Mathematics Asked on December 18, 2021
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?
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 $SDsucceq0$. 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 $s_id_ige0$ for each $i$.
Answered by user1551 on December 18, 2021
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