Prove that the eigenvalues of a product of positive semidefinite matrices are nonnegative

Question

Show that if $A,B \in \mathcal M_n(\mathbb{R})$ are positive semidefinite and $\lambda$ is an eigenvalue of $AB$ then $\lambda \geq 0$.

I don't really know what to do here. If $AB$ was semidefinite positive we would be done but the product of positive semidefinite matrices doesn't have to be positive semidefinite itself.

Answer 1

If $ABx = \lambda x$, $x\neq 0$, then $\langle ABx,Bx\rangle = \lambda \langle x,Bx\rangle$. Now, both $\langle ABx,Bx\rangle$ and $\langle x,Bx\rangle$ are non-negative. Hence, if $\langle x,Bx\rangle > 0$, it follows that $\lambda\ge 0$. If $\langle x,Bx\rangle = 0$, then $Bx = 0$ and thus $\lambda = 0$.

Answer 2

Let $A$ and $B$ be positive semidefinite real symmetric matrices.
Denote $\sqrt{A}$ the square root of $A$, i.e. \begin{equation} A = \sqrt{A}\sqrt{A} \end{equation} where $\sqrt{A}$ is also PSD. Notice that $$\sqrt{A} B \sqrt{A}$$ is symmetric and positive semidefinite. Also, $$AB = \sqrt{A}(\sqrt{A} B)$$ and $$(\sqrt{A}B)\sqrt{A}$$ have the same nonzero eigenvalues.