Matrix-Norm inequality

Question

Given a matrix $A \in \mathbb{R}^{n \times n}$ such that $$\sum_{i=1}^n \lvert a_{ij} \rvert \le C $$ $$j=1,2,\dots,n$$ Show that , for every $x\in \mathbb{R}^{n \times n}$ $$\sum_{i=1}^n \lvert (Ax)_i \rvert \le C\lVert x\rVert_1$$ Also find a non-zero vector $x$ for which the equality holds I tried to take the given inequality and multiply with the norm-1 of the vector $x$ but cannot go on since the multiplication of the summads cannot be fused into one.Any help?

Answer 1

Switching the order of summations for the 3rd equality below, we have: $$ \sum_{i=1}^n \lvert (Ax)_i \rvert=\sum_{i=1}^n\sum_{j}|a_{ij}x_j|=\sum_{i=1}^n\sum_j|a_{ij}||x_j|=\sum_j|x_j|\left(\sum_i|a_{ij}|\right)\leq C\sum_j|x_j| $$ which gives you the desired inequality.

A non-zero $x$ to make the inequality an equality does not generally exist. For example, take $n=1$, $A=(1)$, and $C=2$.