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Let a real (square) matrix $\mathbf A$ is Hurwitz (i.e., all the eigenvalues of $\mathbf A$ have negative real parts). And let $\mathbf P$ is a real symmetric positive definite matrix. What will be the condition(s) (if at all) for the product $\mathbf {PA}$ to be a skew-symmetric matrix?

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There is no such pair $P,A$. We note the following: $$ (PA)^* = -PA \implies\\ A^*P = -PA \implies\\ P^{-1}A^*P = -A $$ So, in order for $PA$ to be skew-symmetric, $-A$ needs to be similar to $A^*$. This is impossible if $A$ is Hurwitz since the eigenvalues of $A^*$ have negative real parts whereas the eigenvalues of $-A$ have positive real parts.

In fact, the above precludes the existence of any satisfactory (invertible) symmetric/self-adjoint matrix $P$ given a Hurwitz $A$. A more sophisticated version precludes even singular, non-zero $P$ that are self-adjoint.

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