TITLE: $B$ is Hermitian, $A$ is positive semidefinite $ \Rightarrow $ $tr(AB)$ is real
QUESTION [2 upvotes]: Let $A, B\in M_n$. 

$B$  is  Hermitian.
$A$ is positive semidefinite.

why is $tr(AB)$ real?
(We know $tr(AB) = tr(A^{1/2}BA^{1/2}$)

REPLY [1 votes]: 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.
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