Suppose $W$ is finite dimensional and $T_1, T_2 \in L(V,W).$ Prove that null($T_1$) $=$ null($T_2$) if and only if there exists and invertable operator $S\in L(W)$ such that $T_1 = S\circ T_2$.
I'm quite sure I have to make use of the rank-nullity theorem here. I started with:
$ dim(v) = dim(null(S\circ T_2)) + dim(Range(S\circ T_2)) $
$=dim(null(T_1)) + dim(Range(S \circ T_2)) $
$=dim(null(S \circ T_2) = dim(Range(nullT_1))$
$=dim(nullT_1) + dim(Range(T_1)) $
$= dim(nullT_1), + dim(Range(S \circ T_2))$
$=dim(Range(T_1)) = dim(Range(S \circ T_2))$
...I'm not sure how to take this further and put everything together.