It's known that if $A$ is a positive definite (symmetric) matrix and $B$ is positive semi-definite (not necessarily symmetric), then $AB$ has non-negative eigenvalues. It follows from knowing that $A$ has a unique SPD square root $A^{1/2}$ and that
$$AB = A^{1/2} A^{1/2} B = A^{1/2} (A^{1/2} BA^{1/2})A^{-1/2}$$
Therefore, $AB$ is similar to $A^{1/2}BA^{1/2}$. The latter is positive definite since
$$x^TA^{1/2}B A^{1/2}x = (A^{1/2}x)^TB(A^{1/2}x) \geq 0$$ I want to know if the following generalization is true:
If $A$ is positive semi-definite (symmetric) and $B$ has non-negative eigenvalues, then $AB$ has non-negative eigenvalues.
The proof above does not go through because $A^{1/2}$ is no longer necessarily invertible and (more importantly) $B$ is no longer positive semi-definite.
It is not true. Consider $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & t \\ 0 & 0 \end{bmatrix}$, where $t < 0$.