orthogonal sets and basis
TRANSCRIPT
![Page 1: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/1.jpg)
Announcements
Ï Quiz 4 (last quiz of the term) tomorrow on sec 3.3, 5.1 and5.2.
Ï Three problems on tomorrow's quiz(Cramer's rule/adjugate,�nding eigenvector(s) given one or more eigenvalue(s), �ndingchar polynomial/eigenvalues of a 2×2 or a nice 3×3 matrix.)
Ï You must show all relevant work on the quiz. Calculatoranswers are not acceptable.
![Page 2: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/2.jpg)
Chapter 6 Orthogonality
Objectives
1. Extend the idea of simple geometric ideas namely �length�,�distance� and �perpendicularity� from R2 and R3 into Rn
2. Useful in "�tting" experimental data of a system Ax= b. If x1is an acceptable solution, we want the distance between b andAx1 to be minimum (or minimize the error)
3. The solution above is called the "least-squares" solution and iswidely used where experimental data is scattered over a widerange and you want to "�t" a straight line.
![Page 3: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/3.jpg)
Chapter 6 Orthogonality
Objectives
1. Extend the idea of simple geometric ideas namely �length�,�distance� and �perpendicularity� from R2 and R3 into Rn
2. Useful in "�tting" experimental data of a system Ax= b. If x1is an acceptable solution, we want the distance between b andAx1 to be minimum (or minimize the error)
3. The solution above is called the "least-squares" solution and iswidely used where experimental data is scattered over a widerange and you want to "�t" a straight line.
![Page 4: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/4.jpg)
Chapter 6 Orthogonality
Objectives
1. Extend the idea of simple geometric ideas namely �length�,�distance� and �perpendicularity� from R2 and R3 into Rn
2. Useful in "�tting" experimental data of a system Ax= b. If x1is an acceptable solution, we want the distance between b andAx1 to be minimum (or minimize the error)
3. The solution above is called the "least-squares" solution and iswidely used where experimental data is scattered over a widerange and you want to "�t" a straight line.
![Page 5: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/5.jpg)
Inner product
Let u and v be two vectors in Rn.
u=
u1u2...un
,v=
v1v2...vn
Both u and v are n×1 matrices.
uT = [u1 u2 . . . un
]This is an 1×n matrix. Thus we can de�ne the product uTv as
uTv= [u1 u2 . . . un
]
v1v2...vn
![Page 6: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/6.jpg)
Inner product
Let u and v be two vectors in Rn.
u=
u1u2...un
,v=
v1v2...vn
Both u and v are n×1 matrices.
uT = [u1 u2 . . . un
]
This is an 1×n matrix. Thus we can de�ne the product uTv as
uTv= [u1 u2 . . . un
]
v1v2...vn
![Page 7: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/7.jpg)
Inner product
Let u and v be two vectors in Rn.
u=
u1u2...un
,v=
v1v2...vn
Both u and v are n×1 matrices.
uT = [u1 u2 . . . un
]This is an 1×n matrix. Thus we can de�ne the product uTv as
uTv= [u1 u2 . . . un
]
v1v2...vn
![Page 8: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/8.jpg)
1×n n×1
Match
Size of uTv
The product will be a 1×1 matrix or it is just a number (not avector) and is given by
u1v1+u2v2+ . . .+unvn
Nothing but sum of the respective components multiplied.
![Page 9: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/9.jpg)
1×n n×1
Match
Size of uTv
The product will be a 1×1 matrix or it is just a number (not avector) and is given by
u1v1+u2v2+ . . .+unvn
Nothing but sum of the respective components multiplied.
![Page 10: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/10.jpg)
1×n n×1
Match
Size of uTv
The product will be a 1×1 matrix or it is just a number (not avector) and is given by
u1v1+u2v2+ . . .+unvn
Nothing but sum of the respective components multiplied.
![Page 11: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/11.jpg)
Inner Product
1. The number uTv is called the inner product of u and v.
2. Inner product of 2 vectors is a number.
3. Inner product is also called dot product (in Calculus II)
4. Often written as u �v
![Page 12: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/12.jpg)
Example
Let
w= 4
12
,x= 5
0−3
Find w �x, w �w and w�x
w�w
w �x=wTx= [4 1 2
] 50−3
= (4)(5)+ (1)(0)+ (2)(−3)= 14
w �w=wTw= [4 1 2
] 412
= (4)(4)+ (1)(1)+ (2)(2)= 21
w �xw �w
= 14
21= 2
3.
