Let $A, B\in M_n$.
- $B$ is Hermitian.
- $A$ is positive semidefinite.
why is $tr(AB)$ real?
(We know $tr(AB) = tr(A^{1/2}BA^{1/2}$)
2026-03-31 19:14:20.1774984460
$B$ is Hermitian, $A$ is positive semidefinite $ \Rightarrow $ $tr(AB)$ is real
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Since $A$ is positive semi-definite, $A^{1/2}$ exists and is Hermitian too. Since $tr(AB)=tr(BA)$ $$ tr(AB)=tr(A^{1/2}A^{1/2}B)=tr(A^{1/2}BA^{1/2}) $$ Since $$ (A^{1/2}BA^{1/2})^*=(A^{1/2})^*B^*(A^{1/2})^*=A^{1/2}BA^{1/2} $$ it is Hermitian and thus all its eigenvalues are real. So $tr(A^{1/2}BA^{1/2})$ is real.