3
$\begingroup$

$\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.

$\endgroup$
2
  • 1
    $\begingroup$ Yes, it's correct. $T_2x=\sum_ic_iT_2u_i$ because $T_2u_i$ was chosen to be a basis of range of $T$. $\endgroup$
    – Berci
    Commented Sep 25, 2021 at 22:35
  • $\begingroup$ @Berci: Thanks a lot for reviewing the proof. 😊 $\endgroup$
    – Koro
    Commented Sep 26, 2021 at 6:08

1 Answer 1

4
$\begingroup$

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 $V$.

$\endgroup$
5
  • $\begingroup$ 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 $S$? $\endgroup$
    – Koro
    Commented Sep 26, 2021 at 6:15
  • $\begingroup$ 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? $\endgroup$
    – Koro
    Commented Sep 26, 2021 at 6:30
  • 1
    $\begingroup$ 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. $\endgroup$
    – D_S
    Commented Sep 26, 2021 at 15:07
  • 2
    $\begingroup$ Provided $M$ is finite dimensional, that is. Otherwise this is clearly false. $\endgroup$
    – D_S
    Commented Sep 26, 2021 at 15:13
  • $\begingroup$ 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. $\endgroup$
    – Hamilton
    Commented Jun 25, 2022 at 5:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .