If $A$ is an $n \times n$ matrix and $AA^*=AA,$ how to prove $A$ is Hermitian?
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An arbitrary square matrix $A$ can be written as $B + i C$ where $B = (A + A^*)/2$ and $C = (A - A^*)/(2i)$ are Hermitian. The given condition says $AC = 0$, i.e. $B C = -iC^2$. But $CBC = -iC^3$, and the left side is Hermitian while the right side is only Hermitian if $C = 0$. Therefore $C = 0$ which means $A$ is Hermitian.
Robert Israel
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Nicely done ${}{}$ – Ben Grossmann Jul 22 '15 at 19:39
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@Robert Israel I see why $C^3 = O$, but can you explain why $C = O$? – Open Season Jul 22 '15 at 23:05
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@OpenSeason any nilpotent Hermitian matrix is $0$. – Ben Grossmann Jul 22 '15 at 23:44
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1@OpenSeason quickest proof of this: note that for any matrix $A$, $A$ and $A^A$ have the same rank. So, $C$ and $C^2 = C^C$ have the same rank. It follows that $C$ and $C^3$ have the same rank. So, if $C^3 = 0$, then $C = 0$. – Ben Grossmann Jul 22 '15 at 23:47
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@OpenSeason Maybe quicker: any Hermitian matrix can be diagonalized. It's obvious for diagonal matrices.. – Robert Israel Jul 23 '15 at 00:15