
10
1621
26
4221
=
==5
0
1 1
1 2
16
17
=
x
x
55
021
5
521
5
5
=
mod mod =
.
Exercise Set 11.15 687
4. (b) From (a) we have that every point of the form (m/101, n/101) returns to its startingposition after 25 iterations, and so every such point is a periodic point with a periodthat must divide 25. The only integers that divide 25 are 1, 5, and 25, and so everysuch point must have period 1, 5, or 25.
(c) We have that
Because all five iterates are different, the period of the periodic point (1/101, 0) mustbe greater than 5.
(d) From part (b) all points of the form (m/101, n/101) must have period 1, 5, or 25. Frompart (c) the point (1/101, 0) has period greater that 5, and so must have period 25.Because the least common multiple of 1, 5, and 25 is 25, it follows that
5. If 0 ≤ x < 1 and 0 ≤ y < 1, then T(x, y) = (x + 5/12, y) mod 1, and so T 2(x, y) = (x + 10/12, y) mod 1, T 3(x, y) = (x + 15/12, y) mod 1, . . . , T12(x, y) = (x + 60/12, y) mod 1 =(x + 5, y) mod 1 = (x, y). Thus every point in S returns to its original position after
( ) .101 25=∏
Γ1
101
0
1 1
1 2
1
=
1101
0
1
1
101
1
101
=
mod
=
,
Γ21
101
0
1 1
1 2
=mod
1
101
1
101
1
2
101
3
101
,
Γ31
101
0
=
1 1
1 2
2
101
3
101
mod11
5
101
108
101
1
101
0
4
=
,
Γ
=
1 1
1 2
5
101
8
101
=
mod1
13
101
21
101
,,
Γ51
101
0
1 1
1 2
=
=mod
13
101
21
101
1
34
101
55
1001
,
688 Exercise Set 11.15
12 iterations and so every point in S is a periodic point with period at most 12. Becauseevery point is a periodic point, no point can have a dense set of iterates, and so the mappingcannot be chaotic.
6. (a) The matrix of Arnold’s cat map, is one in which (i) the entries are all
integers, (ii) the determinant is 1, and (iii) the eigenvalues, …
and …, do not have magnitude 1. The three conditions of anAnosov automorphism are thus satisfied.
(b) The eigenvalues of the matrix are ±1, both of which have magnitude 1. By
part (iii) of the definition of an Anosov automorphism, this matrix is not the matrix ofan Anosov automorphism.
The entries of the matrix are integers; its determinant is 1; and neither of its
eigenvalues, has magnitude 1. Consequently, this is the matrix of an Anosovautomorphism.
The eigenvalues of the matrix are both equal to 1, and soboth have magnitude
1. By part (iii) of the definition, this is not the matrix of an Anosov automorphism.
The entries of the matrix are integers; its determinant is 1; and neither of its
eigenvalues, has magnitude 1. Consequently, this is the matrix of an Anosovautomorphism.
The determinant of the matrix is 2, and so by part (ii) of the definition, this
is not the matrix of an Anosov automorphism.
(c) The eigenvalues of the matrix are ±i; both of which have magnitude 1.
By part (iii) of the definition, this cannot be the matrix of an Anosov automorphism.
0 1
1 0−
6 2
5 2
4 15,±
5 7
2 3
1 0
0 1
2 3,±
3 2
1 1
0 1
1 0
( ) .3 5 2 0 3819+ =
( ) .3 5 2 2 6180+ =
1 1
1 2
,
Exercise Set 11.15 689
Starting with an arbitrary point in the interior of S, (that is, with 0 < x < 1
and 0 < y < 1) we obtain
Thus every point in the interior of S is a periodic point with period at most 4. The
geometric effect of this transformation, as seen by the iterates, is to rotate each point
in the interior of S clockwise by 90° about the center point of S.
Consequently, each point in the interior of S has period 4 with the exception of the
center point which is a fixed point.
For points in the interior of S, we first observe that the origin is a fixed
point, which can easily be verified. Starting with a point of the form with 0 < x < 1,
we obtain a 4-cycle if ,
otherwise we obtain the 2-cycle1 2
0
0
1 2
1 2
0
→
→
.
x ≠ 1 2x
x
x
0
0
1
1
0
0
→
→
→−
−
xx
x
→
0
x
0
0
0
1 2
1 2
1 2
1 2
0 1
1 01
−
=
=
−
mod mod
x
y
y
x11
1
0 1
1 0 1
,
mo
=−
−
=
−
y
x
y
xdd mod ,1
11
1
1
0 1
1 0
=−−
=
−−
−
x
y
x
y
=
−−
=
−− +
mod
1
11
1
1
x
y
y
xmod ,1
1
0 1
1 0
1
=−
−
=
−
y
x
y
x =
− +
=
mod mod .1
11
x
y
x
y
x
y
690 Exercise Set 11.15
Similarly, starting with a point of the form with 0 < y < 1, we obtain a 4-cycle
if , otherwise we obtain the
2-cycle . Thus every point not in the interior of S is a
periodic point with point 1, 2, or 4. Finally, because all points in S can have a dense setof iterates and so the mapping cannot be chaotic.
7. That Arnold’s cat map is a one-to-one mapping over the unit square S and that its range isS is fairly obvious geometrically. An analytical proof would proceed as follows: To showthat it is one-to-one, suppose that (x1, y1) and (x2, y2), with 0 ≤ x
i< 1 and 0 ≤ y
i< 1, are
two points in S such that Γ(x1, y1) = Γ(x2, y2). From the definition of Γ it follows that
(x1 + y1) mod 1 = (x2 + y2) mod 1
(x1 + 2 y1) mod 1 = (x2 + 2 y2) mod 1.
From the definition of “mod 1”, there are integers r1, s1, r2, and s2 such that
x1 + y1 – r1 = x2 + y2 – r2 (A)
x1 + 2y1 – s1 = x2 + 2y2 – s2. (B)
Subtracting the first equation from the second gives
y1 – s1 + r1 = y2 – s2 + r2
or
y1 – y2 = s1 – s2 + r2 – r1. (C)
The right-hand side of Eq. (C) is an integer and the left-hand side satisfies –1 < y1 – y2 < 1.Consequently, the integer on the right-hand side must be zero, and so y1 = y2. Putting y1 = y2 into Eq. (A) gives x1 – r1 = x2 – r2 and a similar argument shows that x1 = x2. Thus(x1, y1) = (x2, y2), which shows that the mapping is one-to-one.
0
1 2
1 2
0
0
1 2
→
→
y ≠ 1 2
0
0
0
1
l
y
y
y
→
→−
→−−
→
y
y0
0
0
y
Exercise Set 11.15 691
To show that the range of Γ is S, let (u, v) be any point in S, so that 0 ≤ u < 1 and 0 ≤ v < 1. We want to show that there exists a point (x, y) in S such that Γ(x, y) = (u, v);that is, such that
(x + y) mod 1 = u
(x + 2y) mod 1 = v.
From the definition of “mod 1”, this means that we must find integers r and s that
x + y – r = u
x + 2y – s = v
and for which x and y lie in [0, 1). Solving for x and y in this linear system gives
x = (2u – v) + (2r – s)
y = (–u + 2 v) + (–r + s).
Let a and b be integers such that (2u – v) + a and (–u + 2v) + b lie in the interval [0, 1).Solving the system
2r – s = a
–r + s = b
gives r = a + b and s = a + 2b, both of which are integers. This choice of r and s thenprovides the desired values of x and y.
