Trace inequality involving $A^TA-AA^T$

Question

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:

1. $g(A)\geq 0$ for all $A$, follows directly from Cauchy-Schwarz inequality.
2. $g(sI+A)=g(A)$, 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.

Answer 1

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}

Answer 2

$\newcommand{\Tr}{\operatorname{Tr}}$ I believe $A = \begin{pmatrix}0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$ is a counter-example, since in this case we have $\Tr(A) = 0$ and $\Tr(A^TA) = 5$, so that $g(A) = \sqrt{5/4}$. On the other hand, we have $B = \begin{pmatrix}1/2 & 0 & 0 & 0 \\ 0 & 3/2 & 0 & 0\\ 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$, so that $\Tr(B) = 0$ (this, by the way, always holds) and $\Tr(B^TB) = 1/4 + 9/4 + 4 = 13/2$, so that $g(B) = \sqrt{13/8} > 5/4$ (because $(5/4)^2 = 25/16 < 26/16$).

---

Just to give some more intuition on what's happening there, consider some matrix $A$; let $\lambda_1, \ldots, \lambda_n \in \mathbb{C}$ denote the eigenvalues of $A$, and let $\sigma_1, \ldots, \sigma_n \in \mathbb{R}$ denote the singular values of $A$. Since $\Tr(A^TA) = \sum_{i=1}^n \sigma_i^2$, we have $g(A) = \sqrt{\frac{1}{n}\sum_i \sigma_i^2 - \left(\frac{1}{n} \sum_i \lambda_i \right)^2}$.

On the other hand, one can show that $$g(B) = \sqrt{\frac{1}{n}\Tr(B^2)} = \sqrt{\frac{1}{2n}\left(\sum_i \sigma_i^4 - \sum_i \rho_i^2\right)},$$ where $\rho_i$ denote the singular values of $A^2$. There seems to be no obvious relation between the two formulas, and in fact the inequality you proposed does not hold.