TITLE: Trace of product of two Hermitian matrices
QUESTION [7 upvotes]: 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 $\operatorname{tr}(AB)=\sum_{i}c_{ii}>0$.

REPLY [3 votes]: You can't just say that because the trace is over the diagonal elements of $(AB)_{ii}$. Instead $\lambda = tr(AB)=\sum A_{ij}B_{ji}$ then $\lambda^* = tr(A^*B^*)=tr(A^TB^T)=\sum A^T_{ij}B^T_{ji} = \sum A_{ji}B_{ji}$ which is clearly $tr(AB)$ under renaming dummy indicies, so $\lambda=\lambda^*$