8. Let (x, y) lie in S, so that 0 ≤ x < 1 and 0 ≤ y < 1. We must show that Γ–1(Γ(x, y)) = (x, y). From the definition of Γ–1 and Γ we obtain
Γ–1(Γ(x, y)) = Γ–1(x + y mod 1, x + 2y mod 1)
= Γ–1(x + y – r, x + 2y – s), for certain integers r and s,
= (2(x + y – r) – (x + 2y – s), – (x + y – r) + (x + 2y – s)) mod 1
= (x – 2r + s, y + r – s) mod 1
= (x, y),
where the last step follows because –2r + s and r – s are integers and 0 ≤ x, y < 1.
692 Exercise Set 11.15
Exercise Set 11.15 693
9. As per the hint, we wish to find the regions in S that map onto the four indicated regionsin the figure below.
We first consider region I′ with vertices (0, 0), (1/2, 1), and (1, 1). We seek points (x1, y1),(x2, y2), and (x3, y3), with entries that lie in [0, 1], that map onto these three points under
the mapping for certain integer values of a and b to be
determined. This leads to the three equations
The inverse of the matrix We multiply the above three matrix
equations by this inverse and set Notice that c and d must be
integers. This leads to
c
d
2
1
=−
−
.1
1
a
b
1
1is
21
2
1
1 1
−
−
.
1
1
1
2
1
1
+
x
y
a
b==
,0
0
1
2
2
2
1
1
x
y
/,+
=
a
b
1 2
1
1
2
1
1
+
=x
y
a
b
3
3
1
1
.
x
y
x
y
a
b
→
+
1 2
1 2
I'
II'
III'
IV'
(1, 1)(0, 1)
(0, 0) (1, 0)( , 0)
( , 1)
12
12
The only integer values of c and d that will give values of xI
and yi
in the interval [0, 1]
are c = d = 0. This then gives a = b = 0 and the mapping that
maps the three points (0, 0), (0, 1/2), and (1, 0) to the three points (0, 0), (0, 1/2), and(1, 1), respectively. The three points (0, 0), (0, 1/2), and (1, 0) define the triangular regionlabeled I in the diagram below, which then maps onto the region I′ above.
For region II′, the calculations are as follows:
1
1
1
2
1
1
+
x
y
a
b==
,1 2
0
1
2
2
2
1
1
x
y++
=
,
a
b
1
1
I
IIIII
IV
(1, 1)(0, 1)
(0, 0) (1, 0)
(1, )12(0, )1
2
x
y
x
y
→
1 1
1 2
x
y
1
1
2 1
1 1
0
0
=−
−
−
= −
,c
d
c
d
x
y
2
2
=−
−
/2 1
1 1
1 2
1 /−
=
−
c
d
c
d
0
1 2
=−
−
,
x
y
3
3
2 1
1 1
1
1
−
=
−c
d
1
0.
c
d
694 Exercise Set 11.15
Only c = 1 and d = –1 will work. This leads to a = 0, b = –1 and the mapping
that maps region II with vertices (0, 1/2), (0, 1), and
(1, 0) onto region II′.
For region III′, thew calculations are as follows:
1
1
1
2
1
1
+
x
y
a
b==
+
,0
0
1
2
2
2
1
1
x
y,
a
b
=
1 2
1
1
2
1
1
x
y
a
b
3
3
0
1
+
=
=−
−
;
x
y
1
1
1
1 1
0
0
2
−
= −
c
d
c
d
x
y
2
2
=−
−
−2 1
1 1
1 2
1
c
d
=
−
,0
1 2
3
3
c
d
x
y
=−
−
−2 1
1 1
0
1
cc
d
c
d
=−
−
.1
1
x
y
x
y
→
+
−
1 1
1 2
0
1
1
1
1
2
1
03
3
+
=
x
y
a
b
=
−−
;
x
y
1
1
1
1 1
1 2
0
2
−
=
−
−
/
c
d
c
d
1
1 2
=
−−
,
x
y
2
2
1
1 1
1
1
2
−
=
−
c
d
c
d
0
1,
x
y
3
3
1
1 1
1
0
=
−−
2.−
=
−
−
c
d
c
d
2
1
Exercise Set 11.15 695
Only c = –1 and d = 0 will work. This leads to a = –1, b = –1 and the mapping
that maps region III with vertices (1, 0), (1, 1/2), and
(0, 1 ) onto region III′.
For region IV′, the calculations are as follows:
Only c = 0 and d = –1 will work. This leads to a = –1, b = –2 and the mapping
that maps region IV with vertices (0, 0), (1, 1), and
(1/2, 0) onto region IV′.
x
y
x
y
→
+1 1
1 2
−−−
1
2
1
1
1
2
1
1
+
x
y
a
b==
+
,0
0
1
2
2
2
1
1
x
y,
a
b
=
1
1
1
2
1
1
x
y
a
b
3
3
1 2
0
+
=
=−
−
;
x
y
1
1
1
1 1
0
0
2
−
= −
c
d
c
d
x
y
2
2
=−
−
−
2 1
1 1
1
1
c
d
=
−
,1
0
3
3
c
d
x
y
=−
−
−2 1
1 1
1 2
0
c
dd
c
d
−
−
.1
1 2
x
y
x
y
→
+1 1
1 2
−−
−
1
1
696 Exercise Set 11.15
10. We first establish the result for n = 2. Given (x0, y0) we have
and
where r0, s0, r1 and s1 are integers. This last expression is the same as
mod 1, which is what we wanted to show.
A similar argument establishes the result for any integer n.
11. As per the hint, we want to show that the only solution of the matrix equation
with 0 ≤ x0 < 1, 0 ≤ y0 < 1, and integer r and s is x0 = y0 = 0. This matrix equation isequivalent to the system
x0 = x0 + y0 – r
y0 = x0 + 2y0 – s.
The first equation is y0 = r, which can only have the solution y0 = r = 0 with the givenconstraints on y0 and r. Putting y0 = 0 into the second equation yields x0 = s, which likewisehas only the solution x0 = s = 0.
x
y
x
y
r
s
0
0
0
0
1
1 2
=
−
1
1.
1
1 2
20
0
x
y
x
y
x
y
2
2
1
1
1
1 21
=
mod
1,=
−
=
1
1
1
1 2
1
1
1
1
1
1
x
y
r
s
22
1 1
1 20
0
0
0
−
−
x
y
r
s
−
=
r
s
x
1
1
201
1 2
1
yy
r s r
r s s0
0 0 1
0 0 12
−
+ ++ +
x
y
x
y
1
1
0
0
1
1 11
=
mod
1==
−
1 1
1 20
0
0
0
x
y
r
s
Exercise Set 11.15 697
12. As per the hint, we want to find all solutions of the matrix equation
where 0 ≤ x0 < 1, 0 ≤ y0 < 1, and r and s are nonnegative integers. This equation can berewritten as
which has the solution First trying r = 0 and s = 0, 1,
2, …, then r = 1 and s = 0, 1, 2, …, etc., we find that the only values of r and s that yield
values of x0 and y0 lying in [0, 1) are:
r = 1 and s = 2, which give x0 = 2/5 and y0 = 1/5;
r = 2 and s = 3, which give x0 = 1/5 and y0 = 3/5;
r = 2 and s = 4, which give x0 = 4/5 and y0 = 2/5;
r = 3 and s = 5, which give x0 = 3/5 and y0 = 4/5.
We can then check that (2/5, 1/5) and (3/5, 4/5) form one 2-cycle and (1/5, 3/5) and (4/5,2/5) form another 2-cycle.
