$\newcommand{\null}{\operatorname{null}}
\newcommand{\span}{\operatorname{span}}$$\newcommand{\range}{\operatorname{range}}$
Suppose $W$ is a finite dimensional vector space and $T_1,T_2\in L(V,W)$, then $\null T_1=\null T_2$ implies that there exists an invertible linear operator $S$ on $W$ such that $T_1=ST_2$.
I tried to prove it like this:
Since range $T_2$ is finite dimensional, let $\{T_2u_i\}_{i=1}^m$ be a basis of range $T_2$. This basis can be extended to a basis of $W$ by adding $w_i$'s such that $\{Tu_i\}_{i=1}^m, w_{m+1},\dots,w_n$ is a basis of $W$. It can be shown here that $u_i$'s are linearly independent (LI) in $V$.
For any $x\in V$, there exist scalars $c_i$'s such that $T_2x=\sum_{i=1}^mc_iT_2u_i$. It follows that $T_2(x-\sum c_iu_i)=0\implies x-\sum c_iu_i\in \null T_2$. It follows that $V=\null T_2+ \span (u_1,…,u_m)$. Now if $y\in \null T_2\cap \span(u_1,…,u_m)$ then there exist scalars $c_i$' such that $T_2y=0=\sum c_i'u_i\implies c_i'=0$ as $u_i$'s are LI. It follows that $$V=\null T_2 \oplus \span (u_1,…,u_m)=\null T_1\oplus \span (u_1,…,u_m)\tag 1$$
It is clear from $(1)$ that $T_1u_i$'s span $\range T_1$. Suppose for any scalars $a_i$'s, $\sum a_iT_1u_i=0$. $$T_1(\sum a_iu_i)=0\implies \sum a_iu_i\in \null T_1\implies \sum a_iu_i=0\implies a_i=0 $$
Hence $T_1u_i$'s is a basis of $\range T_1$. Extending it to a basis of $W$ gives $\{T_1u_i\}_{i=1}^m, w_{m+1}',…,w_n'$.
Now, let $S(T_2u_i):=T_1u_i$ for all $1\le i\leq m$
and $S(w_i):=w_i'$ for all $m\lt i\leq n$. $S:W\to W$ is a linear map.
Now, if for any $w\in W, Sw=0$, then there exist scalars $d_i$'s such that $w=\sum d_i T_2 u_i+\sum d_iw_i$. $0=Sw= \sum T_1u_i+ \sum d_i Sw_i=\sum T_1u_i+ \sum d_i w_i'$. On RHS is linear combination of basis of $W$ hence all $d_i$'s are $0$. It follows that $w=0$ hence $\null S=\{0\}$. So, $S$ is injective. It follows that $S$ a bijection (as $W$ is finite dimensional).
Is my proof correct? Thanks.
Best Answer
Proof looks good. Here is a shorter one (relies on some general facts about how linear maps may extend). Let $K$ be the kernel of $T_1$ and $T_2$, and let $P$ be any complement to $K$ in $W$, so that $V = K \oplus P$. Then $T_1$ and $T_2$ are completely determined by their effects on $P$, and they each restrict to isomorphisms of $P$ onto their images $T_i(P)$. There exists therefore an invertible linear map $S$ from $T_2(P)$ onto $T_1(P)$. Extend $S$ in any way to an isomorphism of $W$ onto itself. Then $S$ is an invertible linear operator on $W$ for which
$$ST_2v = T_1v$$
for all $v \in P$. But also $ST_2v = T_1v = 0$ for all $v \in K$, so in fact $ST_2 = T_1$ as operators on $V$.