Let $A$ and $B$ be two Hermitian complex matrices. (a) Prove that
$\operatorname{tr}(AB)$ is real. (b) Prove that if $A, B$ are
positive, then $\operatorname{tr}(AB)>0$.
(a) The trace of Hermitian matrix is a real number, since $a_{ii} = \bar{a}_{ii}$, that also means that all eigenvalues are real. I can't proceed to conclusion that $\operatorname{tr}(AB)$ is real, since $\operatorname{tr}(AB) \neq \operatorname{tr}A\cdot\operatorname{tr}B$ and product of two Hermitian matrices is also Hermitian only if these matrices commute, which is not the case for arbitrary Hermitian matrices.
(b) Am I missing something or the question is indeed so easy? If all entries $a_{ij}>0$ and $b_{ij}>0$, then all $c_{ij}=\sum_{m}a_{im}\cdot b_{mj}>0$ too, finally $\operatorname{tr}(AB)=\sum_{i}c_{ii}>0$.
Best Answer
Since $A^T = \overline{A}$ and using basic properties of trace :
$$ tr(AB) = tr((AB)^T) = tr(B^T A^T) = tr(\overline{B} \ \overline{A})= tr(\overline{A} \ \overline{B}) $$
Therefore :
$$ tr(AB) = tr(\overline{A} \ \overline{B}) = tr(\overline{AB}) = \overline{tr(AB)} $$
Finally $tr(AB) \in \mathbb R$.