1
$\begingroup$

Suppose $W$ is finite-dimensional and $T_1, T_2 \in L(V,W)$. Prove that $\text{null} T_1 \subset \text{null} T_2$ if and only if there exists $S \in L(W,W)$ such that $T_2 = ST_1$.

If $\dim V<\infty $ then so is $\dim \text{null} T_1$ and then the problem becomes easy but here we have $\dim W<\infty $ .

How can we proceed here?Please help.

$\endgroup$

1 Answer 1

1
$\begingroup$

Hint: Define $S$ on $\operatorname{im} T_1$ using $T_2 = ST_1$. Prove that this is well defined using $\operatorname{null} T_1 \subseteq \operatorname{null} T_2$. Extend $S$ to $W$ arbitrarily on a complement of $\operatorname{im} T_1$.

$\endgroup$
4
  • $\begingroup$ I need to show $T_2=ST_1$ .Then how can I use that $\endgroup$
    – Learnmore
    Commented Oct 4, 2017 at 14:22
  • $\begingroup$ @Maths_Student, you force $T_2=ST_1$ to be true on $\operatorname{im} T_1$. $\endgroup$
    – lhf
    Commented Oct 4, 2017 at 15:26
  • $\begingroup$ How to show that $S$ is well- defined? $\endgroup$
    – Hopmaths
    Commented Mar 5, 2022 at 2:11
  • $\begingroup$ @Hopmaths, ask a separate question describing your efforts and where you got stuck $\endgroup$
    – lhf
    Commented Mar 10, 2022 at 0:49

You must log in to answer this question.

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