Suppose WW is finite-dimensional and T1,T2∈L(V,W)T_1,T_2 \in \mathcal L(V,W). Prove that null T1=null T2\text{null }T_1 = \text{null }T_2 if and only if there exists an invertible operator S∈L(W)S \in \mathcal L(W) such that T1=ST2T_1 = ST_2.

(In other words, find an isomorphism SS between the output ranges such that T1=ST2T_1 = ST_2.)

Suppose null T1=null T2\text{null }T_1 = \text{null }T_2. Let NN denote the nullspace, by 2.34 there exists a subspace UU such that U⊕N=VU \oplus N = V.

Let R1=range T1,R2=range T2R_1 = \text{range }T_1, R_2 = \text{range }T_2. Both T1T_1 and T2T_2 are invertible when treated as maps from UU to their range (restricting inputs to UU makes them injective, treating the outputs as the range makes them surjective).
I'll denote the restricted maps as T~1∈L(U,R1)\tilde T_1 \in \mathcal L(U, R_1) and T~2∈L(U,R2)\tilde T_2 \in \mathcal L(U, R_2).

If w=T~2uw = \tilde T_2u then T~1T~2−1w=T~1u\tilde T_1 \tilde T_2^{-1} w = \tilde T_1 u. This leads us to define S=T~1T~2−1S = \tilde T_1 \tilde T_2^{-1}, clearly S∈L(R2,R1)S \in \mathcal L(R_2, R_1) is invertible as it's the product of two invertible maps.

We've found the desired isomorphism between the ranges, now we just need to extend SS to be invertible on all of WW, this is straightforward since 2.34 gives us subspaces R1′,R2′R_1',R_2' such that W=R1⊕R1′=R2⊕R2′W = R_1\oplus R_1'=R_2\oplus R_2'. Since dimensions match 3.59 allows us to find an isomorphism S′:R1′→R2′S' : R_1' \to R_2' so we can finish our definition of S∈L(W)S \in \mathcal L(W) with

Sw=S(r1+r1′)=T~1T~2−1r1+S′r1′Sw = S(r_1 + r_1') = \tilde T_1\tilde T_2^{-1}r_1 + S'r_1'

Verifying T1=ST2T_1 = ST_2 is straightforward, write each v∈Vv \in V as v=u+nv = u + n to get

ST2(u+n)=T~1T~2−1T2u=T~1T~2−1T~2u=T~1u=T1(u+n)ST_2(u + n) = \tilde T_1\tilde T_2^{-1}T_2u = \tilde T_1 \tilde T_2^{-1}\tilde T_2u = \tilde T_1u = T_1(u + n)

todo: explain why direct sums are needed to ensure maps are well defined (currently left implicit)
todo: add diagram of what SS looks like

(Old, bad proof. left here for it's historic value)

Suppose null T1=null T2\text{null }T_1 = \text{null }T_2. Let u1,…,unu_1,\dots,u_n be a basis for null T1\text{null }T_1 then extend it to a basis u1,…,un,v1,…,vru_1,\dots,u_n,v_1,\dots,v_r of VV. By 3.22 we know r=dim⁡range T1=dim⁡range T2r = \dim \text{range }T_1 = \dim \text{range }T_2, define wk=T1vkw_k = T_1v_k and wk′=T2vkw_k' = T_2v_k. Since any w∈range T1w \in \text{range }T_1 can be written

w=T1v=T1(a1u1+⋯+anun+b1v1+⋯+brvr)=b1w1+⋯+brwr\begin{aligned} w = T_1v &= T_1(a_1u_1 + \dots + a_nu_n + b_1v_1+\dots+b_rv_r) \\ &= b_1w_1 + \dots + b_rw_r \end{aligned}

We have that w1,…,wrw_1,\dots,w_r spans range T1\text{range }T_1, since r=dim⁡range T1r = \dim \text{range }T_1 2.42 implies w1,…,wrw_1,\dots,w_r is a basis of range T1\text{range }T_1. Exactly the same logic shows that w1′,…,wr′w_1',\dots,w_r' is a basis of range T2\text{range }T_2.
Extend to bases w1,…,wmw_1,\dots,w_m and w1′,…,wm′w_1',\dots,w_m' of WW then define Swk′=wkSw_k' = w_k. SS is clearly an isomorphism as we're just relabeling one basis into another.
To see T1=ST2T_1=ST_2 consider v=a1u1+⋯+anun+b1v1+⋯+brvrv = a_1u_1+\dots+a_nu_n+b_1v_1+\dots+b_rv_r and

T1v=b1w1+⋯+brwr=S(b1w1′+⋯+brwr′)=ST2vT_1v = b_1w_1 + \dots + b_rw_r = S(b_1w_1' + \dots + b_rw_r') = ST_2v

Now suppose T1=ST2T_1 = ST_2. If v∈null T1v \in \text{null }T_1 then T1v=0=ST2=0T_1 v = 0 = ST_2 = 0 implies T2=0T_2 = 0 since SS is injective. and if v∈null T2v \in \text{null }T_2 then clearly T1v=0T_1v = 0.

I spent more then 2 hours finding a short, rigorous, somewhat understandable proof. I did compromise a bit at the end, assuming the reader can make reasonable inferences on their own.

edit: I found a better proof, I'll leave the other one for reference