If $A$ and $B$ are real-valued symmetric positive definite matrices, and if $AB$ is normal, i.e., $AB(AB)^T=(AB)^TAB$ (alternatively, $AB^2A=BA^2B$), then how does one prove that $AB=BA$ (i.e., $AB$ is symmetric).
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It works for $A,B$ positive semi-definite. You have, since $XY$ and $YX$ have the same spectrum, $$ \sigma(AB)=\sigma(A^{1/2}A^{1/2}B)=\sigma(A^{1/2}BA^{1/2})\subset[0,\infty). $$ So $AB$ is normal with real spectrum, thus selfadjoint.
The matrix $A^{1/2}BA^{1/2}$ is positive definite, since $$ v^*A^{1/2}BA^{1/2}v=(A^{1/2}v)^*BA^{1/2}v\geq0. $$
Martin Argerami
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