How to prove that the matrix norm $||A||_{1,2} = \max_{j} \, \left( \sum_{i=1}^p a_{i,j}^2 \right)^{\frac{1}{2}}$?

Question

$A \in \mathbb{C}^{p\times q}$,

$||A||_{1,2} = \max_{x \in \mathbb{C}^q } \big\{ ||Ax||_2 \; \big| \; ||x||_1 = 1 \big\}$.

I tried using Holder's Inequality and extremum of quadratic function but did not solve it.

Answer 1

Let $a_1,\dots,a_q$ denote the columns of the matrix $A$, and take $x = (x_1,\dots,x_q) \in \Bbb C^n$. Let $m = \max_j \|a_j\|_2$. We find that \begin{align} \|Ax\|_2 &= \|x_1 a_1 + \cdots + x_n a_n\| \\ & \leq |x_1| \cdot \|a_1\| + \cdots + |x_n|\cdot \|a_n\| \\ & \leq |x_1| \cdot m + \cdots + |x_n| \cdot m = m\|x\|_1. \end{align} With this inequality, we deduce that $\|A\|_{1,2} \leq m$. On the other hand, if $j$ is such that $\|a_j\|_2 = m$, then if we take $x$ to be the vector with a $1$ as its $j$th and $0$'s elsewhere, then $\|Ax\|_2 = \|a_j\|_2 = m$. Thus, we may deduce that $\|A\|_{1,2} \geq m$. The conclusion follows.

A similar process can be used to derive a formula for $\|A\|_{1,p}$ for arbitrary values of $p \in [1,\infty]$.