Positive Definiteness problem Consider the positive definite matrix $B \succ 0$ and the matrix ( not necessarily square) $A$. what can we say abut the positive definiteness of: $$ A^\prime B A$$ My hunch is that this is positive semi definite due to the lack of constraints on $A$ meaning there is a vector $x \neq 0$ such that $Ax=0$. \begin{align} x^\prime A^\prime B A x &=(Ax)^\prime B(Ax) \\\ &\rightarrow trace(B||Ax||^2) \ge 0 \end{align} the eigenvalues of this are greater than or equal to zero.

You are right, but there is an easier way to see it. Why not denote $Ax$ as $y$ so you don't know $y$ equals 0 or not, but you can always have $y'Ay\geq 0$.