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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.