![Page 13: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/13.jpg)
Example
Let
w= 4
12
,x= 5
0−3
Find w �x, w �w and w�x
w�w
w �x=wTx= [4 1 2
] 50−3
= (4)(5)+ (1)(0)+ (2)(−3)= 14
w �w=wTw= [4 1 2
] 412
= (4)(4)+ (1)(1)+ (2)(2)= 21
w �xw �w
= 14
21= 2
3.
![Page 14: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/14.jpg)
Example
Let
w= 4
12
,x= 5
0−3
Find w �x, w �w and w�x
w�w
w �x=wTx= [4 1 2
] 50−3
= (4)(5)+ (1)(0)+ (2)(−3)= 14
w �w=wTw= [4 1 2
] 412
= (4)(4)+ (1)(1)+ (2)(2)= 21
w �xw �w
= 14
21= 2
3.
![Page 15: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/15.jpg)
Example
Let
w= 4
12
,x= 5
0−3
Find w �x, w �w and w�x
w�w
w �x=wTx= [4 1 2
] 50−3
= (4)(5)+ (1)(0)+ (2)(−3)= 14
w �w=wTw= [4 1 2
] 412
= (4)(4)+ (1)(1)+ (2)(2)= 21
w �xw �w
= 14
21= 2
3.
![Page 16: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/16.jpg)
Properties of Inner Product
1. u �v= v �u
2. (u+v) �w= u �w+v �w
3. (cu) �v= u � (cv)
4. u �u≥ 0, and u �u= 0 if and only if u=0
![Page 17: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/17.jpg)
Length of a Vector
Consider any point in R2, v=[a
b
]. What is the length of the line
segment from (0,0) to v?y
x0
|a|
|b|
(a,b)
pa2+b2
![Page 18: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/18.jpg)
Length of a Vector
Consider any point in R2, v=[a
b
]. What is the length of the line
segment from (0,0) to v?y
x0 |a|
|b|
(a,b)
pa2+b2
![Page 19: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/19.jpg)
Length of a Vector
Consider any point in R2, v=[a
b
]. What is the length of the line
segment from (0,0) to v?y
x0 |a|
|b|
(a,b)
pa2+b2
![Page 20: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/20.jpg)
Length of a Vector
Consider any point in R2, v=[a
b
]. What is the length of the line
segment from (0,0) to v?y
x0 |a|
|b|
(a,b)
pa2+b2
![Page 21: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/21.jpg)
Length of a Vector
Consider any point in R2, v=[a
b
]. What is the length of the line
segment from (0,0) to v?y
x0 |a|
|b|
(a,b)
pa2+b2
![Page 22: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/22.jpg)
Length of a Vector
We can extend this idea to Rn, where v=
v1v2...vn
.De�nition
The length (or the norm) of v is the nonnegative scalar ‖v‖ de�nedby
‖v‖ =pv �v=
√v21+v2
2+ . . .+v2n
Since we have sum of squares of the components, the square root isalways de�ned.
![Page 23: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/23.jpg)
Length of a Vector
If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.
De�nition
A vector of length 1 is called a unit vector.
If we divide a vector v by its length ‖v‖ (or multiply by 1
‖v‖), we getthe unit vector u in the direction of v.
The process of getting u from v is called normalizing v.
![Page 24: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/24.jpg)
Length of a Vector
If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.
De�nition
A vector of length 1 is called a unit vector.
If we divide a vector v by its length ‖v‖ (or multiply by 1
‖v‖), we getthe unit vector u in the direction of v.
The process of getting u from v is called normalizing v.
![Page 25: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/25.jpg)
Length of a Vector
If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.
De�nition
A vector of length 1 is called a unit vector.
If we divide a vector v by its length ‖v‖ (or multiply by 1
‖v‖), we getthe unit vector u in the direction of v.
The process of getting u from v is called normalizing v.
![Page 26: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/26.jpg)
Length of a Vector
If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.
De�nition
A vector of length 1 is called a unit vector.
If we divide a vector v by its length ‖v‖ (or multiply by 1
‖v‖), we getthe unit vector u in the direction of v.
