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I have a brief question regarding the infinity matrix norm.

The subordinate matrix infinity norm is defined as:

$\|A\|_{\infty} =\max_{1 \leq i \leq n}\sum_{j=1}^{n}|a_{ij}|.$

This is derived from the general definition of a subordinate matrix norm which is defined as:

$\|A\| = \max \left\{\frac{\|Ax\|}{\|x\|} : x \in K^{n}, x \neq 0\right\}.$

I wanted to try this out in an example. So say we define the matrix:

$A = \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 2 \\ 4 & 4 & 3 \end{bmatrix}$

and

$x = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$

Now if we use the first definition, it is easy to see that $\|A\|_{\infty} = 11$

But if we use the general definition, we get:

$\|A\|_{\infty} =\max \left\{\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} : x \in K^{n}, x \neq 0\right\}.$

Now, we have:

$Ax = \begin{bmatrix} 15 \\ 11 \\ 21 \end{bmatrix}.$

Since the infinity vector norm is defined as:

$\|x\|_{\infty} =\max_{1 \leq i \leq n} |x_i|$

it follows that:

$\|Ax\|_{\infty} = 21$

and:

$\|x\|_{\infty} = 3$

But then we have:

$\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} = \frac{21}3 = 7$

that does not correlate with the fact that we previously found that $\|A\|_{\infty} = 11$.

If anyone can explain to me what is wrong with my reasoning here, I would appreciate it!

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    Hm, good point :). Perhaps this is what I've overlooked. But is there a way to find out the vector which will give the supremum value?2012-09-09

1 Answers 1

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The supremum occurs at $x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$

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    Thanks a lot! I know how to use the definitions for the various norms to find $||A||_1, ||A||_{\infty}$, etc. I just wanted to try to relate these to the overarching definition for matrix norms. As I mentioned, this is new territoriy for me, so I really appreciate the help!2012-09-09