Suppose $A$ is a matrix with positive eigenvalues ($A$ is not Hermitian), and $P$ is a projector matrix. Is the inequality $\operatorname{Tr} (PA)\leq \operatorname{Tr}(A)$ correct?
When $A$ and $P$ commute, it is straightforward because $\operatorname{Tr} (PA)$ is equal to the summation of some of eigenvalues of $A$, and since all eigenvalues of $A$ are positive the inequality holds. But what happens when $A$ and $P$ do not commute?
When $A$ and $P$ do not commute, the inequality can fail. Counterexample:
$$P = \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}\quad\text{ and }\quad A = \begin{bmatrix}5 & -3\\3 & -1\end{bmatrix}$$ It is clear $P$ is a projection operator. It is easy to check $2$ is an eigenvalue of $A$ with multiplicity $2$. However $${\rm Tr}(PA) = 5 > 4 = {\rm Tr}(A)$$