Suppose $V$ is finite-dimensional and $T_1,T_2 \in \mathcal L(V,W)$. Prove that $\text{range }T_1 = \text{range }T_2$ if and only if there exists an invertible operator $S \in \mathcal L(V)$ such that $T_1 = T_2S$.

(In other words, find an isomorphism $S$ that reshuffles the inputs so $T_1 = T_2S$)

Suppose $\text{range }T_1 = \text{range }T_2$.
Notice the nullspaces have the same dimension by 3.22 and the ranges being equal, meaning we can let $u_1,\dots,u_n$ be a basis for $\text{null }T_1$ and $u_1',\dots,u_n'$ be a basis of $\text{null }T_2$ (notice they're both of length $n$).
Extend $u_1,\dots,u_n$ to a basis $u_1,\dots,u_n,v_1,\dots,v_r$ of $V$.
Because the ranges are equal we can define $v_1',\dots,v_r'$ such that $T_1v_k = T_2v_k'$.
To see that $u_1',\dots,u_n',v_1',\dots,v_r'$ is a basis of $V$ we need only show it's independent, since it's the right length 2.39 will imply it's a basis. (Note this argument is similar to the proof of 3.22)

Thus we've shown $a_k = 0$, meaning the original equation becomes $c_1u_1' + \dots + c_nu_n' = 0$
Which obviously implies $c_k = 0$ since $u_1',\dots,u_n'$ is a basis for $\text{null }T_2$.

Define $Su_k = u_k'$ and $Sv_k = v_k'$. $S$ is clearly an isomorphism (with inverse $S^{-1}u_k' = u_k$, $S^{-1}v_k' = v_k$) as we're just moving from one basis of $V$ to another.
We have $T_1 = T_2S$ because they agree on the basis vectors ($T_1u_k = T_2Su_k$ and $T_1v_k = T_2Sv_k$). This completes the forward direction.

Now suppose $T_1 = T_2S$. If $w = T_1v$ then $w = T_2(Sv)$ implying $\text{range }T_1 \subseteq \text{range }T_2$ likewise if $w = T_2v$ then $w = T_1(S^{-1}v)$ implying $\text{range }T_2 \subseteq \text{range }T_1$.

todo: diagram / graphic for the proof idea (what linear map looks like). also describe the general strategy of preranges, postnullspaces etc and how you construct them