Prove that $\operatorname{rank}(B) = \operatorname{rank}(AB) + \operatorname{dim}\left(N(A) \cap C(B)\right)$. I am studying for midterm exam, and solving many problems on the textbook. I want to prove the following statement, but I failed to prove it. Please someone help me how to prove the following statement.

If $A$ is an $n\times m$ matrix and $B$ is an $m \times n$ matrix with $m\le n$, then $$\operatorname{rank}(B) = \operatorname{rank}(AB) + \operatorname{dim}\left(N(A) \cap C(B)\right),$$
  where $N$ and $C$ is space of null and column.

I cannot find how to start.... so there is nothing of what I have done. I just draw many images and tried many examples with real numbers.... However, I cannot find any method to prove it.

I consider $A$ and $B$ as linear maps. Denote by $A|_{C(B)}$ the restriction of the linear map $A$ to the subspace $C(B)$. Then $\operatorname{Im}(A|_{C(B)}) = A(C(B)) = \operatorname{Im}(AB)$. Hence, $\operatorname{rank}(AB) = \operatorname{rank}(A|_{C(B)})$. By the rank-nullity theorem, applied to $A|_{C(B)}$, we get
\begin{align*}
\operatorname{rank}(B)
&= \dim C(B) = \operatorname{rank}(A|_{C(B)}) + \dim N(A|_{C(B)})\\
&= \operatorname{rank}(AB) + \dim(N(A)\cap C(B)).
\end{align*}
