Let $A$ and $B$ be two real symmetric PSD matrices and they commute. Simplify: $\big(\text{tr}(\sqrt{AB})\big)^{2}$. I know that $\text{tr}(AB)\ne \text{tr}(A)\text{tr}(B)$.
Note that the matrix $B$ might be singular. Hence, the square root of a singular matrix might not be defined. How can I proceed?