13. In the matrix equation mod 1, let the matrix be
denoted by , all of whose entries are positive integers. The matrix equation can
then be written as
where r and s are integers. By Cramer’s rule, this last equation has the solution
xd r b s
a d b cy
an n
n n n n
n0 0
1
1 1
1=
− −− − −
=−( )
( )( )
( )and
rr c r
a d b c
n
n n n n
−− − −( )( )
.1 1
x
y
a b
c d
x
y
n n
n n
0
0
0
0
=
−
−
−
r
s
a b
c d
x
y
n n
n n
or1
1
0
00
=
r
s
a b
c d
n n
n n
,
1 1
1 2
nx
y
x
y
n
0
0
0
0
1 1
1 2
=
xr s
yr s
0 04 3
5
3
5.=
− +=
−and
1 3
3 4
0
0
=
,x
y
r
s
x
y
x
y
0
0
0
0
2 3
3 5
=
−rr
s
698 Exercise Set 11.15
These two expressions identify x0 and y0 as the quotients of integers, and hence as rational
numbers. One point remains: to verify that the determinant of the matrix
is not equal to zero, so that Cramer’s rule is valid. This is equivalent to showing that the
matrix does not have the eigenvalue 1. Now, the eigenvalues of this matrix are
λ1n and λ2
n, where λ1 and λ2 are the eigenvalues of Because these eigenvalues are
it is not possible that their
nth powers can be 1 for n = 1, 2, …
λ λ1 3 5 2 2 6180 3 5 2( / ) . ( /= + = = −… and 2 )) . ,= 0 3819 …
1 1
1 2
.
a b
c d
n n
n n
a b
c d
n n
n n
−−
1
1
Exercise Set 11.15 699
EXERCISE SET 11.16
1. First we group the plaintext into pairs and add the dummy letter T:
DA RK NI GH TT
or equivalently from Table 1:
4 1 18 11 14 9 7 8 20 20
(a) For the enciphering matrix we have
1 3
2 1
4
1
7
9
1 3
2 1
=
G
I
=
=
18
11
51
47
25
21
=
=
Y
U
1 3
2 1
14
9
41
37
115
1126
1 3
2 1
7
8
O
K(mod )
=
=
31
22
5
22
1 3
2 1
E
V
=
=
20
20
80
60
2
8
B
H
A =
1 3
2 1
701
The Hill cipher is
GIYUOKEV BH
1. (b) For the enciphering matrix we have
The Hill cipher is
SFANEFZW JH
2. (a) For , we have det(A) = 18 – 7 = 11, which is not divisible by 2 or 13.
Therefore by Corollary 2, A is invertible. From Eq. (2):
A =
9 1
7 2
4 3
1 2
4
1
19
6
4 3
1 2
=
S
F
=
=
18
11
105
40
1
14
=
A
N
4 3
1 2
14
9
83
32==
5
626
4 3
1 2
7
8
E
F(mod )
=
=
52
23
0
23
4 3
1 2
Z
W
=
=
20
20
140
60
10
8
J
H
A =
4 3
1 2
702 Exercise Set 11.16
Checking:
(b) For , we have det(A) = 9 – 5 = 4, which is divisible by 2. Therefore
by Corollary 2, A is not invertible.
(c) For , we have det(A) = 72 – 11 = 61 = 9 (mod 26), which is not
divisible by 2 or 13. Therefore by Corollary 2, A is invertible. From (2):
A− −=
−
−
=−
−
1 199 11
1 83
9 11
1 8( )
==−
−
=
27 33
3 24
1 19
23 242(mod 66).
A =
8 11
1 9
A =
3 1
5 3
AA− =
=
1 9
7
1
2
12 7
23 15
131 78
130 79
=
=
−
1 0
0 126
12 7
23 15
91
(mod )
A A77
1
2
157 26
312 53
1 0
0 1
=
=
(mod 226) .
A− −=
−
−
=−
−
=
1 1112 1
7 919
2 1
7 9
3
( )
88 19
133 171
12 7
23 1526
−
−
=
(mod )).
Exercise Set 11.16 703
Checking:
2. (d) For , we have det(A) = 14 – 1 = 13, which is divisible by 13. Therefore
by Corollary 4, A is not invertible.
(e) For , we have det(A) = 6 – 6 = 0, so that A is not invertible by
Corollary 2.
(f) For , we have det(A) = 3 – 8 = –5 = 21 (mod 26), which is not
divisible by 2 or 13. Therefore by Corollary 2, A is invertible. From (2):
Checking:
AA− =
=11
1
8
3
15 12
21 5
183 52
78 27
=
=
−
1 0
0 126
15 12
21 5
1
(mod )
A A
=
=1
1
8
3
27 156
26 183
1 0
00 126
(mod ).
A− −=
−
−
=−
−
=
1 1213 8
1 15
3 8
1 1
15
( )
−−
−
=
40
5 5
15 12
21 526(mod ).
A =
1 8
1 3
A =
3 1
6 2
A =
2 1
1 7
AA− =
=18 11
1 9
1 19
23 24
261 4116
208 235
1 0
0 126
1
=
−
(mod )
A AA =
=1 19
23 24
8 11
1 9
27 182
208 4469
1 0
0 126
=
(mod ).
704 Exercise Set 11.16
3. From Table 1 the numerical equivalent of this ciphertext is
19 1 11 14 15 24 1 15 10 24
Now we have to find the inverse of .
Since det(A) = 8 – 3 = 5, we have from Corollary 2:
To obtain the plaintext, we multiply each ciphertext vector by A–1:
The plaintext is thus
WE LOVE MATH
16 5
15 6
19
1
309
291
23
=
=55
16 5
15 6
11
14
246
2
=
W
E
449
12
15
16 5
15 6
15
2
=
L
O
44
360
369
22
6
=
=
V
E(mod )26
16 5
15 6
1
15
91
105
=
=
13
1
16 5
15 6
10
24
M
A
=
=
280
294
20
8
T
H
A− −=
−
−
=−
−
1 15
2 1
3 421
2 1
3 4( )
=
−
−
=
42 21
63 84
16 5
15 6 (mod ).26
A =
4 1
3 2
Exercise Set 11.16 705
4. From Table 1 the numerical equivalent of the known plaintext is
AR MY
1 18 13 25
and the numerical equivalent of the corresponding ciphertext is
SL HK
19 12 8 11
so the corresponding plaintext and ciphertext vectors are
We want to reduce
to I by elementary row operations and simultaneously apply these operations to
The calculations are as follows:
19
8
12 1
11 13
18
25
1
8
132 11
11 13
198
25
P
t
t.=
=
pp
pp
1
2
1 18
13 25
C
t
t=
=
cc
cc
1
2
19 12
8 11
pp cc
pp
1 1
2
=
← → =
=
1
18
19
12
113
25
8
11
← → =
cc2
706 Exercise Set 11.16
We multiplied the first row by19–1 = 11 (mod 26) .
We formed the matrix .C P
Thus so the deciphering matrix is
(mod 26).
Since (mod 26), we have
for the enciphering matrix.
A A= =−
−
=
−−
− − −( ) ( )1 1 1235 15
6 717
5 15
6 7
=−
−
=
85 255
102 119
7 5
2 1526(mod )
A−( ) = − = − =1 35 90 55 23
A− =
1 7 15
6 5
( )At− =
1 7 6
15 5
− − −
1
8
2 11
11 13
16
25
1
0
2 11
5 75
16
103
1
0
2 11
21 3
16
1
1
0
2 11
1 15
16
5
−
1
0
0 19
1 15
6
5
1
0
0 7
1 15
6
5
Exercise Set 11.16 707
We added –8 times the first rowto the second.
We replaced the entries in thesecond row by their residuesmodulo 26.
We replaced 132 and 198 bytheir residues modulo 26 .
