# Matrix norms

1. Aug 4, 2003

### meemoe_uk

Duh, I cant calculate matrix norms using the formula....

||A|| = max || Ax || where || x || = 1

This is how I try to calculate them, what am I doing wrong?

e.g. Find norm 2 of A

A = 1 1
0 1

First find A's eigen system....
Solve characteristic polynomial....
( 1 - k ) ( 1 - k )
k = 1 - eigen value of A
Get eigen vector....
A - k I = 0
0 1 = 0
0 0 = 0

eigen vector = 1
0

As || Ax || is at a maximum when x is A's eigen vector, we can now calculate ||A||.
Ax = 1 1 * 1 = 1
= 0 1 0 = 0
Therefore
|| A || = || 1 || = 1
|| 0 ||

Bah. I can do it for norm 1 and infinity, but not any number inbetween. I'm not allowed to use that traspose matrix ,spectral radius formula. What's the secret??? Please help.

I cant seem to display a matrix nicely on my post either. sos

Last edited: Aug 4, 2003
2. Aug 4, 2003

### Hurkyl

Staff Emeritus
Hrm.

Are you sure you have the right matrix? Based on the correct answer, my guess is that it's supposed to be

Code (Text):

A = /0 1\
\1 1/

3. Aug 5, 2003

### meemoe_uk

Hi Hurkyl,
The matrix is the one from Burden - Faires Numerical Analysis 4th Edition Ex Set 7.2 Q 1 b)
If I use the spectral radius formula I get the right answer.

Here' another eg. Q 1 d)

A =
2 1 1
2 3 2
1 1 2

Solve characteristic polynomial
- k^3 + 7k^2 - 11k + 5
( k - 1 ) ^2 ( k - 5 )
k = 1 , 5

Get eigenvectors
For k = 1
A - kI = 0 =
1 1 1
2 2 2
1 1 1

solution space vectors =
1
-1
0

1
0
-1

For k = 5
A - kI = 0 =
-3 1 1
2 -2 2
1 1 -3

solution space vector =
1
2
1

|| Ax || is at maximum when x is eigen vector corisponding to largest eigen value so k=5 and
x =
1
2
1
/ Sqr 6 , to nomalize || x || = 1

Calculate Ax

2 1 1 * 1
2 3 2 * 2
1 1 2 * 1 / Sqr 6

=
5
10
5 / Sqr 6

Get Norm...
= Sqr ((25 + 100 + 25) / 6)
= Sqr ( 150 / 6 )