Proving $A^T B A$ is positive definite or semidefinite, when $B$ is positive definite

Question

Let's assume we have a positive definite matrix called $B$. (Here I assume the definition of the positive definite matrix stating $B$ should be symmetric.)

How could we conclude that $A^T B A$ is positive definite or semidefinite (that is, all eigenvalues are nonnegative) for any matrix $A$?

I tried to use the theorem which states that "for any vector $v \neq 0$ and positive definite or semidefinite matrix $A$, always $v^T A v \ge 0$ " and put $A^T B A$ in place of $A$ in the theorem, but I wasn't successful in reaching a conclusion.

Answer 1

Set $w=Av,$ then

$$v^T(A^TBA)v=(Av)^TB (Av)=w^TBw \ge 0.$$