## 每週問題 October 6, 2014

Let $A$ and $B$ be $n\times n$ Hermitian matrices. Show that if $A$ or $B$ is positive definite, then $AB$ is diagonalizable.

$\displaystyle AB=\sqrt{A}\left(\sqrt{A}B\sqrt{A}\right)\sqrt{A}^{-1}=\sqrt{A}(UDU^\ast)\sqrt{A}^{-1}=\left(\sqrt{A}U\right)D\left(\sqrt{A}U\right)^{-1}$

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