If $Q\in\mathbb R^{n\times n}$ (not necessarily symmetric) is positive semi-definite means for all $x\in\mathbb R^n$, $$ x Qx^T\ge0. $$ Can I have the following result?
If the eigenvalues of matrix $A\in\mathbb R^{n\times n}$ are all negative, and we have $W\in\mathbb S^n$ such that $$ AW\succeq 0\quad (\text{AW is positive semi-definite})\\ WA^T\succeq 0\quad (\text{$WA^T$ is positive semi-definite}) $$
Can I claim that $W$ is negative semi-definite?