$\operatorname{rank}(A\circ B) = \operatorname{rank}(B)$ iff $\operatorname{Im}(B)\cap \operatorname{Ker}(A)= \{0\}$

Question

Let $V$, $W$, $U$ be vector spaces with scalars $K$, and $B\colon V\to W$ and $A\colon W\to U$ be linear transformations.

Prove that $\operatorname{rank}(A\circ B) = \operatorname{rank}(B)$ if and only if $\operatorname{Im}(B)\cap \operatorname{Ker}(A)= \{0\}$

I have a problem right at the beginning: how can I show that $\operatorname{rank}(A\circ B) = \operatorname{rank}(B)$? I think that's the clue.

Answer 1

Adding to the comments : since one inclusion is quite clear from the comments I give the second inclusion.

Let $$x\in~Null(AB)\implies AB(x)=0\implies Bx \in~Null(A)$$ But also $$Bx\in Im(B)$$ But the intersection is trivial, which in turn $$\implies Bx=0\implies x\in~Null(B)$$ Hence the Null spaces are equal.