Prove that $\left \| A \right \| = \max_{\left \| x \right \|=1}\left \| Ax \right \|$

Question

I've been given this problem:

Prove that a subordinate matrix norm is a matrix norm, i.e., if $\left \| \cdot \right \|$ is a vector norm on $\mathbb{R}^{n}$, then $$\left \| A \right \| = \max_{\left \| x \right \| = 1} \left \| Ax \right \|$$ is a matrix norm

I don't even understand the question, and a explanation on what the problem is asking me to do would be much appreciated. Specifically, what does $\max\limits_{\left \| x \right \|=1}\left \| Ax \right \|$ mean?

Answer 1

A mapping $\|.\|$ from $\mathbb R^{n\times n}$ to $\mathbb [0,\infty)$ is a norm if, for all matrices $A, B$ and all scalars $\alpha$:

• $\|A\|=0$ if and only if $A=0$
• $\|\alpha A\| = |\alpha|\|A\|$
• $\|A+B\|\leq \|A\|+\|B\|$
• $\|AB\|\leq \|A\|\|B\|$

So the task is asking of you to verify all those properties are true in your case.

Answer 2

A Matrix Norm has to fullfill certain requirements. First, is has to be a norm, so it has to fullfill the three basic norm-Requirements.

1) Positivity

2) Linear wrt. to a constant

3) Triangular inequality

These are quite easy to show here.

Some Sources need a fourth thing, submultiplicativity. It means $\|A\| \cdot \|B\| \geq \|AB\|$ for all $A$, $B$. This, again, is easy to show, when you substitute your definition.