Inequality for the Schatten norm of symmetric plus antisymmetric matrix

Question

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.

Answer 1

Old thread, but this was never proven, so I give a proof below.

Let $\Sigma_{Y}$ be an $n\times n$ diagonal matrix containing the singular values of $Y$ in the usual ordering of largest to smallest.

claim:
$ \Sigma_{S}\preceq_w \Sigma_{A+S}$
where $\preceq_w$ denotes weak majorization

for $r\in\{1,2...,n\}$
where $Q$ is a real orthogonal matrix and $D_r$ is a diagonal matrix with the last $n-r$ entries all zero and the first $r$ entries $\in \{-1,1\}$
$\sum_{k=1}^r\sigma_{k}^{(S)}$
$= \max_Q \text{trace}\Big(QD_rQ^T S\Big) $
$= \text{trace}\Big(QD_rQ^T S\Big)+0$
$= \text{trace}\Big(QD_rQ^T S\Big)+\text{trace}\Big(QD_rQ^T A\Big)$
$= \text{trace}\Big(QD_rQ^T \big(S+A\big)\Big)$
$\leq \sum_{k=1}^r\sigma_{k}^{(S+A)}$
by the von-Neumann trace Inequality

This proves
$ \Sigma_{S}\preceq_w \Sigma_{A+S}$
and since the Schatten p norm is convex and increasing when applied to $\Sigma_Y$ this gives us
$ \big \Vert S\big \Vert_{S_p}=\big \Vert \Sigma_{S}\big \Vert_{S_p}\leq \big \Vert \Sigma_{A+S}\big \Vert_{S_p}=\big \Vert A+ S\big \Vert_{S_p}$

as for the other inequality: by a nearly identical argument we get
$ \Sigma_{A}\preceq_w \Sigma_{A+S} \implies \big \Vert S\big \Vert_{S_p}\leq \big \Vert A+ S\big \Vert_{S_p}$

Proving the weak majorization goes as follows
with $D_r$ a diagonal matrix with the last $n-r$ entries all zero and the first $r$ entries $\in \{-i,i\}$

$\sum_{k=1}^r\sigma_{k}^{(A)}$
$= \max_Q \text{real}\Big(\text{trace}\Big(QD_rQ^T A\Big)\Big) $
$= \text{trace}\Big(QD_rQ^T A\Big)+0$
$= \text{trace}\Big(QD_rQ^T A\Big)+\text{real}\Big(\text{trace}\Big(QD_rQ^T S\Big)\Big)$
$\leq \Big \vert \text{trace}\Big(QD_rQ^T A\Big)+\text{trace}\Big(QD_rQ^T S\Big)\Big \vert$
$= \Big \vert\text{trace}\Big(QD_rQ^T \big(S+A\big)\Big)\Big \vert$
$\leq \sum_{k=1}^r\sigma_{k}^{(S+A)}$
by the von-Neumann trace Inequality

Answer 2

This is not a complete answer yet, but the first steps that I took to prove the result (which I still believe to be true):

• 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...