Rank(AB)=Rank(A) if and only if Null Space (A) ∩ Image(B)={0}

Question

For any matrices A and B of conformable dimensions. I think I got the "if" part, but I'm stuck in the "only if". I tried a counterpositive argument but I failed to find a contradiction.

Answer 1

Consider $A$ and $B$ as linear maps say $\phi_A$ and $\phi_B$ respectively. Say that $\phi_A : F^n \rightarrow F^m$ and $\phi_B : F^m \rightarrow F^p$. Then you can apply the dimension theorem to $\phi_B$ restricted to $\phi_A(F^n)$. This gives $$dim(\phi_B |_{\phi_A(F^n)}) = dim(N(\phi_B |_{\phi_A(F^n)})) + dim(R(\phi_B |_{\phi_A(F^n)}))$$ $$dim(N(\phi_B |_{\phi_A(F^n)})) = dim(N(\phi_A) \cap R(\phi_B)) + dim(R(\phi_A\phi_B)$$

So in particular $$dim(R(\phi_A\phi_B)) = dim(R(\phi_B |_{\phi_A(F^n)})) - dim(N(\phi_A) \cap R(\phi_B)) $$

Therefore $$Rank(AB) = Rank(A)- \underbrace{dim(N(A) \cap R(B))}_{0}$$