TITLE: Proof of matrix norm property: submultiplicativity
QUESTION [12 upvotes]: I've been searching for the definition of the submultiplicative (I think it has multiple names from what I've seen) property in proof form. Some books define it as part of the properties that define matrix norms, and some include it as an additional property. I still haven't been able to work it out for myself or find it anywhere.
Let $A$ and $B$ be $n\times m$ and $m\times l$ matrices respectively, prove that:
$$\begin{align}
\|AB\| \le \|A\|\|B\| 
\end{align}$$

REPLY [11 votes]: If you have a norm on $K^n$ and if you define a matrix norm (the induced matrix norm) 
$$\lVert A\rVert=\sup\limits_{\lVert x\rVert =1}\{\lVert Ax\rVert :x\in K^n\}$$
then one can readily check $$\lVert Ax\rVert \leqslant \lVert A\rVert \lVert x\rVert $$
There is no problem that $A,B$ aren't square, as long as $AB$ makes sense. In such a case
$$\lVert ABx\rVert \leqslant \lVert A\rVert \lVert Bx\rVert \leqslant \lVert A\rVert \lVert B\rVert  \lVert x\rVert $$
You can check Rudin's Princples Chapter 9, which has this on the very first few pages.