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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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 If and only if they are commuting... – Simon Henry Mar 30 2012 at 16:48
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 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. – Henry Cohn Mar 30 2012 at 21:00
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 This looks like homework – Fernando Muro Apr 14 2012 at 11:55
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 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). – MTS Apr 14 2012 at 18:33
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 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 $AB=(AB)^t=B^tA^t=BA$. – Richard Stanley Apr 14 2012 at 20:54
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closed as too localized by Yemon Choi, Ryan Budney, Vladimir Dotsenko, Chris Godsil, Scott Morrison♦ May 26 2012 at 20:48

1 Answer

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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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5 
 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. – Denis Serre Mar 31 2012 at 6:48

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