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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?

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  • $\begingroup$ If we replace every complex number $a+bi$ an $M$ with the real block matrix $\begin{pmatrix} a & -b \\ b & a\end{pmatrix}$ then we get a real matrix $M'$ with the same spectral norm as $M$. We need to show that $\left|M'\right|$ has larger spectral norm than $M$, but that's true for all real matrices, since $\left|M\right| \left|v\right|$ has larger norm than $M v$. $\endgroup$ – Anton Malyshev Apr 7 '14 at 16:57
  • $\begingroup$ @Anton Malyshev: Thanks, but it is not exactly as you said. Using the principle indicated by you, from the $n \times n$ complex matrices $C = A + jB$ and $D = |A| + j|B|$ we construct the $2n \times 2n$ real matrices $C' = \left[ {\begin{array}{*{20}{c}} A & { - B}\\ B&A \end{array}} \right]$ and $D' = \left[ {\begin{array}{*{20}{c}} {|A|}&{ - |B|}\\ {|B|}&{|A|} \end{array}} \right]$. Now the problems comes to proving that $||C'|{|_2} \le ||D'|{|_2}$. Note that $D'$ is not $|C'|$. Any other idea? $\endgroup$ – Hanah Apr 7 '14 at 19:37
  • $\begingroup$ It is clear that $\| A +jB\|_{2}^{2} \leq \sqrt{\|A\|_{2}^{2} + \|B \|_{2}^{2}}.$ $\endgroup$ – Geoff Robinson Apr 10 '14 at 20:06
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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.

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  • $\begingroup$ Oops! Clearly I was misled too, $\endgroup$ – Geoff Robinson Apr 10 '14 at 21:20
  • $\begingroup$ Generally, for a complex matrix $C$, when we write $C=A+jB$, it refers to the Cartesian decomposition of $C$. That is $A, B$ are Hermitian. Then the answer to your question is yes, if $|A|$ is defined as $(A^*A)^{1/2}$. $\endgroup$ – Russel Apr 10 '14 at 23:59

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