Matrix norms

  • Thread starter meemoe_uk
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  • #1
meemoe_uk
125
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Duh, I can`t 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 ||

Actual answer = 1.618

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 can`t seem to display a matrix nicely on my post either. sos
 
Last edited:

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,967
19
Hrm.

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

Code:
A = /0 1\
    \1 1/
 
  • #3
meemoe_uk
125
0
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 )
= 5 My answer

Actual Answer = 5.2035

I get the eigen system correct, but it's the matrix norm calculation where I go wrong I think.
 

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