The process of getting u from v is called normalizing v.
![Page 27: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/27.jpg)
Example 10, sec 6.1
Find a unit vector in the direction of v= −6
4−3
.
To compute the length of v, �rst �nd
v �v= (−6)2+42+ (−3)2 = 36+16+9= 61
Then,‖v‖ =
p61
The unit vector in the direction of v is
u= 1
‖v‖v=1p61
−64−3
=
−6/p61
4/p61
−3/p61
![Page 28: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/28.jpg)
Example 10, sec 6.1
Find a unit vector in the direction of v= −6
4−3
.
To compute the length of v, �rst �nd
v �v= (−6)2+42+ (−3)2 = 36+16+9= 61
Then,‖v‖ =
p61
The unit vector in the direction of v is
u= 1
‖v‖v=1p61
−64−3
=
−6/p61
4/p61
−3/p61
![Page 29: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/29.jpg)
Example 10, sec 6.1
Find a unit vector in the direction of v= −6
4−3
.
To compute the length of v, �rst �nd
v �v= (−6)2+42+ (−3)2 = 36+16+9= 61
Then,‖v‖ =
p61
The unit vector in the direction of v is
u= 1
‖v‖v=1p61
−64−3
=
−6/p61
4/p61
−3/p61
![Page 30: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/30.jpg)
Distance in Rn
In R (the set of real numbers), the distance between 2 numbers iseasy.
The distance between 4 and 15 is |4−14| = |−10| = 10 or|14−4| = |10| = 10.
Similarly the distance between -5 and 5 is |−5−5| = |−10| = 10 or|5− (−5)| = |10| = 10
Distance has a direct analogue in Rn.
![Page 31: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/31.jpg)
Distance in Rn
In R (the set of real numbers), the distance between 2 numbers iseasy.
The distance between 4 and 15 is |4−14| = |−10| = 10 or|14−4| = |10| = 10.Similarly the distance between -5 and 5 is |−5−5| = |−10| = 10 or|5− (−5)| = |10| = 10
Distance has a direct analogue in Rn.
![Page 32: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/32.jpg)
Distance in Rn
De�nition
For any two vectors u and v in Rn, the distance between u and vwritten as dist(u,v) is the length of the vector u-v.
dist(u,v)= ‖u-v‖
![Page 33: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/33.jpg)
Example 14, sec 6.1
Find the distance between u= 0
−52
and v= −4
−18
.
To compute the distance between u and v, �rst �nd
u−v= 0
−52
− −4
−18
= 4
−4−6
Then,
‖u-v‖ =p16+16+36=
p68
![Page 34: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/34.jpg)
Orthogonal Vectors
0
-v
v
u‖u-v‖
‖u-(-v)‖
If the 2 green lines are perpendicular, u must have the samedistance from v and -v
![Page 35: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/35.jpg)
Orthogonal Vectors
0
-v
v
u
‖u-v‖
‖u-(-v)‖
If the 2 green lines are perpendicular, u must have the samedistance from v and -v
![Page 36: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/36.jpg)
Orthogonal Vectors
0
-v
v
u‖u-v‖
‖u-(-v)‖
If the 2 green lines are perpendicular, u must have the samedistance from v and -v
![Page 37: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/37.jpg)
Orthogonal Vectors
‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares
‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)
= u � (u+v)+v � (u+v)
= u �u+u �v+v �u+v �v
= ‖u‖2+‖v‖2+2u �v
Interchange -v and v and we get
‖u-v‖2 = ‖u‖2+‖v‖2−2u �v
![Page 38: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/38.jpg)
Orthogonal Vectors
‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares
‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)
= u � (u+v)+v � (u+v)
= u �u+u �v+v �u+v �v
= ‖u‖2+‖v‖2+2u �v
Interchange -v and v and we get
‖u-v‖2 = ‖u‖2+‖v‖2−2u �v
![Page 39: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/39.jpg)
Orthogonal Vectors
‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares
‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)
= u � (u+v)+v � (u+v)
= u �u+u �v+v �u+v �v
= ‖u‖2+‖v‖2+2u �v
Interchange -v and v and we get
‖u-v‖2 = ‖u‖2+‖v‖2−2u �v
![Page 40: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/40.jpg)
Orthogonal Vectors
‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares
‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)
= u � (u+v)+v � (u+v)
= u �u+u �v+v �u+v �v
= ‖u‖2+‖v‖2+2u �v
Interchange -v and v and we get
‖u-v‖2 = ‖u‖2+‖v‖2−2u �v
![Page 41: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/41.jpg)
Orthogonal Vectors
‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares
‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)
= u � (u+v)+v � (u+v)
= u �u+u �v+v �u+v �v
= ‖u‖2+‖v‖2+2u �v
Interchange -v and v and we get
‖u-v‖2 = ‖u‖2+‖v‖2−2u �v
![Page 42: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/42.jpg)
Orthogonal Vectors
Equate the 2 expressions,
‖u‖2+‖v‖2+2u �v= ‖u‖2+‖v‖2−2u �v
=⇒ 2u �v=−2u �v
=⇒ u �v= 0
If u and v are points in R2, the lines through these points and (0,0)are perpendicular if and only if
u �v= 0
Generalize this idea of perpendicularity to Rn. We use the wordorthogonality in linear algebra for perpendicularity.
