I am confused by conflicting answers to this question. In some books, submultiplicativity of matrix norms is defined only for square matrices while other books don't mention such a restriction.

On this website too, I found the following conflicting opinions:

So what should I conclude from all this ? I am confused.

  • 1
    $\begingroup$ For product of non-square matrices you need to specify 3 norms. That's all. $\endgroup$
    – user251257
    Mar 3, 2016 at 5:25

1 Answer 1


Generally, you don't need to restrict yourself to square matrices.

Suppose you have normed vector spaces $V, W$ with vector norms $\|.\|_V$ and $\|.\|_W$ as well as a linear map $\ f\colon V \to W$. Then (assumed that $V \neq \{0\}$) the operator norm with regard to the vector norms is: $$\|f\| = \sup_{x \in V\setminus\{0\}} \frac{\|f(x)\|_W}{\|x\|_V} = \sup_{\|x\|_V=1} \|f(x)\|_W$$

$\|f\|$ depends on the choice of $\|.\|_V$ and $\|.\|_W$!

Now suppose you have three normed vector spaces $V, W, X$ as well as linear maps $f\colon V \to W$ and $g\colon X \to V$. Then the operator norms satisfy submultiplicativity, that is:

$$\|f \circ g\| \leq \|f\|\cdot\|g\|$$

Proof: For all $x \in V$ we obviously have: $$\|f(x)\|_W \leq \|f\|\cdot\|x\|_V$$ Thus $\forall x \in X$ $$\|(f \circ g)(x)\|_W \leq \|f\|\cdot\|g(x)\|_V \leq \|f\|\cdot\|g\|\cdot\|x\|_X$$
So we get $\forall x \in X\setminus\{0\}:$ $$\frac{\|(f \circ g)(x)\|_W}{\|x\|_X} \leq \|f\|\cdot\|g\|$$

Building the supremum provides the desired result.

Now, as each linear map between two finite-dimensional vector spaces can be expressed as a matrix, the above holds for matrix norms and is not restricted to square matrices.
Or, as put shortly in the commments by user251257, you need to specify 3 norms. That's all.


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