Under what conditions is the product of a PSD matrix and another matrix itself PSD?

Question

Suppose I have a symmetric PSD matrix $P$ and another square matrix $A$. What conditions on $A$ are necessary to ensure that $PA$ is symmetric PSD?

Answer 1

Here is a partial answer. In the special case where $A$ is also symmetric, if $PA$ is positive semidefinite, $PA$ must be symmetric. Hence $PA=(PA)^T=AP$, i.e. $P$ commutes with $A$. Since each of them is also orthogonally diagonalisable, they are simultaneously orthogonally diagonalisable. It follows that there exist an orthogonal matrix $Q$ and two diagonal matrices $S$ and $D$ such that $P=QSQ^T,\,A=QDQ^T$ and $SD\succeq0$. This clearly is also a sufficient condition.

In other words, when both $P$ and $A$ are symmetric, $PA$ is positive semidefinite if and only if they share a common orthogonal eigenbasis $\{v_1,v_2,\ldots,v_n\}$ such that $Pv_i=s_iv_i$ and $Av_i=d_iv_i$ with $s_id_i\ge0$ for each $i$.