## 每週問題 February 18, 2013

Let $A$ and $B$ be $n\times n$ real symmetric matrices. If $AB$ is symmetric, show that every eigenvalue $\lambda$ of $AB$ can be written as $\lambda=\alpha\beta$, where $\alpha$ is an eigenvalue of $A$ and $\beta$ is an eigenvalue of $B$.

$Q^TABQ=(Q^TAQ)(Q^TBQ)=\mathrm{diag}(\alpha_1\beta_1,\ldots,\alpha_n\beta_n)$

PowSol-Feb-18-13

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