reduced echelon form
DESCRIPTION
Reduced echelon form. Because the reduced echelon form of A is the identity matrix, we know that the columns of A are a basis for R 2. Return to outline. Matrix equations. Because the reduced echelon form of A is the identity matrix:. Return to outline. Return to outline. - PowerPoint PPT PresentationTRANSCRIPT
Reduced echelon form Reduced echelon form Reduced echelon form
Matrix equations Matrix equations Matrix equations
Null space Null space Null space
Range Range Range
Determinant Determinant Determinant
Invertibility Invertibility Invertibility
Similar matrices Similar matrices Similar matrices
Eigenvalues Eigenvalues Eigenvalues
Eigenvectors Eigenvectors Eigenvectors
Diagonabilty Diagonabilty Diagonabilty
Power Power Power
72
41A
1100
1100
5311
4211
B
142
274
3106
C
72
41A Reduced echelon form
10
012411
10
41
10
4122
150
41
150
411222
72
41
151
rrr
rr
rrr
Because the reduced echelon form of A is the identity matrix,we know that the columns of A are a basis for R2
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72
41A Matrix equations
Because the reduced echelon form of A is the identity matrix:
solutionuniqueahascxA
solutionuniqueahasxA
0
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72
41A
0
0
Aofspacenullthe
solutionuniqueahasxABecause
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72
41A
Every vector in the range of A is of the form:
7
4
2
1
72
41yx
y
x
Is a linear combination of the columns of A.
The columns of A span R2 = the range of A
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72
41A
The determinant of A = (1)(7) – (4)(-2) = 15
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72
41A
Because the determinant of A is NOT ZERO, A is invertible (nonsingular)
151
152
154
157
1
10
01
1072
0141toreduces
AIIA
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72
41A
If A is the matrix for T relative to the standard basis,what is the matrix for T relative to the basis:
?1
2,
1
3
11
23
72
41
11
23
34
051
1 PAPQ
Q is similar to A.Q is the matrix for T relative to the basis, (columns of P)
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72
41A
053158871
72
41det)det(
2
AI
The eigenvalues for A are 3 and 5
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72
41A
1
2
42
42
732
413
3
ofnullspace
ofnullspace
AIofnullspace
1
1
22
44
752
415
5
ofnullspace
ofnullspace
AIofnullspace
1
1,
1
2rseigenvectoofbasisa Return to outline
50
03
72
41tosimilarisA
1
1,
1
2
50
03
rseigenvectoofbasisthe
torelativeTfirmatrixtheis
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11
12
72
41
11
12
50
031
72
41A
A square root of A =
72373
732732
11
12
70
03
11
121
1
10
10
11
12
70
03
11
12
A10 =
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1100
1100
5311
4211
B
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The reduced echelon form of B =
0000
0000
1100
2011
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0
0
0
0
1100
1100
5311
4211
0
z
y
x
w
iffxB
zz
zy
xx
zxw
2
0
0
0
0
0000
0000
1100
2011
1
1
0
2
,
0
0
1
1
nullspacetheforbasisa
4
3
2
1
1100
1100
5311
4211
c
c
c
c
z
y
x
w
cxB
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124
123
12
1
4
3
2
1
0000
0000
1100
4211
1100
1100
5311
4211
ccc
ccc
cc
c
toreduces
c
c
c
c
00 124123
cccandccc
ifonlysolutionahascxB
1100
1100
5311
4211
B
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The range of B is spanned by its columns. Because its null spacehas dimension 2 , we know that its range has dimension 2.(dim domain = dim range + dim null sp).Any two independent columns can serve as a basis for the range.
1
1
3
2
,
0
0
1
1
Bofrangetheforbasisa
0
0000
0000
1100
4211
det
1100
1100
5311
4211
det
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Because the determinant is 0, B has no inverse. ie. B is singular
1100
1100
5311
4211
B
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If P is a 4x4 nonsingular matrix, then B is similar toany matrix of the form P-1 BP
0)2(
110
110
421
det)1(
110
110
531
det)1(
1100
1100
5311
4211
det)det(
3
BI
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The eigenvalues are 0 and 2.
1100
1100
5311
4211
B
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The null space of (0I –B)= the null space of B.
The eigenspace belonging to 0= the null space of the matrix
1
1
0
2
,
0
0
1
1
The null space of (2I –B)=
The eigenspace belonging to 2
0
0
1
1
1100
1100
5311
4211
B
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There are not enough independent eigenvectors to make a basis for R4 .
The characteristic polynomial root 0 is repeated three times, but the eigenspace belonging to 0 is two dimensional.
B is NOT similar to a diagonal matrix.
142
274
3106
C
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The reduced echelon form of C is
000
010
01 21
142
274
3106
C
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becausesolutionnohasxC
1
1
1
1
6
0
010
000
132
1
1
1
142
274
3106
toreduces
toreduces
0
0
0
142
274
3106
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zz
y
zx
0
0
0
0
000
010
01 21
21
A basis for the null space is:
2
0
1
142
274
3106
C
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The columns of the matrix span the range.
The dimension of the null space is 1.
Therefore the dimension of the range is 2.
Choose 2 independent columns of C to form a basis for the range
1
2
3
,
4
7
10
142
274
3106
C
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The determinant of C is 0.
Therefore C has no inverse.
0
142
010
3106
detdet
142
010
3106
3 (-2)row 2 row with 2 row replace
142
274
3106
2
C
142
274
3106
C
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For any nonsingular 3x3 matrix P, C is similar to P-1 CP
0)1)(1(
)13(2)210(4)16)(6(
142
274
3106
det)det(
3
2
CI
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The eigenvalues are: 1, -1, and 0
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042
284
3105
1 CI
Its null space =
0
1
2
242
264
3107
1 CI
Its null space =
1
1
1
142
274
3106
0 CI
Its null space =
2
0
1
000
010
001
142
274
3106
tosimilarisC
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1
1
210
011
112
000
010
001
210
011
112
142
274
3106
PDPC
The columns of P are eigenvectors and the diagonal elements of D are eigenvalues.
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133
133
210
011
112
000
010
001
210
011
112
142
274
3106
PDPC
C