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I have the following matrix norm:

$$\Vert A \Vert = \max_{1\leq i, j\leq n} \vert a_{ij} \vert \>.$$

I have to decide if this is a subordinate matrix norm or not. I have tried to use the definition of a subordinate matrix norm, but it doesn't lead to anything. So, is there someone who can give a hint?

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A subordinate matrix norm is submultiplicative, namely $\lVert AB\rVert\leq \lVert A\rVert\lVert B\rVert$. What hapens if we take $A=B$, the matrix whose all entries are $1$. – Davide Giraudo Sep 24 '11 at 20:59
Thank you so much. I also have to prove that it defines a norm on the vectorspace of all n x n matrices - My idea is to show ||x||>0, ||cx|| = |c| ||x|| and the triangle inequality. But I'm not sure if this is enough. – user1839 Sep 24 '11 at 22:18
Yes, it's enough, since you check the definition of a norm. – Davide Giraudo Sep 24 '11 at 22:29
up vote 4 down vote accepted

First we show the general following fact:

If the matrix norm $\lVert\cdot\rVert$ is such that $\displaystyle\lVert A\rVert=\sup_{x\in\mathbb R^n,x\neq 0}\frac{N(Ax)}{N(x)}$ for all $A\in\mathcal M_n(\mathbb R)$ (where $N(\cdot)$ is a norm over $\mathbb R^n$, then for all $A,B\in\mathcal M_n(\mathbb R)$ we have $\lVert AB\rVert\leq \lVert A\rVert\lVert B\rVert$.

Indeed, $\displaystyle\sup_{x\in\mathbb R^n,x\neq 0}\frac{N(Ax)}{N(x)}$ is well defined since the map $x\mapsto Ax$ is continuous and the unit sphere of $\mathbb R^n$ is compact, and for any $x\in\mathbb R^n$ we have $N(Ax)\leq \lVert A\rVert N(x)$. Hence for $x\in\mathbb R^n$ we have $N(ABx)=N(A(Bx))\leq \lVert A\rVert N(Bx) \leq \lVert A\rVert \lVert B\rVert N(x)$. We get the result dividing by $N(x)$ for $x\neq 0$ and taking the supremum over $x\in\mathbb R^n\setminus\{0\}$.

Now, for $n\geq 2$, consider the matrix whose all entries are $1$. Then $A^2=nA$ and if $\displaystyle\lVert A\rVert =\sup_{1\leq i,j\leq n}|a_{i,j}|$, we have $\lVert A\rVert= 1$ and $\lVert A^2\rVert= n$. Since $\lVert A^2\rVert =n>\lVert A\rVert^2=1$, $\lVert \cdot\rVert$ cannot be a subordinate norm. (of course, for $n=1$ it's a subordinate norm)

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Well, actually it is well known that $$ \|A\| = \|A\|_{1\rightarrow\infty} = \max_{x\ne 0}\frac{\|Ax\|_{\infty}}{\|x\|_1}\,. $$ Thus, in particular, it is consistent with the corresponding norms: $$ \|Ax\|_{\infty} \le \|A\| \|x\|_1\,. $$

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