2
$\begingroup$

I am confused by conflicting answers to the question. Some books seem to define submultiplicativity of matrix norms only for square matrices while some books dont mention such a restriction.

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

This one does not demand restriction to square matrices: show operator norm submultiplicative

This one does, and says so too. $\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$ and What are some usual norms for matrices?

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

$\endgroup$
  • 1
    $\begingroup$ For product of non-square matrices you need to specify 3 norms. That's all. $\endgroup$ – user251257 Mar 3 '16 at 5:25
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.