TITLE: An inequality involving the spectral norm of a complex matrix
QUESTION [5 upvotes]: Let $A,B \in {M_n}(R)$ be real $n \times n$ matrices and let matrices $|A|$ and $|B|$ contain the absolute values of the elements of $A$ and $B$ respectively. Construct the complex matrices $C = A + j \cdot B$ and $D = |A| + j \cdot |B|$, where $j$ is the imaginary unit.
For a square complex matrix $M \in {M_n}(C)$ its spectral norm is given by
$||M|{|_2} = \mathop {\max }\limits_{z \ne 0} \frac{{||Mz|{|_2}}}{{||M|{|_2}}} = \sqrt {{\lambda _{\max }}({M^*}M)} = {\sigma _{\max }}(M)$
Through random generation in MATLAB of 100000 matrices of different orders I obtain that the following inequality holds
$||C|{|_2} \le ||D|{|_2}$.
I have tried to prove it but I cannot succeed.
Have you encountered this inequality somewhere or can you prove it?

REPLY [3 votes]: Thank you all for your interest. On ResearchGate Prof. Leonid Gurvits just pointed out an excellent 2 x 2 counterexample to the inequality I intended to prove. 
I was tricked by the fact that the inequality was numerically satisfied for a huge number of 5 x 5, 7 x 7, 10 x 10 randomly generated matrices. I went straight to big dimensions, overlooking the test of 2 x 2 matrices. Running again the MATLAB program for 2 x 2 matrices I noticed that it cannot be true.
The 2 x 2 counterexample is:
$C(1,1) = C(2,2) = 1$ and $C(1,2) = a + j b$, $C(2,1) = a - jb$, where $a^2 + b^2 = 1$; $a,b > 0$.
Therefore
$D(1,1) = D(2,2) = 1$ and $D(1,2) = D(2,1) = a + jb$.
We get $||C|{|_2} = 2$ and $||D|{|_2} = \sqrt{2 + 2 a}$; since $0 < a < 1$, $||D|{|_2} < ||C|{|_2}$.
Sorry for bothering you all.