Let $V$ be a finite dimensional complex inner product space. Let $\rho$ be self-adjoint, positive semidefinite and $\operatorname{tr}\rho = 1$. Let $A,B \in \mathrm{End} (V)$. I want to show that
$\langle A,B \rangle_\rho := \operatorname{tr}(A^\star B \rho)$
is a positive semidefinite Hermitian form on $\mathrm{End}(V)$. So firstly, I want to show that $\langle A,A \rangle_\rho \geq 0$ holds. Now if $A$ or $A^\star A$ were positive semidefinite, it would be easy since the product of positive semidefinite matrices is positive semidefinite as well. But I don't see why that should be the case and if not, how I would show this. Could anyone give me a hint?