We added –2 times the secondrow to the first.
We replaced –19 by its residuemodulo 26.
We multiplied the second rowby 21–1 = 5 (mod 26).
5. From Table 1 the numerical equivalent of the known plaintext is
AT OM
1 20 15 13
and the numerical equivalent of the corresponding ciphertext is
JY QO
10 25 17 15
The corresponding plaintext and ciphertext vectors are:
We want to reduce
to I by elementary row operations and simultaneously apply these operations to
The calculations are as follows:
10
17
25 1
15 15
20
13
27
17
40 16
15 15
33
13
P =
1 20
15 13.
C =
10 25
17 15
pp cc
pp
1 1
2
1
20
10
25
15
13
=
← → =
=
← → =
cc2
17
15
708 Exercise Set 11.16
We added the second row tothe first (since 10–1 does notexist mod 26).
We formed the matrix .C P
Thus , and so the deciphering matrix is
From Table 1 the numerical equivalent of the given ciphertext is
LN GI HG YB VR EN JY QO
12 14 7 9 8 7 25 2 22 18 5 14 10 25 17 15
A− =
1 24 5
19 14.
At−( ) =
1 24 19
5 14
− − −
1
17
14 16
15 15
7
13
1
0
14 16
223 257
7
1106
1
0
14 16
11 3
7
24
1
0
14 16
1 57
7
4
556
1
0
14 16
1 5
7
14
1
0
0 54
1 5
189
1
− −
44
1
0
0 24
1 5
19
14
Exercise Set 11.16 709
We replaced the entries in thefirst row by their residuesmodulo 26.
We replaced the entries in thesecond row by their residuesmodulo 26.
We added –17 times the firstrow to the second.
We replaced the entries in thesecond row by their residuesmodulo 26.
We multiplied the second rowby 11-1 = 19 (mod 26).
We replaced –54 and –189 bytheir residues modulo 26.
We added –14 times the secondrow to the first.
To obtain the plaintext pairs, we multiply each ciphertext vector by A–1:
which yields the message
THEY SPLIT THE ATOM
6. Since we want a Hill 3-cipher, we will group the letters in triples. From Table 1 thenumerical equivalent of the known plaintext are
I H A V E C O M E
9 8 1 22 5 3 15 13 5
24 5
19 14
12
14
358
424
=
=220
8
24 5
19 14
7
9
21
=
T
H
33
259
5
25
24 5
19 14
=
E
Y
88
7
227
250
19
16
2
=
=
S
P
44 5
19 14
25
2
610
503
1
=
=22
926
24 5
19 14
22
18
(mod )L
I
=
=
618
670
20
20
24 5
19
T
T
114
5
14
190
291
8
5
=
=
=
H
E
24 5
19 14
10
25
365
540
=
1
20
24 5
19 14
17
15
A
T
=
=
483
533
15
13
O
M
710 Exercise Set 11.16
and the numerical equivalents of the corresponding ciphertext are
H P A F Q G G D U
8 16 1 6 17 7 7 4 21
The corresponding plaintext and ciphertext are
We want to reduce
to I by elementary row operations and simultaneously apply these operations to
The calculations are as follows:
8
6
7
16
17
4
1 9
7 22
21 15
8
5
13
1
3
5
P =
9 8 1
22 5 3
15 13 5
.
C =
8 16 1
6 17 7
7 4 21
pp cc
pp
1 1
2
=
← → =
=
9
8
1
8
16
1
22
5
3
← → =
=
cc
pp
2
3
6
17
7
15
13
5
← → =
cc3
7
4
21
Exercise Set 11.16 711
We formed the matrix .C P
15
6
7
20
17
4
22 24
7 22
21 15
21
5
13
6
3
5
1
6
7
140
17
4
154 168
7
222
21 15
147
5
13
42
3
5
1
6
7
1
00
17
4
24 12
7 22
21 15
17
5
13
16
3
5
−
−
− −
− −
−
−
1
0
0
10
43
66
24 12
137 50
147 69
17
97
1106
16
93
107
1
0
0
10
9
12
24 12
19
−
−
22
9 9
17
7
24
16
11
23
1
0
0
10
11
12
24 12
57 6
9 9
17
21
24
16
33
23
1
0
0
10
1
12
24 12
5 6
9 9
17
21
24
16
7
23
− −
− −
−1
0
0
0
1
0
26 48
5 6
51 63
1993
21
228
54
7
61−
−
−
712 Exercise Set 11.16
We added the third row to thefirst since 8–1 does not existmodulo 26.
We replaced the entries in thefirst row by their residuesmodulo 26.
We replaced the entries in thesecond and third rows by theirresidues modulo 26.
We added –6 times the first rowto the second and –7 times thefirst row to the third.
We replaced the entries in thesecond row by their residuesmodulo 26.
We added –10 times the secondrow to the first and –12 timesthe second row to the third.
We multiplied the first row by15–1 = 7 (mod 26).
We multiplied the second rowby 9–1 = 3 (modulo 26).
Thus,
and so the deciphering matrix is
From Table 1 the numerical equivalent of the given ciphertext is
H P A F Q G G D U G D D
8 16 1 6 17 7 7 4 21 7 4 4
H P G O D Y N O R
8 16 7 15 4 25 14 15 18
A− =
1
4 9 15
15 17 6
24 0 17
.
At−( ) =
1
4 15 24
9 17 0
15 6 17
1
0
0
0
1
0
0 4
5 6
11 15
15
21
6
24
7
17
1
0
0
0
1
00
0 4
0 69
1 15
15
9
6
24
78
17
1
0
0
− − −
0
1
0
0 4
0 9
1 15
15
17
6
24
0
17
Exercise Set 11.16 713
We replaced the entries in thefirst and second row by theirresidues modulo 26.
We replaced the entries in thesecond row by their residuesmodulo 26.
We added –5 times the thirdrow to the second.
To obtain the plaintext triples, we multiply each ciphertext vector by A–1:
Finally, the message is
I HAVE COME TO BURY CAESAR
4 9 15
15 17 6
24 0 17
8
16
1
191
3
= 998
209
9
8
1
4 9 15
15 17 6
2
=
I
H
A
44 0 17
6
17
7
282
421
263
=
=
22
5
3
4 9 15
15 17 6
24 0 17
V
E
C
=
=
7
4
21
379
299
525
155
13
5
4 9 15
15 17 6
24 0 17
7
O
M
E
44
4
124
197
236
20
15
2
=
=
T
O
B
(mod )26
4 9 15
15 17 6
24 0 17
8
166
7
281
434
311
21
18
25
=
=
U
R
Y
4 9 15
15 17 6
24 0 17
15
4
25
=
=
471
443
785
3
1
5
C
AA
E
4 9 15
15 17 6
24 0 17
14
15
18
==
=
461
573
642
19
1
18
S
A
R
714 Exercise Set 11.16
7. Since 29 is a prime number, by Corollary 1 a matrix A with entries in Z29 is invertible if andonly if det(A) ≠ 0 (mod 29).
