Suppose $V$ and $W$ are finite-dimensional and $T_1,T_2 \in \mathcal L(V,W)$. Prove that there exist invertible operators $R \in \mathcal L(V)$ and $S \in \mathcal L(W)$ such that $T_1 = ST_2R$ if and only if $\dim \text{null }T_1 = \dim \text{null }T_2$.

Suppose $\dim \text{null }T_1 = \dim\text{null }T_2$. Let $R : \text{null }T_1 \to \text{null }T_2$ be an isomorphism then extend $R$ to all of $V$ in such a way that it remains invertible (see 3d/3). This gives

Now let $v_1,\dots,v_r$ be independent vectors in $V$ such that $w_k = T_1v_k$ is a basis for $\text{range }T_1$.
Define $w_k' = T_2Rv_k$, $w_1',\dots,w_r'$ are independent since

Define $Sw_k' = w_k$

as desired, $R$ is an isomorphism between nullspaces and $S$ is an isomorphism between ranges.

Now suppose $T_1 = ST_2R$. let $u_1,\dots,u_n$ be a basis for $\text{null }T_1$ and notice how $Ru_1,\dots,Ru_n$ is a basis for $\text{null }T_2$ ($S$ can be ignored as it's injective so $ST_2Rv = 0$ iff $T_2Rv = 0$.)