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Let $A$, $B$ $\in$ $M_n\mathbb{(C)}$ be self-adjoint, and assume that $A$ is positive semidefinite; prove that all eigenvalues of $AB$ are real.

I've seen similar questions ($T,U$ self-adjoint, $U$ positive definite, then $TU$ has only real eigenvalues) but it seems that their proofs required the eigenvalues of $A$ to be strictly positive, and I don't konw how to deal with this issue here.

Left Hand
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  • It should be noted exactly where the issue of $A$ being positive semi-definite becomes an issue in the linked source. It's not in computing the matrix square root of $A$: If self-adjoint $A$ has nonnegative eigenvalues, then $A=QDQ^$ for unitary $Q$ and nonnegative diagonal $D$. So $A^{1/2}=Q D^{1/2}Q^$ will be valid. But this can be singular, so the decomposition $AB=A^{-1/2} (A^{1/2}B A^{1/2})A^{1/2}$ isn't possible. Thus the impediment lies in step 2 of the linked answer. – Semiclassical Aug 26 '21 at 06:53

1 Answers1

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First note that eigenvalues are invariant under cyclic permutations, i.e, $\sigma(AB) = \sigma(BA)$ where $\sigma(X)$ denotes the set of eigenvalues of $X$. Then all we need to do is note $$ \sigma(AB) = \sigma(A^{1/2}A^{1/2} B) = \sigma(A^{1/2} B A^{1/2}) $$ but $\sigma(A^{1/2} B A^{1/2}) \in \mathbb{R}$ as $A^{1/2} B A^{1/2}$ is self-adjoint.

Rammus
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