# How to prove that $\langle P,A^2 \rangle \le 0$ for every positive $P$ and skew-symmetric $A$?

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).

• After your last sentence, it suffices to observe that every positive definite $P$ has positive diagonal entries (because $e_i^TPe_i > 0$). – Rahul Feb 6 at 8:17
• 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. – Rahul Feb 6 at 8:26
• 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?) – Asaf Shachar Feb 6 at 8:32
• 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. – Rahul Feb 6 at 8:51

## 1 Answer

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, $$\langle P,A^2\rangle=-\text{tr}(A^TPA)\le 0.$$

• 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. – Asaf Shachar Feb 6 at 8:36
• Anyway, there is a suggested proof outlined in the comments above. But I am interested to know if your approach works. – Asaf Shachar Feb 6 at 8:37
• @Asaf: $X^TPX \succeq 0$ for any $X$, because for any vector $y$ we have $y^T(X^TPX)y=(Xy)^TP(Xy) \ge 0$. – Rahul Feb 6 at 8:45
• @Rahul Thanks! I understand now. Thank you for all your comments. – Asaf Shachar Feb 6 at 8:47