# An inequality involving the spectral norm of a complex matrix

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?

• 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 $M v$. – Anton Malyshev Apr 7 '14 at 16:57
• @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? – Hanah Apr 7 '14 at 19:37
• It is clear that $\| A +jB\|_{2}^{2} \leq \sqrt{\|A\|_{2}^{2} + \|B \|_{2}^{2}}.$ – Geoff Robinson Apr 10 '14 at 20:06

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 $0 < a < 1$, $||D|{|_2} < ||C|{|_2}$.
• 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 $(A^*A)^{1/2}$. – Russel Apr 10 '14 at 23:59