# Inequality for the Schatten norm of symmetric plus antisymmetric matrix

I consider $\mathbb{R}^N$ equipped with the standard inner product $\langle\cdot,\cdot\rangle$, which induces the Schatten $p$-norm on the space of linear maps $M: \mathbb{R}^N\to\mathbb{R}^N$ defined by \begin{align} \lVert M\rVert_p=\left(\sum_{i}\sigma_i^p\right)^{1/p}=\left(\mathrm{Tr}\,(M^\intercal M)^{p/2}\right)^{1/p}\,, \end{align} where $\sigma_i$ are the singular values of $M$ given by the eigenvalues of $\sqrt{M^\intercal M}$.

Given two matrices $A$ and $S$ that are antisymmetric and symmetric respectively, namely \begin{align} A=-A^\intercal\,,\qquad S=S^\intercal\,, \end{align} I would like to prove the inequalities \begin{align} \lVert A+S\rVert_p&\geq\lVert A\rVert_p\,,\\ \lVert A+S\rVert_p&\geq\lVert S\rVert_p\,, \end{align} for $p\geq 1$. For $p=2$, this is obviously true because $A$ and $S$ are orthogonal with respect to the standard matrix inner product $\langle M,N\rangle=\mathrm{Tr}\,(M^\intercal N)$, which induces the Frobenius norm $\lVert\cdot\rVert_{p=2}$.

I believe that these inequalities also hold for arbitrary $p\geq 1$ (based on numerical tests and by plotting the unit $p$-sphere for low $N$), but I'm not sure how to prove the statements in general.

An alternative statement could also be to prove in general that for two arbitrary matrices $A,B$ that are orthogonal with respect to the standard matrix inner product satisfy the inequalities.

• I have a hunch that the Hölder-von Neumann inequality $$|\operatorname{Tr}(PQ)| \leq \|P\|_p \|Q\|_q$$ (where $1/p + 1/q = 1$) would be useful here. Just a hunch, though. – Omnomnomnom Mar 20 '18 at 16:26
• I'm aware of the Hölder inequalities, but I really wasn't sure how to use them as they are product inequalities. Also, I don't want to relate specific $p$ and $q$, but prove the statement for all $p\geq 1$. Thank you for the suggestion though, I'll think about it if there is a way to use them! – LFH Mar 20 '18 at 17:38

• It is NOT true that for any two matrices $A,S$ with $\langle A,S\rangle=0$, we have $\lVert A+S\rVert_p\geq \lVert A\rVert_p$. For two random matrices, $A,B$ the unit circle in the plane spanned by the two matrices will look like this (here: $p=1.3$):
• However, if the matrices $A$ and $S$ are anti-symmetric and symmetric respectively, the unit sphere in the plane spanned by the two matrices looks like this:
• Only for $p<1$, the unit sphere may be deformed to not be convex anymore and the inequality won't be satisfied:
Based on this small numerical study (I ran it many times to get different random matrices), I believe that the singular values are symmetric, meaning $\pm A\pm S$ all have the same norm for arbitrary sign choices. I still would like to prove the general inequalities and the numerical results support my belief that the splitting in symmetric and antisymmetric matrices is the correct one...