How to prove $\||A|\|_2 \le \sqrt{n} \|A \|_2$?

Question

How can we prove $$ \||A|\|_2 \le \sqrt{n} \|A \|_2$$ for any $A \in {\bf C}^{m\times n}$ with $m\ge n$ ?

Answer 1

\begin{align} \||A|\|_2 &= \sup_{\|x\|_2 \le 1} \||A|x\|_2 \quad \mbox{(by definition of $\|\cdot\|_2$ for matrices)} \\ &= \sup_{\|x\|_2 \le 1} \sqrt{\sum_{i=1}^m ||A_i|^Tx |^2} \quad \mbox{($A_i^T$ is $i$-th row of $A$)} \\ &\le \sup_{\|x\|_2 \le 1} \sqrt{\sum_{i=1}^m \|A_i\|_2^2\|x\|_2^2} \quad \mbox{(by Cauchy-Schwarz inequality)} \\ &\le \sqrt{\sum_{i=1}^m \|A_i\|_2^2} \\ &= \sqrt{\sum_{i=1}^m\sum_{j=1}^n |A_{ij}|^2} \\ &= \sqrt{\mbox{trace}(A^*A)} \\ &= \sqrt{\sum_{i=1}^r \sigma_i^2} \quad \mbox{(by properties of trace, $\sigma_i$ is singular value of $A$)} \\ &\le \sqrt{r} \, \sigma_{\max} \quad \mbox{($\sigma_{\max}$ is maximum singular value of $A$)} \\ &= \sqrt{r} \, \|A\|_2, \quad \mbox{(follows from definition of $\sigma_{\max}$ and $\|\cdot\|_2$)} \end{align} where $r$ is the rank of $A$, which in this case satisfies $r \le n$.