If $B^{m\times m}$ is positive semidefinite, is $A^TBA$ is positive semidefinite?

Question

If $B^{m\times m}$ is positive semidefinite, is it true that $A^TBA$ is positive semidefinite with $A^{m\times n}$? I think it is, because all of my counterexamples have failed, but I don't know how to prove it. I know that $x^TBx\ge 0$ if B is positive semidefinite. Is this the same? Do I need to prove it or is it well-known?

Answer 1

First, note that $M := A^\top B A$ is an $n$-by-$n$ square matrix, and further it is symmetric: $$ M^\top = (A^\top B A)^\top = A^\top B^\top A = A^\top B A = M $$ since $B$ is symmetric. Now, indeed, a (real) symmetric $n$-by-$n$ matrix $M$ is psd if, and only if, $x^\top M x \geq 0$ for every $x\in\mathbb{R}^n$. In our case, we can check that, for every $x\in\mathbb{R}^n$, $$ x^\top M x = x^\top A^\top B A x = (Ax)^\top B (Ax) \geq 0 $$ since $Ax\in\mathbb{R}^m$ and $B$ itself is psd, so $y^\top B y \geq 0$ for every $y\in \mathbb{R}^m$.