Does AB = 0 mean that BZA = 0? Simple linear algebra question, but I can't offer a rigorous proof. Given $A$ and $B$ non-square matrices and knowing $ AB = 0$ does this imply that $BZA = 0$, where $Z$ is an arbitrary matrix with dimensions such that $BZA$ is conformable for multiplication? I think it does because each row nonzero row in A must be multiplied by a column of zeros in B, so the product will vanish regardless of the $Z$. But I'm having trouble being more rigorous or finding a property that proves this explicitly. Thanks in advance.

Try $A=\left(\begin{smallmatrix}0&1\\\0&0\end{smallmatrix}\right)$, $B=\left(\begin{smallmatrix}1&0\\\0&0\end{smallmatrix}\right)$, $Z=I$.