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We are given real matrices $S$ and $A$. We know that $S$ is symmetric positive definite and that $SA$ is symmetric. Is A necessarily symmetric then?

I've figured out that if $A$ is symmetric, then $S$ and $A$ must commute. I've tried finding a $2 \times 2$ and a $3 \times 3$ counterexample, as it seemed to me that this is not generally true, but I couldn't find any.

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  • $\begingroup$ Is $S$ also assumed symmetric? $\endgroup$ Commented Jun 5, 2021 at 18:34
  • $\begingroup$ @BartMichels Yes. I will add that to my question. $\endgroup$ Commented Jun 5, 2021 at 18:40
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    $\begingroup$ Start by finding a symmetric positive definite $S$ and symmetric $B$ that don't commute. Then let $A=S^{-1}B$. $\endgroup$
    – user932138
    Commented Jun 5, 2021 at 18:53
  • $\begingroup$ Yes, following O.Peters suggestion, you can ''easily'' find a counterexample. He's $B$ is yours $SA$ $\endgroup$
    – nonuser
    Commented Jun 5, 2021 at 19:04

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Let $S = \begin{bmatrix}2 & 0 \\ 0 & 1 \end{bmatrix}$, $A = \begin{bmatrix}1 & 1/2 \\ 1 & 2 \end{bmatrix}$. Then $SA = \begin{bmatrix}2 & 1 \\ 1 & 2 \end{bmatrix}$ is symmetric, but $A$ is not.

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