Download engineering 212 98a

Chapter 11
Matrix Inverse
and Condition
Matrix Inverse
How do we get inverse [A] 1?
One way is through LU decomposition and backsubstitution
Solve [A]{xi} = {bi} with
1 
0 
 
b

 1   ,
0 
0
0 
1 
 
b

 2   ,
0 
0
0 
0 
 
b

 3   ,
1 
0
0 
0 
 
b

 4  
0 
1 
Sequentially!
Put together x’s to get [A]-1 ={x1,x2,x3,x4}
Matrix Inverse
[A] [A] 1 = [ I ]
Use LU factorization to find [X] = [A]1
LU X   I 

 LY   I 
 

 U X   Y 
LU decomposition:
[A] = [L][U]
Forward substitution: [L][Y] = [ I ]
Back substitution:
[U][X] = [Y]
Matrix Inverse
Forward Substitution [L][Y] = [ I ]
 l 11
l
 21
 l 31

 l 41
 l 11
l
 21
 l 31

 l 41
0
l 22
l 32
l 42
0
l 22
l 32
l 42
0   y 11  1
0
0   y 21  0 
    ;
l 33
0   y 31  0 

l 43 l 44   y 41  0 
 l 11
l
 21
 l 31

 l 41
0   y 13  0 
0
0   y 23  0 
    ;
l 33
0   y 33   1
   
l 43 l 44   y 43  0 
 l 11
l
 21
 l 31

 l 41
0
0
0
l 22
l 32
l 42
0
l 22
l 32
l 42
0   y 12  0 
0
0   y 22  1
  
l 33
0   y 32  0 

l 43 l 44   y 42  0 
0
0   y 14  0 
0
0   y 24  0 
  
l 33
0   y 34  0 
   
l 43 l 44   y 44  1
0
Matrix Inverse
back Substitution [U][X] = [ Y ]
 u11 u12 u13
 0 u
u 23
22

 0
0 u 33

0
0
 0
u14   x 11   y 11 
u 24   x 21   y 21 
    ;
u 34   x 31   y 31 

u44   x 41   y 41 
 u11 u12 u13 u14   x 12   y 12 
 0 u
 x   y 
u
u
 22 
22
23
24   22 


   
 0
0 u 33 u 34   x 32   y 32 

   
0
0
0
u
44   x 42 

 y 42 
 u11 u12 u13
 0 u
u 23
22

 0
0 u 33

0
0
 0
u14   x 13   y 13 
u 24   x 23   y 23 
    ;

u 34  x 33   y 33 
   
u44   x 43   y 43 
 u11 u12 u13
 0 u
u 23
22

 0
0 u 33

0
0
 0
 A
1
u14   x 14   y 14 
u 24   x 24   y 24 
  

u 34  x 34   y 34 
   
u44   x 44   y 44 
 x1 , x2  , x3 , x4 
function x
% Function
%
L -->
%
U -->
%
B -->
= LU_Solve_Gen(L, U, B)
to solve the equation L U x = B
Lower triangular matrix (1's on diagonal)
Upper triangular matrix
Right-hand-side matrix
[n n2] = size(L); [m1 m] = size(B);
% Solve L d = B using forward substitution
for j = 1 : m
d(1,j) = B(1,j);
for i = 2 : n
d(i,j) = B(i,j) - L(i, 1:i-1) * d(1:i-1,j);
end
end
% SOlve U x = d using back substitution
for j = 1 : m
x(n,j) = d(n,j) / U(n,n);
for i = n-1 : -1 : 1
x(i,j) = (d(i,j) - U(i,i+1:n) * x(i+1:n,j)) /
U(i,i);
end
end
Example: Matrix Inverse
» A=[-19 20 -6;-12 13 -3;30 -30 12]
» B=eye(3)
A =
B =
-19
-12
30
20
13
-30
-6
-3
12
1
0
0
0
1
0
0
0
1
» [L,U]=LU_factor(A);
» x=LU_solve_gen(L,U,B)
L =
x =
1.0000
0.6316
-1.5789
0
1.0000
4.2857
0
0
1.0000
-10.0000
-8.0000
5.0000
3.0000
2.5000
-1.1667
» inv(A)
U =
-19.0000
0
0
11.0000
9.0000
-5.0000
20.0000
0.3684
0.0000
-6.0000
0.7895
-0.8571
ans =
11.0000
9.0000
-5.0000
MATLAB function
-10.0000
-8.0000
5.0000
3.0000
2.5000
-1.1667
Error Analysis and System Condition
 The matrix inverse provides a means to
discern whether a system is ill-conditioned
1. Normalize rows of matrix [A] to 1. If the
elements of [A]1 are >> 1, the matrix is likely
ill-conditioned
2. If [A]1[A] is not close to [ I ], the matrix is
probably ill-conditioned
3. If ([A]1)1 is not close to [A], the matrix is illconditioned
Vector and Matrix Norms
 Norm: provide a measure of the size or “length” of multicomponent mathematical entities such as vectors and matrices
Euclidean norm of
a vector (3D space)
F   a b c 
F
e
 a2  b2  c 2
 X   x1 x 2 x 3  xn 
X
e

