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?

  • 1
    $\begingroup$ 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 $1$. $\endgroup$ Sep 24, 2011 at 20:59
  • $\begingroup$ 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. $\endgroup$
    – user1839
    Sep 24, 2011 at 22:18
  • $\begingroup$ Yes, it's enough, since you check the definition of a norm. $\endgroup$ Sep 24, 2011 at 22:29

2 Answers 2


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)


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: $$ \|Ax\|_{\infty} \le \|A\| \|x\|_1\,. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.