$BAA' = CAA'$ implies $BA = CA$

Question

Suppose that $BAA' = CAA'$ holds, where $A$ is a $n\times k$ matrix and $C,B$ are (arbitrary) matrices of size $n\times n$. Then it follows that $$BA = CA.$$ I don't think it's true but I could not come up with a counter example. If I set $n=k=1$, the above obviously holds. For $n=k=2$ and possible non-singular choices of $A$, $B$ and $C$, I couldn't come up with a working example because I usually end up with $0 = 0$ implies $0 = 0$... $n=k\leq 2$ couldn't help either. Any hints?

Answer 1

The statement is equivalent to: $XAA^T = 0$ implies $XA=0$ (just take $X=B-C$).

If $XAA^T=0$, then $XAA^TX^T=0$, and therefore $Trace((XA)(XA)^T) = Trace(XAA^TX^T)=0$. But for any matrix $M$, $Trace(MM^T)$ is the sum of the squares of the entries of $M$. Therefore $XA=0$.