n
2
x
 i
i 1
Matrix Norm
For n  n matrix, the Frobenius norm
Frobenius norm provides a single value to quantify
the size of matrix [A]
Ae 
n
n
2
a
 ij
i 1 j 1
Other alternatives – p norms for vectors

p
   xi 
 i 1

n
X
p
1/ p
p = 2 : Euclidean norm
Vector and Matrix Norms
Vector norms ( p-norms)
p  1,
p  ,
n
X
X
1
  xi
Sum of absolute values
 max x i
Maximum-magnitude or
uniform-vector norm

i 1
1 i  n
Matrix norms
p  1,
p  ,
n
A 1  max  aij
1 j  n
i 1
n
A

 max  aij
1i  n
j 1
Column-sum norm
Row-sum norm
Matrix Condition Number
 Condition number defined in terms of matrix norms
Cond[ A]  A  A
1
1
 Cond[A] is great than or equal to 1
 Matrix A is ill-conditioned if Cond[A] >> 1
X
X
 Cond[ A]
A
A
Relative error of the norm
 If [A] has 107 rounding error and Cond[A] = 105, then the
solution [X] may be valid to only 102
Matrix Condition Evaluation
Hilbert matrix – notoriously ill-conditioned

1
1

2
[ A]   1
3

1

n
1
2
1
3
1
4

1
n1
1
3
1
4
1
5

1
n 2





1 
1

1

n 
2


1
2
 Normalization  1
n1 
3


[ A]  
1 
3
1

n 2 
4







1
n
1


2n  1
n1

1
3
2
4
3
5

n
n 2





Row-sum norm (last row has the largest sum)
A
n
n
n
 1


n1 n 2
2n  1
1 
n 
2 

n1 
3 
n 2 
 
n 

2n  1
Matrix Condition Evaluation
5  5 Hilbert matrix
1
1

[ A]   1

1
 1
1/ 5
2 / 3 2 / 4 2 / 5 2 / 6

3 / 4 3 / 5 3 / 6 3 /7

4 / 5 4 / 6 4 / 7 4 / 8
5 / 6 5 / 7 5 / 8 5 / 9 
1/ 2
1/ 3
1/ 4
[ A]1
25  150
350
 350
126 

  300
2400
 6300
6720  2520 


  1050  9450
26460  29400
11340 



1400
13440

39200
44800

17640


 630  6300
18900  22050
8820 
Condition number
A   1
A 1

5 5 5 5
    3.7282
6 7 8 9
  1400  13440   39200  44800   17640  116480
Cond[ A]  A   A  1

 ( 3.7282 )  ( 116480 )  434260.736
MATLAB Norms and Condition Numbers
>> A=[1 1/2 1/3 1/4 1/5; 1 2/3 2/4 2/5 2/6; 1 3/4 3/5 3/6 3/7;
1 4/5 4/6 4/7 4/8; 1 5/6 5/7 5/8 5/9];
>> Ainv = inv(A)
Ainv =
25
-150
350
-350
126
-300
2400
-6300
6720
-2520
1050
-9450
26460
-29400
11340
-1400
13440
-39200
44800
-17640
630
-6300
18900
-22050
8820
>> norm(A,inf)
ans =
3.7282
>> norm(Ainv,inf)
ans =
1.1648e+005
>> cond(A,inf)
ans =
4.3426e+005
>> norm(X,p)
>> cond(A,'fro')
p-norm
ans =
>> cond(X,p)
2.7944e+005
>> cond(A)
Frobenius norm
>>
Cond(A,’fro’)
ans =
2.7746e+005
Spectral norm
>> cond(A)
CVEN 302-501
Homework No. 7
 Chapter 10
 Prob. 10.4(20), 10.5(20) both using LU decomposition
(hand calculation only, hand partial pivoting if necessary),
Prob. 10.8 a)&b) (20).
 Chapter 11
 Prob. 11.2 (15), 11.3 a) & b) (20) (Hand Calculations)
 Prob. 11.6 (25) (Hand Calculations; Check your solutions
using MATLAB)
 Due Monday 10/13/08 at the beginning of the period