Does $\operatorname{rank}(AB) = \operatorname{rank}(A) $ imply $k_1 A = k_2 A$ if $ k_1 AB = k_2 AB $?

Question

If I know that these are true

$ \begin{cases} \operatorname{rank}(AB) = \operatorname{rank}(A) \\ k_1 AB = k_2 AB \end{cases} $

where $k_1$ and $k_2$ are constants; edit: $k_1 \ne k_2$

I need to say that $k_1 A = k_2 A$.

I know that since $ \operatorname{rank}(AB) =\operatorname{rank}(A) \le \operatorname{rank}(B) $, no matter what $B$ is, when $A$ is multiplied by $B$ its rank won't increase. So when $ k_1 AB = k_2 AB $ I can deduce that $ rank(k_1A) = rank(k_2A)$ but I can't seem to grasp how $ k_1A = k_2A $ can be true.

Answer 1

If $k_1=k_2$, there is nothing to prove. If $k_1\ne k_2$, then $AB=0$. Since $A$ has the same rank as $AB$, $A$ is zero too. Therefore $k_1A=k_2A\,(=0)$.