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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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  • $\begingroup$ How do you know that the supremum is reached at this vector? $\endgroup$ Sep 9, 2012 at 19:21
  • $\begingroup$ 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? $\endgroup$
    – Kristian
    Sep 9, 2012 at 19:23

2 Answers 2

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

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    $\begingroup$ Thanks a lot! But how do you know this? I must admit that this theory is very new to me, so it hasn't quite sunk in yet. $\endgroup$
    – Kristian
    Sep 9, 2012 at 19:24
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    $\begingroup$ Because I went through the proof, decades ago. In order to have the maximum row sum happen, all the entries in $x$ are $\pm 1.$ The $\pm$ signs are chosen to agree with the $\pm$ signs in the best row, so as to get all absolute value. If you negated the middle column in your $A,$ you would also negate the middle entry in $x.$ Note that we are always forcing $\parallel x \parallel_\infty = 1.$ $\endgroup$
    – Will Jagy
    Sep 9, 2012 at 19:30
  • $\begingroup$ Thanks a lot! I will look for the proof, so I also can see this! Appreciate it a lot! $\endgroup$
    – Kristian
    Sep 9, 2012 at 19:35
  • $\begingroup$ @Kristian, note that for the $1$-norm, the matrix norm becomes the best column sum of absolute values, and this is achieved with an $x$ that has $n-1$ entries $0,$ and a single entry $1$ corresponding to the best column. See en.wikipedia.org/wiki/Matrix_norm#Induced_norm for part of this. $\endgroup$
    – Will Jagy
    Sep 9, 2012 at 19:55
  • $\begingroup$ 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! $\endgroup$
    – Kristian
    Sep 9, 2012 at 20:09
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It's wrong that you asume that $\|Ax\|_\infty=21$ because you need to make thee product of $Ax= ( 15, 22, 33)$ Then $\|Ax\|_\infty=33$ After that we have $\|x\|_\infty=3$

Finally you have $$\|A\|_\infty=\frac{33}3=11$$

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