Can someone please explain what does this Matrix operator norm means? I'm trying to understand what does this matrix operator norm means and what it does to matrix A.  $${{\left\| A \right\|}_{1,\,\infty }}:={{\max }_{{{\left\| x \right\|}_{\infty }}=1}}{{\left\| Ax \right\|}_{1}}$$  Can somebody help with the explanation and maybe an example?
 A: It is $$\max_{||x||_\infty =1} ||Ax||_1= \max \{ ||Ax||_1 : ||x||_\infty =1\}.$$
We take the max first norm of $Ax$, where $x$ has length $1$ in infinity norm.
A: This definition may seem strange on its own, but it's natural in functional analysis.  If $V$ is a normed vector space (a vector space over $\mathbb R$ or $\mathbb C$ with a norm $|| -||$ satisfying:

*

*$||v|| = 0$ if and only if $v = 0$


*$||\lambda v|| = |\lambda| \cdot ||v||$ for all scalars $\lambda$ and vectors $v \in V$


*$||v+w|| \leq ||v|| + ||w||$ for all $v$ and $w$)
Then, if $W$ is another normed space, one can talk about whether a linear map $T: V \rightarrow W$ is continuous with respect to the metrics on $V$ on $W$ coming from this norm.  It turns out that $T$ is continuous if and only if $$\{ ||T(v)||_W : v \in V, ||v||_V = 1\}$$ is a bounded set in $\mathbb R$.  If it is, then we define the norm $||T||$ of $T$ to be the least such upper bound:
$$||T|| = \sup\limits_{||v||_V=1}||T(v)||_W .$$
In your case, $V = \mathbb R^n$ or $\mathbb C^n$ with the infinity norm $||(x_1, ... , x_n)||_{\infty} = \max_{1 \leq i \leq n} |x_i|$, and $W$ is the same space with the $1$-norm $||(x_1, ... , x_n)||_1 = |x_1| + \cdots + |x_n|$.  Any matrix $A$ identifies with a linear operator $V \rightarrow W$, and its norm is
$$||A|| = \sup\limits_{||v||_{\infty}=1} ||Av||_1$$