![Page 43: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/43.jpg)
Orthogonal Vectors
Equate the 2 expressions,
‖u‖2+‖v‖2+2u �v= ‖u‖2+‖v‖2−2u �v
=⇒ 2u �v=−2u �v
=⇒ u �v= 0
If u and v are points in R2, the lines through these points and (0,0)are perpendicular if and only if
u �v= 0
Generalize this idea of perpendicularity to Rn. We use the wordorthogonality in linear algebra for perpendicularity.
![Page 44: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/44.jpg)
Orthogonal Vectors
Equate the 2 expressions,
‖u‖2+‖v‖2+2u �v= ‖u‖2+‖v‖2−2u �v
=⇒ 2u �v=−2u �v
=⇒ u �v= 0
If u and v are points in R2, the lines through these points and (0,0)are perpendicular if and only if
u �v= 0
Generalize this idea of perpendicularity to Rn. We use the wordorthogonality in linear algebra for perpendicularity.
![Page 45: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/45.jpg)
Orthogonal Vectors
De�nition
Two vectors u and v in Rn are orthogonal (to each other) if
u �v= 0
The zero vector 0 is orthogonal to every vector in Rn.
Theorem
Two vectors u and v are orthogonal if and only if
‖u+v‖2 = ‖u‖2+‖v‖2
This is called the Pythagorean theorem.
![Page 46: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/46.jpg)
Orthogonal Vectors
De�nition
Two vectors u and v in Rn are orthogonal (to each other) if
u �v= 0
The zero vector 0 is orthogonal to every vector in Rn.
Theorem
Two vectors u and v are orthogonal if and only if
‖u+v‖2 = ‖u‖2+‖v‖2
This is called the Pythagorean theorem.
![Page 47: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/47.jpg)
Example 16, 18 section 6.1
Decide which pair(s) of vectors are orthogonal
16)u= 12
3−5
,v= 2
−33
u �v= (12)(2)+ (3)(−3)+ (−5)(3)= 24−9−15= 0.
Thus u and v are orthogonal.
18)y=
−3740
,z=
1−815−7
y �z= (−3)(1)+(7)(−8)+(4)(15)+(0)(−7)=−3−56+60−0= 1 6= 0.
Thus y and z are not orthogonal.
![Page 48: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/48.jpg)
Example 16, 18 section 6.1
Decide which pair(s) of vectors are orthogonal
16)u= 12
3−5
,v= 2
−33
u �v= (12)(2)+ (3)(−3)+ (−5)(3)= 24−9−15= 0.
Thus u and v are orthogonal.
18)y=
−3740
,z=
1−815−7
y �z= (−3)(1)+(7)(−8)+(4)(15)+(0)(−7)=−3−56+60−0= 1 6= 0.
Thus y and z are not orthogonal.
![Page 49: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/49.jpg)
Orthogonal Complement
Let W be a subspace of Rn. If any vector z is orthogonal to everyvector in W , we say that z is orthogonal to W .
There could be more than one such vector z which is orthogonal toW .
De�nition
A collection of all vectors that are orthogonal to W is called theorthogonal complement of W .
The orthogonal complement of W is denoted by W⊥ and is read as"W perpendicular" or "W perp".
