TITLE: Trace inequality involving $A^TA-AA^T$
QUESTION [7 upvotes]: Let $A\in\mathbb{R}^{n\times n}$ and $B=(A^TA-AA^T)/2$. Furthermore, let
$$
f(A)=\frac{1}{n}\text{Tr}(A),\qquad g(A)=\sqrt{f(A^TA)-(f(A))^2}
$$
where $\text{Tr}(A)$ denotes the trace of the matrix $A$.
After numerous experiments I think that the following bound holds:
$$
g(B)\leq (g(A))^2
$$
How to prove/disprove this statements?
Some useful properties that I have found:

$g(A)\geq 0$ for all $A$, follows directly from Cauchy-Schwarz inequality.
$g(sI+A)=g(A)$, for all $s\in\mathbb{R}$.

EDIT: I see now that I must have misread my notes, and in fact, the inequality I was after is $C=(A-A^T)/2$ and
$$
g(C^2)\leq(g(A))^2
$$
Sorry for the inconviencience.

REPLY [2 votes]: Edit. Your new inequality $g(C^2)\le g(A)^2$ is still wrong. For a concrete counterexample to your inequality $g(C^2)\le g(A)^2$, consider a large $n$ and $A=C=\pmatrix{0&-1\\ 1&0}\oplus 0_{(n-2)\times(n-2)}$. In this case, $g(A) = \sqrt{\frac2n} \approx g(C^2) =\sqrt{\frac2n - \frac4{n^2}}\approx0$. Therefore $g(C^2)>g(A)^2$.
To correct the wrong inequality, you may multiply $g(A)^2$ by $\sqrt{n}$:
$$
g(C^2)\le\color{red}{\sqrt{n}}\,g(A)^2.\tag{1}
$$
This is equivalent to
$$
n^3g(C^2)^2\le n^4 g(A)^4.\tag{2}
$$
To prove $(2)$, note that for any matrix $X$, we have
$$
n^2g(X)^2 = \|X\|^2\|I\|^2 - \langle X,I\rangle^2,
$$
where $\langle X,Y\rangle=\operatorname{tr}(XY^T)$ and $\|\cdot\|$ is the Frobenius norm induced by this inner product. In general, in this inner product space, symmetric matrices are orthogonal to skew symmetric matrices. Also, for any skew symmetric $C$, we have $\|C\|^4=\langle C^2,I\rangle^2$. And the Frobenius norm is also known to be submultiplicative. So, if $H$ denotes the symmetric part of $A$, then
\begin{align}
n^4g(A)^4
&=\left(\|A\|^2\|I\|^2 - \langle A,I\rangle^2\right)^2\\
&=\left(\|C\|^2\|I\|^2 + \|H\|^2\|I\|^2 - \langle H,I\rangle^2\right)^2\\
&\ge \left(\|C\|^2\|I\|^2\right)^2\\
&=n^2 \|C\|^4\\
&\ge n^2 \|C^2\|^2\\
&\ge n\left(n\|C^2\|^2 - \|C\|^4\right)\\
&=n\left(\|C^2\|^2\|I\|^2 - \langle C^2,I\rangle^2\right)\\
&=n^3g(C^2)^2.
\end{align}