Operator norm increases under taking absolute value of all entries of a matrix

Question

Let $\|A\|:=\sup_{\|v\|=1} \|Av\|$ denote the operator norm induced by the Euclidean distance. If $B$ is a matrix such that $B_{ij} = |A_{ij}|,$ show that

$$\|B\|\geq\|A\|.$$

Answer 1

I just figured it out. It is easy to verify that for any $v,$

$$\|Av\|\leq \|Bu\|,$$

where $u_i = |v_i|.$

Answer 2

To spare the next person some work. Let $v$ be some vector and define $u_i=|v_i|$, then we have

\begin{align} \|Av\|^2 = \Big\|\sum_{i=1}^n a_{i,\cdot} v_i \Big\| = \sum_{j=1}^m \underbrace{\Big|\sum_{i=1}^n a_{i,j} v_i \Big|^2}_{\le \Big(\sum_{i=1}^n |a_{i,j}| |v_i| \Big)^2} \le \sum_{j=1}^m \Big(\sum_{i=1}^n b_{i,j} u_i \Big)^2 = \|Bu\|^2 \end{align}

Since $\|u\| =\|v\|$ as we take the absolute value of the entries for the norm anyway, we get the claim $$\|A\|=\sup_{\|v\|=1}\|Av\|\le\sup_{\|u\|=1}\|Bu\|=\|B\|$$