show $ \|\,|A|\,\|_2 \le \sqrt n\|A\|_2 $ for $ m \ge n $

Question

how can I go about showing that $ \||A|\|_2 \le \sqrt n\|A\|_2 $ when $m$ (rows)$ \ge n $ (columns)

$|A| $ denotes the absolute value of $ |A_{ij}|$ for all $i,j$.

The only thing I can think of is using the matrix norm induced by vector norm definition but apart from that I'm pretty clueless on how to proceed/complete the proof. I get:

$\|\,|A|\,\|_2= \sup \dfrac{\|\,|A|\,x\|_2}{\|x\|_2} $ for $ x \neq0 $

Answer 1

From the Cauchy-Schwarz inequality you get $$ \|\,|A|\,\|_2\le \|A\|_F:=\sqrt{\sum_{ij}|A_{ij}|^2} $$ as $$ \|\,|A|\,x\|_2\le\sum_{j=1}^n\|\,|A|\,e_i\|_2·|x_i| \le\sqrt{\sum_{j=1}^n\|\,|A|\,e_i\|_2^{\;2}}·\|x\|_2=\|A\|_F·\|x\|_2 $$ Then the classical norm equivalence estimate is $$ \|A\|_F^2=\sum_{i=1}^n\|Ae_i\|_2^{\;2} \le\sum_{i=1}^n\left(\|A\|_2·\|e_i\|_2\right)^2 = n\|A\|_2^{\,2}. $$

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One can also cut out the middle man Frobenius and combine both steps as \begin{align} \|\,|A|\,x\|_2 &\le\sum_{j=1}^n\|\,|A|\,e_i\|·|x_i| =\sum_{j=1}^n\|A\,e_i\|_2·|x_i| \\&\le\sum_{j=1}^n\|A\|_2\|e_i\|_2·|x_i| =\|A\|_2\|x\|_1\le\sqrt{n}\|A\|_2\|x\|_2 \end{align}