8. Testing x = 0, 1, 2, … , 25 in 4x = 1 (mod26) gives
4( 0) = 0 ≠ 1
4( 1) = 4 ≠ 1
4( 2) = 8 ≠ 1
4( 3) = 12 ≠ 1
4( 4) = 16 ≠ 1
4( 5) = 20 ≠ 1
4( 6) = 24 ≠ 1
4( 7) = 28 = 2 ≠ 1
4( 8) = 32 = 6 ≠ 1
4( 9) = 36 = 10 ≠ 1
4(10) = 40 = 14 ≠ 1 (mod 26)
4(11) = 44 = 18 ≠ 1
4(12) = 48 = 22 ≠ 1
4(13) = 52 = 0 ≠ 1
4(14) = 56 = 4 ≠ 1
4(15) = 60 = 8 ≠ 1
4(16) = 64 = 12 ≠ 1
4(17) = 68 = 16 ≠ 1
4(18) = 72 = 20 ≠ 1
4(19) = 76 = 24 ≠ 1
4(20) = 80 = 2 ≠ 1
4(21) = 84 = 6 ≠ 1
4(22) = 88 = 10 ≠ 1
4(23) = 92 = 14 ≠ 1
4(24) = 96 = 18 ≠ 1
4(25) = 100 = 22 ≠ 1
Exercise Set 11.16 715
9. (a) We have
where Api
= ci
i = 1, 2, … , n.
Assuming A–1 exists, we can write
pi
= A–1ci
i = 1, 2,…, n
or pit = c
it (A–1)t i = 1, 2,…, n
which in matrix form is
P = C (A–1)t.
(b) From part (a), P = C (A–1)t. Multiply both sides by En… E2E1 from the left to
obtain
En… E2E1P = E
n… E2E1C (A–1)t.
But En…E2E1C = I, so that
En… E2E1P = (A–1)t.
10. (a) We have
where Api
= ci
(mod 26)
or pi
= A–1ci
(mod 26)
or pit = c
it (A–1)t (mod 26), i = 1, 2,…, n
P C
t
t
nt
t
t
nt
=
=
pp
pp
pp
cc
cc
cc
1
2
1
2
P C
t
t
nt
t
t
nt
=
=
pp
pp
pp
cc
cc
cc
1
2
1
2
716 Exercise Set 11.16
which in matrix form is
P = C (A–1)t (mod 26) .
Multiplying both sides by C–1 gives
C–1P = (A–1)t
or
A–1 = (C–1 P)t (mod 26).
(b) From Example 8, we have
Using Eq. (2) to find C–1 we have
and
C P− =
=
1 22 17
25 21
4 5
1 18
1055 416
121 503
1 0
17 9(mod
=
26)) .
C− − −= −
−−
= −
−1 1 118 2852 15
19 9267
2 1( ) ( )
55
19 9
192 15
19 911
2 151
−
=−
−
=
−−
−119 9
22 165
209 99
22 17
25 21
=−
−
=
(mod ),26
P C=
=
4 5
1 18
9 15
19 2, .
Exercise Set 11.16 717
Thus from part (a):
A C Pt− −= =
1 1 1 17
0 9( ) .
718 Exercise Set 11.16
EXERCISE SET 11.17
1. Use induction on n, the case n = 1 being already given. If the result is true for n – 1, thenMn = Mn–1 M = (PDn–1 P–1)(PDP–1) = PDn–1(P–1 P)DP–1 = PDn–1 DP–1 = PDn P–1, provingthe result.
2. Using Table 1 and the notations of Example 1, we derive the following equations:
The transition matrix is thus
The characteristic polynomial of M is det(λI – M) = λ3 – (3/2)λ2 + (1/2)λ = λ(λ – 1)
so the eigenvalues of M are and λ3 = 0. Corresponding λ λ,= =11
22λ ,−
1
2
M =
1
2
1
40
1
2
1
2
1
2
01
4
1
2
.
a a b
b a b c
n n n
n n n
= +
= + +
− −
− −
1
2
1
41
2
1
2
1
2
1 1
1 1 nn
n n nc b c
−
− −= +
1
1 11
4
1
2.
719
eigenvectors (found by solving (λI – M)x = 0) are e1 = [1 2 1]t, e2 = [1 0 –1]t ande3 = [1 –2 1]t. Thus
This yields
Remembering a0 + b0 + c0 = 1, we obtain
Since approaches zero as we obtain as
n → ∞.
anda b cn n n→ → →1
4
1
2
1
4, ,n → ∞.1
2
1
+n
a a b c an
n n
= + + +
−
+ +1
4
1
4
1
4
1
2
1
20 0 0
1
0
1
cc
a c
b a b c
n
n
0
1
0 0
0 0 0
1
4
1
2
1
2
1
2
1
2
= +
−( )
= + + =
+
11
2
1
4
1
4
1
2
1
2
1
20 0 0
1
0c a b c an
n
= + + −
+
+ nn
n
c
a c
+
+
= −
−( )
1
0
1
0 01
4
1
2.
xx( )n
n
n
n
a
b
c
=
=
+
1
4
1
2
nn n
n
+ +
+
−
−
1 1
1
1
4
1
4
1
2
1
2
1
2
1
2
1
4
1
2
1
4
1
44
1
2
1
0
0
0+
+n
a
b
c
.
M PD Pn n= = −
−
−1
1 1 1
2 0 2
1 1 1
1 0 0
01
2
−
−
n
0
1 0 0
1
4
1
4
1
41
20
1
21
4
11
4
1
4
.
720 Exercise Set 11.17
3. Call M1 the matrix of Example 1, and M2 the matrix of Exercise 2. Then x(2n) = (M2M1)n x(0)
and x(2n+1) = M1(M2M1)n x(0). We have
The characteristic polynomial of this matrix is so the eigenvalues are
λ1 = 1, λ3 = 0. Corresponding eigenvectors are e1 = [5 6 1]t, e2 = [–1 0 1]t,
and e3 = [1 –2 1]t. Thus,
Using the notation of Example 1 (recall a0 + b0 + c0 = 1), we obtain
and
a a b c
b a
n n
n n
2 1 0 0 0
2 1 0
2
3
1
6 42 4
1
3
1
6 42
+
+
= +⋅
− +( )
= −⋅
−− +( )=+
b c
c n
0 0
2 1
4
0.
a a b c
b
c
n n
n
n
2 0 0 0
2
2
5
12
1
6 42 4
1
21
12
1
6 4
= +⋅
− +( )
=
= −⋅ nn
a b c2 40 0 0− +( ) .
( )M M PD Pn n
2 11
5 1 1
6 0 2
1 1 1
1
= =
−
−
−
00 0
01
40
0 0 0
1
12
1
12
1
12
−
n
11
3
1
6
2
31
4
1
4
1
4−
λ21
4= ,
λ λ λ3 25
4
1
4− + ,
M M2 1
1
2
1
40
1
2
1
2
1
2
01
4
1
2
11
=
220
01
21
0 0 0
1
2
3
8
1
41
2
1
2
1
2
01
=
88
1
4
.
Exercise Set 11.17 721
4. The characteristic polynomial of M is (λ – 1) so the eigenvalues are λ1 = 1 and
Corresponding eigenvectors are easily found to be e1 = [1 0]t and e2 = [1 –1]t.
From this point, the verification of Eq. (7) is in the text.
5. From Eq. (9), if b0 = .25 = , we get b1 , then b2 = , b3 = ,
and in general bn
= We will reach = .10 in 12 generations. According to
Eq. (8), under the controlled program the percentage would be in 12 generations, or
= .00006 = .006%.
6. P
D P
t
n
−
−
= + −
=
1 0
1 0
1
2
1
20 0
1
2
5
2
1
2
5
2xx
xx
( )
( ) 11
2
1
20 0 2
1 5
22
1 5
2
1 1+
−
+ +n ntt
n
n
PD P−
+
=
− +( ) +
− −(
1 0
11
2
1
23 5
1 5
4
1
23 5
xx( )
)) −
+
+
−
+
+ +
1 5
4
21 5
42
1 5
4
1
1
n
n n 11
1
2
1 5
4
1
2
1 5
4
1
2
1 5
4
+
+
−
+
+
n n
n
11
2
1 5
4
21 5
42
1 5
4
1 1
−
+
+
−
+ +
n
n n
11
2
1
23 5
1 5
4
1
23 5
1 5
4
1
− +( ) +
− −( ) −
+n n++
1
1
16384
1
214
2
20
2
8 + n.
