lecture 7 intersection of hyperplanes and matrix inverse shang-hua teng
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Lecture 7Intersection of Hyperplanes and
Matrix Inverse
Shang-Hua Teng
Elimination Methods for 2 by 2 Linear Systems
• 2 by 2 linear system can be solved by eliminating the first variable from the second equation by subtracting a proper multiple of the first equation and then
• by backward substitution• Sometime, we need to switch the order of the first
and the second equation• Sometime we may not be able to complete the
elimination
Singular Systems versus Non-Singular Systems
• A singular system has no solution or infinitely many solution– Row Picture: two line are parallel or the same– Column Picture: Two column vectors are co-
linear
• A non-singular system has a unique solution– Row Picture: two non-parallel lines– Column Picture: two non-colinear column
vectors
Gaussian Elimination in 3D
• Using the first pivot to eliminate x from the next two equations
10732
8394
2242
zyx
zyx
zyx
Gaussian Elimination in 3D
• Using the second pivot to eliminate y from the third equation
125
4
2242
zy
zy
zyx
Gaussian Elimination in 3D
• Using the second pivot to eliminate y from the third equation
84
4
2242
z
zy
zyx
Now We Have a Triangular System
• From the last equation, we have
84
4
2242
z
zy
zyx
Backward Substitution
• And substitute z to the first two equations
2
4
2242
z
zy
zyx
Backward Substitution
• We can solve y
2
42
2442
z
y
yx
Backward Substitution
• Substitute to the first equation
2
2
2442
z
y
yx
Backward Substitution
• We can solve the first equation
2
2
2482
z
y
x
Backward Substitution
• We can solve the first equation
2
2
1
z
y
x
Generalization
• How to generalize to higher dimensions?
• What is the complexity of the algorithm?
• Answer:
Express Elimination with Matrices
Step 1Build Augmented Matrix
10732
8394
2242
zyx
zyx
zyx
Ax = b
10732
8394
2242
bA[A b]
Pivot 1: The elimination of column 1
1
2
10732
8394
2242
10732
4110
2242
12510
4110
2242
Pivot 2: The elimination of column 2
1
12510
4110
2242
8400
4110
2242
Upper triangular matrix
Backward Substitution 1: from the last column to the first
8400
4110
2242
Upper triangular matrix
2100
4110
2242
2100
2010
2242
2100
2010
6042
2100
2010
2002
2100
2010
1001
Expressing Elimination by Matrix Multiplication
Elementary or Elimination Matrix
• The elementary or elimination matrix
That subtracts a multiple l of row j from row i can be obtained from the identity matrix I by adding (-l) in the i,j position
jiE ,
jiE ,
10
010
001
1,3
l
E
Elementary or Elimination Matrix
3,33,12,32,11,31,1
3,22,21,2
3,12,11,1
3,32,31,3
3,22,21,2
3,12,11,1
3,32,31,3
3,22,21,2
3,12,11,1
1,3
10
010
001
alaalaala
aaa
aaa
aaa
aaa
aaa
laaa
aaa
aaa
E
Pivot 1: The elimination of column 1
12510
4110
2242
1
2
Elimination matrix
10732
8394
2242
10732
4110
2242
10732
8394
2242
100
012
001
12510
4110
2242
10732
4110
2242
101
010
001
The Product of Elimination Matrices
101
012
001
100
012
001
101
010
001
111
012
001
101
012
001
110
010
001
Elimination by Matrix Multiplication
8400
4110
2242
10732
8394
2242
111
012
001
Linear Systems in Higher Dimensions
9
5
2
0
201041
10631
4321
1111
x
Linear Systems in Higher Dimensions
9201041
510631
24321
01111
919930
59520
23210
01111
310300
36300
23210
01111
04000
36300
23210
01111
Linear Systems in Higher Dimensions
04000
36300
23210
01111
01000
36300
23210
01111
01000
30300
20210
00111
01000
10100
20210
00111
01000
10100
00010
10011
01000
10100
00010
10001
310300
36300
23210
01111
919930
59520
23210
01111
1030
0120
0010
0001
Booking with Elimination Matrices
919930
59520
23210
01111
9201041
510631
24321
01111
1001
0101
0011
0001
04000
36300
23210
01111
310300
36300
23210
01111
1100
0100
0010
0001
Multiplying Elimination Matrices
04000
36300
23210
01111
9201041
510631
24321
01111
1131
0121
0011
0001
Inverse Matrices
• In 1 dimension
13333
39393
11
1
xx
Inverse Matrices• In high dimensions
IAAAAA
bAx
bAx
11
1
1
such that? matrix a thereIs
write?Can we
Inverse Matrices• In 1 dimension
0 iff exists existnot does 0
1
1
a a
!!matrices!singular exist?not doesWhen 1A
• In higher dimensions
Some Special Matrices and Their Inverses
nn d
d
d
d
II
/1
/1
1
1
1
1
Inverses in Two Dimensions
ac
bd
bcaddc
ba 11
Ibcad
bcad
bcaddc
ba
ac
bd
bcad
0
011
Ibcad
bcad
bcadac
bd
bcaddc
ba
0
011
Proof:
Uniqueness of Inverse Matrices
CICCBABACACBBIB
CB IACIBA
:Proof
then and
Inverse and Linear System
bAxbAIx
bAAxA
bA
bAxA
1
1
11
1
:Proof
by given solution unique a has
theninvertible is if
Inverse and Linear System
• Therefore, the inverse of A exists if and only if elimination produces n non-zero pivots (row exchanges allowed)
Inverse, Singular Matrix and Degeneracy
Suppose there is a nonzero vector x such that Ax = 0 [column vectors of A co-linear] then A cannot have an inverse
00
:Proof11
xAAxA
Contradiction:So if A is invertible, then Ax =0 can only have the zero
solution x=0
One More Property
Proof
111 ABAB
IBBBAABABAB 11111
So
1111 ABCABC
Gauss-Jordan Elimination for Computing A-1
• 1D1 implies 1 axax
• 2D
10
01then
1
0 and
0
1
22
11
2221
1211
2
1
2221
1211
2
1
2221
1211
yx
yx
aa
aa
y
y
aa
aa
x
x
aa
aa
Gauss-Jordan Elimination for Computing A-1
• 3D
100
010
001then
100
and 010
, 001
333
222
111
333231
232221
131211
3
2
1
333231
232221
131211
3
2
1
333231
232221
131211
3
2
1
333231
232221
131211
zyx
zyx
zyx
aaa
aaa
aaa
zzz
aaa
aaa
aaa
yyy
aaa
aaa
aaa
xxx
aaa
aaa
aaa
Gauss-Jordan Elimination for Computing A-1
• 3D: Solving three linear equations defined by A simultaneously
• n dimensions: Solving n linear equations defined by A simultaneously
11 , AIIAA
Example:Gauss-Jordan Elimination for Computing A-1
100
010
001
210
121
012
X
100
010
001
210
121
012
• Make a Big Augmented Matrix
Example:Gauss-Jordan Elimination for Computing A-1
100
010
001
210
121
012
100
012/1
001
210
12/30
012
13/23/1
012/1
001
3/400
12/30
012
Example:Gauss-Jordan Elimination for Computing A-1
13/23/1
012/1
001
3/400
12/30
012
13/23/1
4/32/34/3
001
3/400
02/30
012
13/23/1
4/32/34/3
2/112/3
3/400
02/30
002
Example:Gauss-Jordan Elimination for Computing A-1
13/23/1
4/32/34/3
2/112/3
3/400
02/30
002
4/32/14/1
2/112/1
4/12/14/3
100
010
002