MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## 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 $||AB||_{max}\leq||A||_{a}||B||_{b}$?

-
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 2010 at 10:25
Well, take a look at en.wikipedia.org/wiki/Matrix_norm . – Wadim Zudilin May 28 2010 at 10:38

The inequality $\|A\|_{\max} \leq \|A\|_{a}\|B\|_{b}$ 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 $\|A\|_{\max} \leq d.\|A\|_{\max}\|B\|_{\max}$. 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.