Is the following statement true: If $A\succeq 0$ and $B \preceq 0$ are two $n\times n$ real-valued matrices, then $AB \preceq 0$.


If not, is it true that $\forall x\ge 0$ (i.e., all vectors in the positive orthant), $x^TABx \le 0$?


If not, is the second statement true if all entries of $B$ are non-positive.


I'd really appreciate help with this.


Thanks!



Product of positive semidefinite and negative semidefinite matrices
No, $AB$ won't even be Hermitian in general. The correct formulation is that $A^{1/2}BA^{1/2} \preceq 0$.


