Given a matrix $A$ yielding a linear operator $L$, how can I express the operator norm in terms of $A$'s norm? As far as I've understood, the operator norm in terms of a matrix is given by
$$
\|L\|_{op} = \sup_{x\neq0}\frac{\|Ax\|_{\infty}}{\|x\|_\infty}
$$
This should mean that
$$
\sup_{x\neq0} \frac{\|Ax\|_{\infty}}{\|x\|_\infty}
= \sup_{x\neq0} \left\|\frac{Ax_{\infty}}{\|x\|_\infty}\right\|_\infty
= \sup_{x\neq0} \left\|A\left(\frac{x_{\infty}}{\|x\|_\infty}\right)\right\|_\infty
$$
However, is there a way to simplify this further, removing any reference to $x$? As I see it, I should be able to reduce $\frac{x_{\infty}}{\|x\|_\infty} \to 1$ somehow. But I don't see quite how?
Best Answer
I wouldn't call your norm "the" operator norm: the operator norm given by the 2-norm is way more natural than the one you are using.
For your norm $$\|A\|_{\rm op}=\sup\left\{\|Ax\|_\infty:\ \|x\|_\infty=1\right\},$$ it is very easy to see that $$ \|A\|_{\rm op}=\max\left\{\sum_{j=1}^n|A_{kj}|:\ k=1,\ldots,n\right\}. $$ So $\|A\|_{\rm op}$ is the maximum of the 1-norms of the rows of $A$.
For what is very commonly named as "the" operator norm, that is $$ \|A\|=\sup\left\{\|Ax\|_2:\ \|x\|_2=1\right\}, $$ one can show (not that easily) that $$ \|A\|=\max\{\lambda^{1/2}:\ \lambda\in\sigma(A^*A)\}, $$ where $\sigma(A^*A)$ denotes the spectrum of $A^*A$, i.e., its list of eigenvalues.