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In generally, the product of two symmetric matrices is not symmetric, so I am wondering under what conditions the product is symmetric.

Likewise, over complex space, what are the conditions for the product of 2 Hermitian matrices being Hermitian?

Thanks!

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    $\begingroup$ If and only if they are commuting... $\endgroup$
    – Simon Henry
    Commented Mar 30, 2012 at 16:48
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    $\begingroup$ Simon's condition is presumably the best possible answer. Incidentally, the "right" product structure on symmetric matrices is the Jordan product $A \circ B = (AB + BA)/2$, which reduces to the ordinary product if and only if $A$ and $B$ commute. $\endgroup$
    – Henry Cohn
    Commented Mar 30, 2012 at 21:00
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    $\begingroup$ This looks like homework $\endgroup$
    – Fernando Muro
    Commented Apr 14, 2012 at 11:55
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    $\begingroup$ Voting to close as too localized (and so the site won't kick it back to the front page, as no answer has been accepted). $\endgroup$
    – MTS
    Commented Apr 14, 2012 at 18:33
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    $\begingroup$ I assume Simon thought that the proof of his answer is too trivial to warrant mention, but for what it's worth, if $A$, $B$, and $AB$ are symmetric, then $AB=(AB)^t=B^tA^t=BA$. $\endgroup$
    – Richard Stanley
    Commented Apr 14, 2012 at 20:54

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Incidentally, every real matrix is the product of two symmetric matrices. (If I remember correctly, I once read about this in a paper by Halmos).

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    $\begingroup$ Right. But not all complex matrices are product of two Hermitian matrices. A necessary and sufficient condition is to be similar to a real matrix. $\endgroup$
    – Denis Serre
    Commented Mar 31, 2012 at 6:48

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