# Submultiplicative matrix norm: Max Norm

Various sources claim that a maximum norm $||A||_{max}=\max_{i,j}|a_{ij}|$ is not submultiplicative, i.e. $||AB||_{max}\not\leq||A||_{max}||B||_{max}$.

Where can I find what norm a,b satisfy $||AB||_{max}\leq||A||_{a}||B||_{b}$?

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It's not difficult to show directly that the max norm is not submultiplicative: just let $A$ and $B$ be $2 \times 2$ matices with all entries equal to $1$. – Ian Morris May 28 '10 at 10:25
Well, take a look at en.wikipedia.org/wiki/Matrix_norm . – Wadim Zudilin May 28 '10 at 10:38

The inequality $\|AB\|_{\max} \leq \|A\|_{a}\|B\|_{b}$ for all $A$, $B$ can be achieved or destroyed just by rescaling the norms $\|\cdot\|_a$ and $\|\cdot\|_b$. Let's suppose that we're considering $d \times d$ matrices. If we just make sure that the two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ are scaled so that both of them have the property $\|C\|_i \geq \sqrt{d}\|C\|_{max}$ for all $d \times d$ matrices $C$, then the desired inequality follows from the elementary inequality $\|AB\|_{\max} \leq d.\|A\|_{\max}\|B\|_{\max}$. Conversely, if the norms are rescaled so that both of them give norm $\frac{1}{2}$ to the identity matrix, then the inequality clearly cannot hold since $\|Id\|_{max}=1$. The fact that such rescalings exist follows from the fact that norms on a finite-dimensional space are pairwise equivalent.

The point of this is that there are a lot of norms on the space of matrices if we don't make any additional requirements on them. Is this the kind of answer you were looking for? Or do you want the two norms to have additional properties?

edited: there was a typo on the main inequality

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This is what I was looking for, thank you. Furthermore, can you refer me to any book on this? – user6358 May 28 '10 at 13:08
To be honest, I actually can't recommend any books on this! In the absence of any other recommendations, you might try the books referred to in the Wikipedia article on matrix norms linked to by Wadim above. In particular, the book by Horn and Johnson seems to be widely appreciated. – Ian Morris May 28 '10 at 13:26

The Hilbert-Schmidt norm, $||A||_F= (\sum_{j=1}^n a_{i,j}^2)^{1/2}$ is clearly always larger than $||A||_{max}$ and is also submultiplicative.

Hence, $||AB||_{max} \leq ||AB||_F \leq ||A||_F ||B||_F$

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