1 5
11 10
2
11=
2 9
10 9
1
5=1 4
9 8
2
9=
1
4
λ21
2= .
λ −
,
1
2
722 Exercise Set 11.17
Since all exponentiated terms above are less than one in absolute value, we have
7. From (13) we have that the probability that the limiting sibling-pairs will be type (A, AA)is
The proportion of A genes in the population at the outset is as follows: all the type (A, AA)genes, 2/3 of the type (A, Aa) genes, 1/3 the type (A, aa) genes, etc. … yielding
8. From an (A, AA) pair we get only (A, AA) pairs and similarly for (a, aa). From either(A, aa) or (a, AA) pairs we must get an Aa female, who will not mature. Thus no offspringwill come from such pairs. The transition matrix is then
9. For the first column of M we realize that parents of type (A, AA) can produce offspring onlyof that type, and similarly for the last column. The fifth column is like the second column,and follows the analysis in the text. For the middle two columns, say the third, note thatmale offspring from (A, aa) must be type a, and females are of type Aa, because of the waythe genes are inherited.
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1
.
a b c d e0 0 0 0 02
3
1
3
2
3+ + + + .
a b c d e0 0 0 0 02
3
1
3
2
3+ + + + .
lim .( )
n
n
t
→∞=
xx1
20 0 0 0
1
2
Exercise Set 11.17 723
EXERCISE SET 11.18
1. (a) The characteristic polynomial of L is λ2 – λ – 3/4, so the eigenvalues are λ1 = 3/2,λ2 = –1/2. The eigenvector corresponding to λ1 is [3 1]t.
(b)
(c)
2. For L in Eq. (4), we obtain
The determinant of this matrix is obtained most easily using cofactors along the first row.Each minor is a lower-triangular matrix. Thus, the characteristic polynomial of L is
(λ – a1)λn–1 – a2b1λn–2 – a3b1b2λ
n–3 – … – anb1b2
… bn–1.
λ
λ
λ
λI L
a a a a a
b
b
n n
− =
− − − ⋅⋅⋅ − −
− ⋅⋅⋅
− ⋅
−1 2 3 1
1
2
0 0 0
0 ⋅⋅ ⋅
⋅ ⋅ ⋅ −
−
0 0
0 0 0 1
bn λ
.
xx xx xx6 51
5857
285
855
287( ) ( ) ( )= =
=
L . λ ..
xx xx xx xx1 0 2 3100
50
175
50( ) ( ) ( ) (= =
=
L , , ))
( ) ( )
=
=
=
250
88
382
125
570
194 5
,
,xx xx11
.
725
3. From the solution in Exercise 2, we see that the derivative of the characteristic polynomialevaluated at λ1 is
nλ1n–1 – (n – 1)a1λn
1–2
– (n – 2)a2b1λ1n–3 – … – a
n–1b1b2… b
n–2.
Recalling that ai
≥ 0 and at least one ai
> 0, and bi
> 0, we see that the above is strictlygreater than
So the derivative is positive and λ1 is a simple root.
4. The product
is [c 0 0 … 0]t, where c is the first element of P–1x(0). But then P[c 0 0 … 0]t returns c times the first column of P which is cx1, x1 being the first eigenvector, i.e., thus the firstcolumn of P.
5. a1 is the average number of offspring produced in the first age period. a2b1 is the numberof offspring produced in the second period times the probability that the female will liveinto the second period, i.e., it is the expected number of offspring per female during thesecond period, and so on for all the periods. Thus, the sum of these, which is the netreproduction rate, is the expected number of offspring produced by a given female duringher expected lifetime.
6. If the population is eventually decreasing, then λ1 < 1 and the characteristic polynomial of L is positive for all λ > λ1, in particular for λ = 1. But the polynomial evaluated at 1
1 0 0 0
0 0 0 0
0 0 0 0
⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅
−
p
11 0xx( )
n a a b a b bn n nn nλ λ λ1
11 1
22 1 1
31 1 2
− − −− −− − − −( ⋅⋅⋅ ⋅ ⋅ ⋅ ))
≥ ≥⋅⋅⋅ −n
a b bn nλ11 1 0.
726 Exercise Set 11.18
(which is 1 minus the net reproduction rate) is positive. Thus the net reproduction rate isless than one. Conversely, if the R < 1 we get that the characteristic polynomial at 1 ispositive; thus λ1 < 1 and the population is eventually decreasing. For R > 1, just switch allinequality symbols above.
7.
8. R = 0 + (.00024)(.99651) + … + (.00240)(.99651) … (.987)
= 1.49611.
9. Let λ be any eigenvalue of L, and write We know λn = rneinθ, so
Since q(λ is real, we can
ignore all the sine terms in the formula.
Thus
i.e., . Since q is a decreasing function, we have r ≤ λ1.q r( ) ≥ 1
1 21 2 12
( ) cos cos= = + + ⋅⋅⋅ +qa
r
a b
r
anλ θ θbb b b
rn
a
r
a b
r
a b b
n
n
1 2 12
1 2 1 1 2
−
≤ + + ⋅⋅⋅ +
cos θ
⋅⋅⋅⋅=−b
rq rn
n
1 ( )
1 1 1 2 1= = − + ⋅⋅⋅+⋅⋅⋅ −q
a
ri
a b b bn n( ) (cos sin )λ θ θrr
n i nn
(cos sin )θ θ−
λ = re r iiθ θ θ= +(cos sin ).
R = +
+
=0 41
23
1
2
1
419 8.
Exercise Set 11.18 727
EXERCISE SET 11.19
1. (a) The characteristic polynomial of L is λ3 – 2λ – 3/8 = (λ – 3/2)[λ2 + (3/2)λ + 1/4], so
λ1 = 3/2. Thus h, the fraction harvested of each age group, is from Eq. (6),
so the yield is of the population. The eigenvector corresponding to λ1 = 3/2 is
[1 1/3 1/18]t; this is the age distribution vector after each harvest.
(b) From Eq. (10), the age distribution vector x1 is [1 1/2 1/8]t. Eq. (9) tells us that h1 = 1 – 1/(19/8) = 11/19, so we harvest 11/19 or 57.9% of the youngest age class.Since Lx1 = [19/8 1/2 1/8]t, the youngest class contains 79.2% of the population.Thus the yield is 57.9% of 79.2%, or 45.8% of the population.
2. The Leslie matrix of Example 1 has b1 = .845, b2 = .975, b3 = .965, etc. This, together withthe harvesting data from Eq. (13) and the formula of Eq. (5) yields
The total of the entries of Lx1 is 7.584. The proportion of sheep harvested is h1(Lx1)2 +h9(Lx1)9 = 1.51, or 19.9% of the population.
3. Using the L of Eq. (3) and the x1 of Eq. (10) we obtain for the first coordinate of Lx1,
a1 + a2b1 + a3b1b2 + … + anb1b2b3
… bn–1.
x
x
1
1
1 845 824 795 755 699 626 532 0 0 0 0= [ ]. . . . . . . .t
L == [ 2 090 845 824 795 755 699 626 532 418 0 0 0. . . . . . . . . ]] t.
