# If $T_1,T_2\in L(V,W)$, then null $T_1=$ null $T_2$ implies that there exists an invertible linear operator $S$ on $W$ such that $T_1=ST_2$.

$$\newcommand{\null}{\operatorname{null}} \newcommand{\span}{\operatorname{span}}\newcommand{\range}{\operatorname{range}}$$ 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.

• Yes, it's correct. $T_2x=\sum_ic_iT_2u_i$ because $T_2u_i$ was chosen to be a basis of range of $T$. Commented Sep 25, 2021 at 22:35
• @Berci: Thanks a lot for reviewing the proof. ðŸ˜Š
– Koro
Commented Sep 26, 2021 at 6:08

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 $$V$$.

• Thanks a lot for reviewing the proof. I'm still learning linear maps. In this answer, I don't understand why there exists $S$ from $T_2(P)$ onto $T_1(P)$. That is the reason why I tried to construct S explicitly in my post. Can you please help me understand how you concluded existence of $S$?
– Koro
Commented Sep 26, 2021 at 6:15
• I think you meant $V$ instead of $W$ in the second line? range $T_1$ and range $T_2$ have the same dimension so they are isomorphic? Hence the existence of $S$ on $T_2(P)$. It is yet to be proven though that the extended $S$ is also 1-1?
– Koro
Commented Sep 26, 2021 at 6:30
• I have edited. Yes that is why $S$ exists on $T_2(P)$. As for why it can be extended to an isomorphism, if $M'$ and $M''$ are isomorphic subspaces of a larger vector space $M$, it is easy to see using bases that any isomorphism from $M'$ to $M''$ can be extended to an isomorphism from $M$ to itself.
– D_S
Commented Sep 26, 2021 at 15:07
• Provided $M$ is finite dimensional, that is. Otherwise this is clearly false.
– D_S
Commented Sep 26, 2021 at 15:13
• V is not necessarily finite-dimensional, so you have to use Zorn's Lemma to guarantee this direct sum decomposition, which is not introduced in that book. Commented Jun 25, 2022 at 5:43