I have stumbled upon the following claim, and I wonder if it has a simple proof:

Let $P$ be a real $n \times n$ symmetric positive definite matrix. Then for every real skew-symmetric matrix $A$, $\langle P,A^2 \rangle \le 0 $.

($\langle \,, \rangle$ is the standard Frobenius inner product of matrices).

Here are some partial observations:

If $P$ and $A^2$ are commuting, then we can orthogonally diagonalize them simultaneously. Since the inner product is orthogonally-invariant, this reduces the problem to the case where $P$ and $A^2$ are both diagonal. In that case, all the (diagonal) elements of $P$ are positive, and those of $A^2$ are non-positive, since they are squares of the eigenvalues of $A$, which must be imaginary, since $A$ is skew-symmetric.

I am not sure what to do in the general case, where $A^2,P$ are not commuting:

In that case, we can orthogonally diagonalize either of them, but I don't see how to continue from that point. For instance, let us diagonalize $P$: Write $P=V \Sigma V^T$; then $$ \langle P,A^2 \rangle = \langle V \Sigma V^T,A^2 \rangle=\langle \Sigma ,V^TA^2V \rangle=\langle \Sigma ,(V^TAV)^2 \rangle. $$ Since $V^TAV$ is also skew-symmetric, we have reduced the problem to the case where $P=\Sigma$ is diagonal, and $A$ is an arbitrary skew-symmetric matrix.

If we start by diagonalizing $A^2$ instead, we reduce the problem to arbitrary symmetric positive-definite $P$ and diagonal $A^2$ (with non-positive values).

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    $\begingroup$ After your last sentence, it suffices to observe that every positive definite $P$ has positive diagonal entries (because $e_i^TPe_i > 0$). $\endgroup$ – Rahul Feb 6 at 8:17
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    $\begingroup$ That said, the way I would go about proving it instead is by noting that $A^2=-A^TA$ is negative semidefinite, and using (or proving) the fact that the inner product of two positive semidefinite matrices is nonnegative. $\endgroup$ – Rahul Feb 6 at 8:26
  • $\begingroup$ Thanks! Regarding your second comment, I guess that you meant to say that the inner product of a negative semidefinite and a positive semidefinite is nonnegative. (And this you would prove along the lines of your first comment I guess: Diagonalize one of them, and then use what you know on the sign of the diagonal elements of the other one, right?) $\endgroup$ – Asaf Shachar Feb 6 at 8:32
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    $\begingroup$ I meant what I wrote (the product in your comment would be nonpositive), but I wrote it confusingly. What I was going for was to state the fact in a general and broadly useful form ($\langle S,T\rangle\ge0$ for all $S,T\succeq 0$), but perhaps forcing it into that shape did more harm than good. And yes, I believe diagonalization would be the easy proof, as you suggest. $\endgroup$ – Rahul Feb 6 at 8:51

Notice that $$ \langle P,A^2\rangle =\text{tr}(PA^2)=\text{tr}(PA\cdot A)=\text{tr}(A\cdot PA)=-\text{tr}(A^TPA). $$ Since $-A^TPA\le O$, all its eigenvalues are non-positive. Hence, $$ \langle P,A^2\rangle=-\text{tr}(A^TPA)\le 0. $$

  • $\begingroup$ Thanks. Can you say why $A^TPA$ is positive semidefinite? I see why it is symmetric. But I don't see why it needs to have non-negative eigenvalues; the eigenvalues of $P$ are non-negative, but it is not clear to me that $A^TPA$ and $P$ are similar. $\endgroup$ – Asaf Shachar Feb 6 at 8:36
  • $\begingroup$ Anyway, there is a suggested proof outlined in the comments above. But I am interested to know if your approach works. $\endgroup$ – Asaf Shachar Feb 6 at 8:37
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    $\begingroup$ @Asaf: $X^TPX \succeq 0$ for any $X$, because for any vector $y$ we have $y^T(X^TPX)y=(Xy)^TP(Xy) \ge 0$. $\endgroup$ – Rahul Feb 6 at 8:45
  • $\begingroup$ @Rahul Thanks! I understand now. Thank you for all your comments. $\endgroup$ – Asaf Shachar Feb 6 at 8:47

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