33 13
%
12
3
1
3− =
729
The other coordinates of Lx1 are, respectively, b1, b1b2, b1b2b3, … , b1b2… b
n–1. Thus,
4. In this situation we have hI
≠ 0, and h1 = h2 = … = hI–1 = h
I+1 = … = hn
= 0. Eq. (4) thentakes the form
a1 + a2b1 + a3b1b2 + aIb1b2
… bI–1(1 – h
I) + a
I+1b1b2… b
I(1 – h
I)
+ … + anb1b2
… bn–1(1 – h
I) = 1.
Factoring out (1 – hI) we get:
(1 – hI)[a
Ib1b2
… bI–1 + a
I+1b1b2… b
I+ … + a
nb1b2
… bn–1]
= 1 – a1 – a2b1 – … – aI–1b1b2
… bI–2.
L
a a b a b b a b b bn n
xx xx1 1
1 2 1 3 1 2 1 2 1
− =
+ + + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − −−
=
−
1
0
0
0
1
0
0
0
R
.
730 Exercise Set 11.19
So
5. Here hJ
= 1, hI
≠ 0, and all the other hk
are zero. Then Eq. (4) becomes
a1 + a2b1 + … + aI–1b1b2
… bI–2 + (1 – h
I)[a
Ib1b2
… bI–1 + … + a
J–1b1b2… b
J–2] = 1.
We solve for hI
to obtain
ha a b a b b b
a b bI
I I
I
=+ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ −
⋅− −1 2 1 1 1 2 2
1 2
1
(
⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅+
=+ +
− −b a b b b
a a b
I n n1 1 2 1
1 2 1
1
⋅⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − + ⋅ ⋅ ⋅ +− − −a b b b a b b bI I I I1 1 2 2 1 2 11 ⋅⋅ ⋅ ⋅ + ⋅ ⋅ ⋅⋅ ⋅ ⋅ + ⋅ ⋅ ⋅
−
−
a b b b
a b b b
n n
I I
1 2 1
1 2 1 ++ ⋅ ⋅ ⋅
=−
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ +
−
−
a b b b
R
a b b b
n n
I I
1 2 1
1 2 1
1
aa b b bn n1 2 1⋅ ⋅ ⋅ −
Exercise Set 11.19 731
ha a b a b b b
a b bI
I I
I
=+ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ −
⋅− −1 2 1 1 1 2 2
1 2
1
⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅+
=+
− − −b a b b b
a a b
I J J1 1 1 2 1
1 2 1
1
++ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ −⋅ ⋅ ⋅ +
− −
−
a b b b
a b b b
J J
I I
1 1 2 2
1 2 1
1
⋅⋅ ⋅ ⋅ + ⋅ ⋅ ⋅− −.
a b b bJ J1 1 2 1
EXERCISE SET 11.20
1. From Theorem 2, we compute
So the least-squares trigonometric polynomial is
2. From Theorem 3, we compute
and
So the least-squares trigonometric polynominal is
T T t
T
t
T
t
T
2 2
23
2 1
4
4 1
9
6 1
16
8+ + + +
ππ π π
cos cos cos cosππ
ππ π
t
T
T t
T
t
T
− + +sin sin sin2 2 1
2
4 1
3
6ππ πt
T
t
T+
1
4
8sin .
bT
tk t
Tdt
T
kk
Tsin .= = −∫
2 222
0
ππ
aT
tk t
Tdt
T
kk
T= =∫
2 222
2 20cos
ππ
aT
t dt TT
02 2
0
2 2
3= =∫ ,
π 2
34 2
4
93+ + +cos cost t tcos .
4 10
22
0
2
kb t kt dtk, ( ) sin .and = − =∫π
ππ
a t dt a t kt dtk02 2 2
0
2
0
21 2
3
1= − = = −∫π
π ππ
ππ
( ) , ( ) cosππ
∫ =
733
3. As in Exercise 1,
(Note the upper limit on second integral),
So the least-squares trigonometric polynomial is
4. As in Exercise 1,
So the least-square trigonometric polynomial is
2 4
1 3
2
3 5
3
5 7π πcos cos cos
−⋅
+⋅
+⋅
+ ⋅⋅⋅t t t
++− +
cos
( )( ).
nt
n n2 1 2 1
a t kt dtk
k = = −−
= −
∫ sin cos( )
1 1
2
4
4 1
4
0
2
2π π
π
π
(( )( ),
sin sin .
2 1 2 1
1 1
20
0
2
k k
b t kt dtk
− +
= =∫ππ
a t dt0 0
21 1
2
4= =∫π π
πsin .
1 1
2
2
32
2
154
π π π+ − −sin cos cos .t t t
a t kt dt
kk kt
k =
=−
∫1
1 1
1
0
2
π
π
πsin cos
[ sin sinn cos cos ]
[ (
t kt t
k
+
=−
+ −
0
2
1 1
10 1
π
π)) ]
( )
k
k
kk
− −
=−
−
1
2
1
0
2
1
if is odd
if isπ
even.
if
= =∫b kt t dtk
11
20ππ
sin sinkk
k
=
1
if 10 >.
a f t dt t dt0 00
21 1 2= = =∫∫π π π
ππ( ) sin
734 Exercise Set 11.20
5. As in Exercise 2,
So the least-squares trigonometric polynomial is:
if n is even; the last term involves n – 1 if n is odd.
6. (a)
(b)
(c)
7. There are several cases to consider. First, the function 1 is orthogonal to all the othersbecause
cos sin .kt dt kt dt0
2
0
20
π π∫ ∫= =
sin sincos
kt kt dtt
dt= =−
=∫ ∫20
2
0
2 1 2
2
π ππ ..
cos coscos
kt kt dtt
dt= =+
=∫ ∫20
2
0
2 1 2
2
π ππ ..
1 20
2= =∫ dt
ππ .
T T t
T
t
T4
8 1
2
2 1
6
6 1
10
12 2 2 2
cos cos cos− + +π
π π 00 1
2
22
π πt
T n
nt
T( )cos+ ⋅⋅⋅ +
aT
f t dtT
t dtT
T t dtTT T
T
T
0 0 0
2 2 2
2
12
12
= = + − =∫ ∫ ∫( ) ( ) ..
cos ( ) cosaT
tk t
Tdt
TT t
k t
Tk
T
T
T= + −∫
2 2 2 20
12
12
π π∫∫
= − − =
dt
T
k
k
T
k
k4
41 1
0
8
2
2 22
ππ
(( ) )
( )
if is even
22
2 2 2
if is oddk
bT
tk t
Tdt
TTk
= + −
.
cos (π
ttk t
Tdt
T
TT)cos .1
2
12 2
00 ∫∫ =
π
Exercise Set 11.20 735
Then consider cos mt cos nt dt = 0 if m ≠ n. Similarly sin mt sin nt dt = 0
if m ≠ n. Finally cos mt sin nt dt = 0 for all values of m and n.
8. is defined for all in the interval immediately implies
and f is defined in the interval [0, T]. is a continuous function of for
in the interval since there it is the composition of two continuous functions, namely
f and Theorem 2 tells us that the trigonometric polynomial that minimizes
But then by making the change of variable we obtain Theorem 3 immediately.tT
=τ
2π
fT
g dτπ
τ τ2
2
−
( ) has coefficients00
2
0
21
2
1
π
π
πτπ π
τ
∫
∫=
=a fT
kr dr b fT
k kcos and220
2
πτ
π
∫ sin .kr d
g( )τg( )τ τπ
=2Τ
.
[ , ]0 2πτ
τfTτ
π2
0 ≤2
≤ ,Τ Ττ
π
[0, 2 ] because 0π τ π≤ ≤ 2τfTτ
π2
0
2π∫
0
2π∫0
2π∫
736 Exercise Set 11.20
EXERCISE SET 11.21
1. (a) Equation (2) is
and Equation (3) is c1 + c2 + c3 = 1. These equations can be written in combinedmatrix form as
This system has the unique solution c1 = 1/5, c2 = 2/5, and c3 = 2/5. Because thesecoefficients are all nonnegative, it follows that v is a convex combination of the vectorsv1, v2, and v3.