![Page 50: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/50.jpg)
Orthogonal Complement
Let W be a subspace of Rn. If any vector z is orthogonal to everyvector in W , we say that z is orthogonal to W .
There could be more than one such vector z which is orthogonal toW .
De�nition
A collection of all vectors that are orthogonal to W is called theorthogonal complement of W .
The orthogonal complement of W is denoted by W⊥ and is read as"W perpendicular" or "W perp".
![Page 51: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/51.jpg)
Orthogonal Complement
Let W be a subspace of Rn. If any vector z is orthogonal to everyvector in W , we say that z is orthogonal to W .
There could be more than one such vector z which is orthogonal toW .
De�nition
A collection of all vectors that are orthogonal to W is called theorthogonal complement of W .
The orthogonal complement of W is denoted by W⊥ and is read as"W perpendicular" or "W perp".
![Page 52: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/52.jpg)
Orthogonal Complement
1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .
2. W⊥ is a subspace of Rn.
3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)
4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)
![Page 53: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/53.jpg)
Orthogonal Complement
1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .
2. W⊥ is a subspace of Rn.
3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)
4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)
![Page 54: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/54.jpg)
Orthogonal Complement
1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .
2. W⊥ is a subspace of Rn.
3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)
4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)
![Page 55: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/55.jpg)
Orthogonal Complement
1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .
2. W⊥ is a subspace of Rn.
3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)
4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)
![Page 56: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/56.jpg)
Section 6.2 Orthogonal Sets
Consider a set of vectors{u1,u2, . . . ,up
}in Rn. If each pair of
distinct vectors from the set is orthogonal (that is u1 �u2 = 0,u1 �u3 = 0, u2 �u3 = 0 etc etc) then the set is called an orthogonalset.
![Page 57: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/57.jpg)
Example 2 section 6.2
Decide whether the set
1−21
,
012
,
−5−21
is orthogonal.
1−21
�
012
= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0
012
�
−5−21
= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0
1−21
�
−5−21
= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0
Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)
![Page 58: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/58.jpg)
Example 2 section 6.2
Decide whether the set
1−21
,
012
,
−5−21
is orthogonal.
1−21
�
012
= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0
012
�
−5−21
= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0
1−21
�
−5−21
= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0
Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)
![Page 59: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/59.jpg)
Example 2 section 6.2
Decide whether the set
1−21
,
012
,
−5−21
is orthogonal.
1−21
�
012
= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0
012
�
−5−21
= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0
1−21
�
−5−21
= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0
Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)
![Page 60: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/60.jpg)
Example 2 section 6.2
Decide whether the set
1−21
,
012
,
−5−21
is orthogonal.
1−21
�
012
= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0
012
�
−5−21
= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0
1−21
�
−5−21
= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0
Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)
![Page 61: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/61.jpg)
Orthogonal set and Linear Independence
Theorem
Let S = {u1,u2, . . . ,up
}be an orthogonal set of NONZERO vectors
in Rn. S is linearly independent and is a basis for the subspace
spanned (generated) by S.
Make sure that the zero vector is NOT in the set. Otherwise theset is linearly dependent.
Remember the de�nition of basis? For any subspace W of Rn, a setof vectors that
1. spans W and
2. is linearly independent
![Page 62: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/62.jpg)
Orthogonal set and Linear Independence
Theorem
Let S = {u1,u2, . . . ,up
}be an orthogonal set of NONZERO vectors
in Rn. S is linearly independent and is a basis for the subspace
spanned (generated) by S.
Make sure that the zero vector is NOT in the set. Otherwise theset is linearly dependent.
Remember the de�nition of basis? For any subspace W of Rn, a setof vectors that
1. spans W and
2. is linearly independent
![Page 63: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/63.jpg)
Orthogonal set and Linear Independence
Theorem
Let S = {u1,u2, . . . ,up
}be an orthogonal set of NONZERO vectors
in Rn. S is linearly independent and is a basis for the subspace
spanned (generated) by S.
Make sure that the zero vector is NOT in the set. Otherwise theset is linearly dependent.
Remember the de�nition of basis? For any subspace W of Rn, a setof vectors that
1. spans W and
2. is linearly independent
![Page 64: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/64.jpg)
Orthogonal Basis
An orthogonal basis for a subspace W of Rn is a set
1. spans W and
2. is linearly independent and
3. is orthogonal
Theorem
Let{u1,u2, . . . ,up
}be an orthogonal basis for a subspace W of Rn.
