# Proof of matrix norm property: submultiplicativity

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}

• There are many different matrix norms - which one are you talking about? – Anthony Carapetis Sep 8 '13 at 23:46
• This is normally not a property of all matrix norms. It just happens to be true, e.g., for operator norms and Frobenius norm. Books where it is included in the definition usually do not bother to go beyond the operator norms. – Algebraic Pavel Sep 9 '13 at 0:22

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.

• Sorry I should include that $A$ and $B$ are $m\times n$ and $m\times l$ respectively, and I thought this only applied to a square matrix? – afrotaint Sep 8 '13 at 23:22
• @afrotaint What is the problem? – Pedro Tamaroff Sep 8 '13 at 23:23
• Sorry, I'm just returning to linear algebra after a few years so I'm trying to reforge my understandings. It's not a problem, it's just this property that I saw in one particular place that differed from all the others I saw, in which the property stated that $A$ and $B$ are $m\times n$ and $m\times l$ respectively, and every other instance showed $n\times n$. I know I'm doing something wrong if it doesn't have a definition/proof, it's probably something simple I should know from the vector norms as you showed, but it's just not clicking. – afrotaint Sep 8 '13 at 23:26
• @afrotaint Could you add the details in your answer? I don't want to confuse you. – Pedro Tamaroff Sep 8 '13 at 23:35
• @PedroTamaroff In your last equation $\lVert ABx\rVert \leqslant \lVert A\rVert \lVert Bx\rVert \leqslant \lVert A\rVert \lVert B\rVert \lVert x\rVert$ I assume that after that you divided everything by $\lVert x\rVert$. However, the $\lVert x\rVert$ in the numerator of $\frac{\lVert A\rVert \lVert B\rVert \lVert x\rVert}{\lVert x\rVert}$ and denomenator are in different normed vector spaces and do not cancel. So I don't get how you can cancel them. – Student Mar 3 '15 at 6:41