Is $A^TBA$ positive semidefinite, for any square matrix $B$?

Question

Im not sure if $B$ must itself be positive semidefinite; or the statement is true in general (I assume positive semidefinite implies symmetric).

Answer 1

It's sufficient: (in the following I assume $B$ is symmetric, and $B\ge0$ means $B$ is positive semidefinite).

$$B\ge 0 \iff \forall x\ne 0:\, x^t B x\ge0 $$ but $$x^t A^t B A x = y^t By\ge0 \quad {\rm with} \quad y=Ax$$ hence $A^t B A\ge 0$.

(notice that if we change positive semidefinite to strictly positive, we need $A$ non singular to preserve the property)