For each y in W , the weights in the linear combination
y= c1u1+c2u2+ . . .+cpup
are given by
c1 = y �u1u1 �u1
,c2 = y �u2u2 �u2
,c3 = y �u3u3 �u3
. . .
![Page 65: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/65.jpg)
Orthogonal Basis
An orthogonal basis for a subspace W of Rn is a set
1. spans W and
2. is linearly independent and
3. is orthogonal
Theorem
Let{u1,u2, . . . ,up
}be an orthogonal basis for a subspace W of Rn.
For each y in W , the weights in the linear combination
y= c1u1+c2u2+ . . .+cpup
are given by
c1 = y �u1u1 �u1
,c2 = y �u2u2 �u2
,c3 = y �u3u3 �u3
. . .
![Page 66: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/66.jpg)
Orthogonal Basis
If we have an orthogonal basis
1. Computing the weights in the linear combination becomesmuch easier.
2. No need for augmented matrix/ row reductions.
![Page 67: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/67.jpg)
Example 8, section 6.2Show that {u1,u2} is an orthogonal basis and express x as a linear
combination of the u's where u1 =[31
],u2 =
[ −26
],x=
[ −63
]
Solution: You must verify whether the set is orthogonal.[31
]�[ −2
6
]= (3)(−2)+ (1)(6)= 0
. So we have an orthogonal set. By the theorem, we also have anorthogonal basis.
To �nd the weights so that we can expressx= c1u1+c2u2, we need
x �u1 =[ −6
3
]�[31
]=−18+3=−15
u1 �u1 =[31
]�[31
]= 9+1= 10
![Page 68: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/68.jpg)
Example 8, section 6.2Show that {u1,u2} is an orthogonal basis and express x as a linear
combination of the u's where u1 =[31
],u2 =
[ −26
],x=
[ −63
]
Solution: You must verify whether the set is orthogonal.[31
]�[ −2
6
]= (3)(−2)+ (1)(6)= 0
. So we have an orthogonal set. By the theorem, we also have anorthogonal basis. To �nd the weights so that we can expressx= c1u1+c2u2, we need
x �u1 =[ −6
3
]�[31
]=−18+3=−15
u1 �u1 =[31
]�[31
]= 9+1= 10
![Page 69: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/69.jpg)
Example 8, section 6.2
c1 = x �u1u1 �u1
= −1510
=−1.5
x �u2 =[ −6
3
]�[ −2
6
]= 12+18= 30
u2 �u2 =[ −2
6
]�[ −2
6
]= 4+36= 40
c2 = x �u2u2 �u2
= 30
40= 0.75
Thusx=−1.5u1+0.75u2.
![Page 70: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/70.jpg)
Example 8, section 6.2
c1 = x �u1u1 �u1
= −1510
=−1.5
x �u2 =[ −6
3
]�[ −2
6
]= 12+18= 30
u2 �u2 =[ −2
6
]�[ −2
6
]= 4+36= 40
c2 = x �u2u2 �u2
= 30
40= 0.75
Thusx=−1.5u1+0.75u2.
![Page 71: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/71.jpg)
Example 8, section 6.2
c1 = x �u1u1 �u1
= −1510
=−1.5
x �u2 =[ −6
3
]�[ −2
6
]= 12+18= 30
u2 �u2 =[ −2
6
]�[ −2
6
]= 4+36= 40
c2 = x �u2u2 �u2
= 30
40= 0.75
Thusx=−1.5u1+0.75u2.