(b) As in part (a) the system for c1, c2 and c3 is
which has the unique solution c1 = 2/5, c2 = 4/5, and c3 = –1/5. Because one of thesecoefficients is negative, it follows that v is not a convex combination of the vectors v1,v2, and v3.
1 3 4
1 5 2
1 1 1
2
4
1
1
2
3
=
c
c
c
1 3 4
1 5 2
1 1 1
3
3
1
1
2
3
=
c
c
c
.
c c c1 2 31
1
3
5
4
2
3
3
+
+
=
737
1. (c) As in part (a) the system for c1, c2 and c3 is
which has the unique solution c1 = 2/5, c2 = 3/5, and c3 = 0. Because these coefficients are all nonnegative, it follows that v is a convex combination of the vectorsv1, v2, and v3.
(d) As in part (a) the system for c1, c2 and c3 is
which has the unique solution c1 = 4/15, c2 = 6/15, and c3 = 5/15. Because thesecoefficients are all nonnegative, it follows that v is a convex combination of the vectorsv1, v2, and v3.
2. For both triangulations the number of triangles, m, is equal to 7; the number of vertexpoints, n, is equal to 7; and the number of boundary vertex points, k, is equal to 5. Equation(7), m = 2n – 2 – k, becomes 7 = 2(7) – 2 – 5, or 7 = 7.
3. Combining everything that is given in the statement of the problem, we obtain:
w = Mv + b = M(c1v1 + c2v2 + c3v3) + (c1 + c2 + c3)b
= c1(Mv1 + b) + c2(Mv2 + b) + c3(Mv3 + b) = c1w1 + c2w2 + c3w3.
4. A triangulation of the points in Figure 11.21.3 in which the points v3, v5, and v6 form thevertices of a single triangle is given in diagram (a), and a triangulation in which the pointsv2, v5, and v7 do not form the vertices of a single triangle is given in diagram (b):
3 2 3
3 2 0
1 1 1
1
0
1
1
2
3
−−
=
c
c
c
3 2 3
3 2 0
1 1 1
0
0
1
1
2
3
−−
=
c
c
c
738 Exercise Set 11.21
5. (a) Let Then the three matrix equations Mvi+ b =
wi, i =1, 2, 3, can be written as the six scalar equations
m11 + m12 + b1 = 4
m21 + m22 + b2 = 3
2m11 + 3m12 + b1 = 9
2m21 + 3m22 + b2 = 5
2m11 + m12 + b1 = 5
2m21 + m22 + b2 = 3
The first, third, and fifth equations can be written in matrix form as
and the second, fourth, and sixth equations as
1 1 1
2 3 1
2 1 1
3
5
3
21
22
2
=
m
m
b
.
1 1 1
2 3 1
2 1 1
4
9
5
11
12
1
=
m
m
b
.
Mm m
m m
b
b=
=
11 12
21 22
1
2and bb .
v1 v2 v1 v2
v7 v7v6 v6
v4 v4
v3 v3
v5 v5
(a) (b)
Exercise Set 11.21 739
740 Exercise Set 11.21
The first system has the solution m11 = 1, m12 = 2, b1 = 1 and the second
system has the solution m21 = 0, m22 = 1, b2 = 2. Thus we obtain
(b) As in part (a), we are led to the following two linear systems:
Solving these two linear systems leads to
(c) As in part (a), we are led to the following two linear systems:
and
Solving these two linear systems leads to
(d) As in part (a), we are led to the following two linear systems:
0 2 1
2 2 1
4 2 1
5 2
711
12
1− −
=m
m
b
22
7 2−
M =
=
−
1 0
0 1
2
3and b .
−
=−2 1 1
3 5 1
1 0 1
221
22
2
m
m
b
22
3−
.
−
=2 1 1
3 5 1
1 0 1
0
511
12
1
m
m
b 33
M =−
=
3 1
1 1
0
1and b .
−
=−2 2 1
0 0 1
2 1 1
8
0
5
11
12
1
m
m
b
−
and
2 2 1
0 0 1
2 1 1
21
22
2
m
m
b
=
1
1
4
.
=
1
2bb .
M =
1 2
0 1and
and
Solving these two linear systems leads to
6. (a) From Section 3.1, a point v lies on the line segment connecting a and b if v =a + t(b – a) for some t with 0 ≤ t ≤ 1. This can be written as v = (1 – t) a + tb orv = c1a + c2b where c1 = 1 – t and c2 = t. We note that c1 and c2 are nonnegativenumbers such that c1 + c2 = 1.
(b) If c1 + c2 = 0, then c1 = c2 = 0 and so c1a + c2b = 0 and we are finished. If c1 + c2 > 0, set
d1 = c1/(c1 + c2) and d2 = c2/(c1 + c2). Then d1 and d2 both lie in the interval [0, 1] and
d1 + d2 = 1. Consequently, the vector w = d1a + d2b lies on the line segment connecting
the vectors a and b. The vector (c1 + c2)w then lies on the line segment connecting the
origin and the vector w since 0 < c1 + c2 ≤ 1. Because this line segment is contained in
the triangle connecting the origin and the tips of the vectors a and b, it follows that the
vector (c1 + c2)w, or c1a + c2b, is also contained in that triangle.
(c) Because c1, c2, and c3 are nonnegative numbers such that c1 + c2 + c3 = 1, it follows
that 0 ≤ c1 + c2 ≤ 1. As per the hint, set a = v1 – v3 and b = v2 – v3. Then from part (b)
the vector w = c1a + c2b lies in the triangle connecting the origin and the tips of the
vectors a and b. The vector w + v3 then lies on the displaced triangle connecting the
vectors v3, a + v3(= v1), and b + v3(= v2). Finally,
w + v3 = c1a + c2b + v3 = c1(v1 – v3) + c2(v2 – v3) + v3
= c1v1 + c2v2 + (1 – c1 – c2)v3 = c1v1 + c2v2 + c3v3.
7. (a) The vertices v1, v2, and v3 of a triangle can be written as the convex combinations
v1 = (1)v1 + (0)v2 + (0)v3, v2 = (0)v1 + (1)v2 + (0)v3, and v3 = (0)v1 + (0)v2 + (1)v3.
In each of these cases, precisely two of the coefficients are zero and one coefficient is
one.
(b) If, for example, v lies on the side of the triangle determined by the vectors v1 and v2then from Exercise 6(a) we must have that v = c1v1 + c2v2 + (0)v3 where c1 + c2 = 1.Thus at least one of the coefficients, in this example c3, must equal zero.
M =
=
−
1 2 1
2 0
1 2
1and bb .
0 2 1
2 2 1
4 2 1
1
321
22
2− −
=−m
m
b −−
9
.
Exercise Set 11.21 741
742 Exercise Set 11.21
(c) From part (b), if at least one of the coefficients in the convex combination is zero,then the vector must lie on one of the sides of the triangle. Consequently, none of thecoefficients can be zero if the vector lies in the interior of the triangle.
8. Consider the vertex v1 of the triangle and its opposite side determined by the vectors v2 andv3. The midpoint vm of this opposite side is (v2 + v3)/2 and the point on the line segmentfrom v1 to vm that is two-thirds of the distance to vm is given by
1
3
2
3
1
3
2
3 2
1
3
1
3
1
31 12 3
1 2vv vv vvvv vv
vv vv vv+ = ++
= + +m 33.