Does the concept of submultiplicative norm make sense for non-square matrix? 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:

*

*Show that the operator norm is submultiplicative does not demand restriction to square matrices


*$\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$ and
What are some usual norms for matrices? do and say so too.
So what should I conclude from all this ? I am confused.
 A: 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.