![Page 72: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/72.jpg)
Example 10, section 6.2
Show that {u1,u2,u3} is an orthogonal basis for R3 and express x asa linear combination of the u's where
u1 = 3
−30
,u2 = 2
2−1
,u3 = 1
14
,x= 5
−31
Solution: You must verify whether the set is orthogonal (check allpairs). 3
−30
�
114
= (3)(1)+ (−3)(1)+ (0)(4)= 0
. 114
�
22−1
= (1)(2)+ (1)(2)+ (4)(−1)= 0
![Page 73: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/73.jpg)
Example 10, section 6.2
3−30
�
22−1
= (3)(2)+ (−3)(2)+ (0)(4)= 0
. So we have an orthogonal set. By the theorem, we also have anorthogonal basis. To �nd the weights so that we can expressx= c1u1+c2u2+c3u3, we need
x �u1 = 5
−31
�
3−30
= 15+9= 24
u1 �u1 = 3
−30
�
3−30
= 9+9= 18
c1 = x �u1u1 �u1
= 24
18= 4
3
![Page 74: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/74.jpg)
Example 10, section 6.2
3−30
�
22−1
= (3)(2)+ (−3)(2)+ (0)(4)= 0
. So we have an orthogonal set. By the theorem, we also have anorthogonal basis. To �nd the weights so that we can expressx= c1u1+c2u2+c3u3, we need
x �u1 = 5
−31
�
3−30
= 15+9= 24
u1 �u1 = 3
−30
�
3−30
= 9+9= 18
c1 = x �u1u1 �u1
= 24
18= 4
3
![Page 75: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/75.jpg)
Example 10, section 6.2
x �u2 = 5
−31
�
22−1
= 10−6−1= 3
u2 �u2 = 2
2−1
�
22−1
= 4+4+1= 9
c2 = x �u2u2 �u2
= 3
9= 1
3
![Page 76: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/76.jpg)
x �u3 = 5
−31
�
114
= 5−3+4= 6
u3 �u3 = 1
14
�
114
= 1+1+16= 18
c3 = x �u3u3 �u3
= 6
18= 1
3
Thus
x= 4
3u1+ 1
3u2+ 1
3u3.
![Page 77: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/77.jpg)
Section 6.2 Orthonormal Sets
Consider a set of vectors{u1,u2, . . . ,up
}. If this is an orthogonal
set (pairwise dot product =0) AND if each vector is a unit vector(length 1), the set is called an orthonormal set. A basis formed byorthonormal vectors is called an orthonormal basis (linearlyindependent by the same theorem we saw earlier).
![Page 78: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/78.jpg)
Example 20 section 6.2
Decide whether the set u= −2/3
1/32/3
,v= 1/3
2/30
is an
orthonormal set. If only orthogonal, normalize the vectors toproduce an orthonormal set.
−2/31/32/3
�
1/32/30
=−2
3+ 2
3+0= 0
The set is orthogonal. Find the length of each vector to checkwhether it is orthonormal.
‖u‖ =pu �u=
√4
9+ 1
9+ 4
9
‖u‖ =pu �u=
√4
9+ 1
9+ 4
9=
√9
9= 1.
Thus u has unit length.
![Page 79: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/79.jpg)
Example 20 section 6.2
Decide whether the set u= −2/3
1/32/3
,v= 1/3
2/30
is an
orthonormal set. If only orthogonal, normalize the vectors toproduce an orthonormal set. −2/3
1/32/3
�
1/32/30
=−2
3+ 2
3+0= 0
The set is orthogonal. Find the length of each vector to checkwhether it is orthonormal.
‖u‖ =pu �u=
√4
9+ 1
9+ 4
9
‖u‖ =pu �u=
√4
9+ 1
9+ 4
9=
√9
9= 1.
Thus u has unit length.
![Page 80: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/80.jpg)
Example 20 section 6.2
Decide whether the set u= −2/3
1/32/3
,v= 1/3
2/30
is an
orthonormal set. If only orthogonal, normalize the vectors toproduce an orthonormal set. −2/3
1/32/3
�
1/32/30
=−2
3+ 2
3+0= 0
The set is orthogonal. Find the length of each vector to checkwhether it is orthonormal.
‖u‖ =pu �u=
√4
9+ 1
9+ 4
9
‖u‖ =pu �u=
√4
9+ 1
9+ 4
9=
√9
9= 1.
Thus u has unit length.
![Page 81: Orthogonal sets and basis](https://reader034.vdocuments.site/reader034/viewer/2022052619/556289fed8b42ad1688b554d/html5/thumbnails/81.jpg)
Example 20 section 6.2
‖v‖ =pv �v=
√1
9+ 4
9+0=
p5p9=
p5
3.
Since this is not of unit length we have to divide each component
of v by its length which isp5
3. This gives
1
3/p5
3
2
3/p5
3
0
=
1p52p5
0