Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A lemma is that:

Let $V$ be a finite dimensional vector space over a field $F$ and let $T\in L(V,V)$. Then there exists a nonzero linear transformation $S\in L(V,V)$ such that $TS=0$ if and only if there exists a nonzero vector $v\in V$ such that $T(v)=0$.

This proof is not difficult. I want to know that whether it is true when $V$ is infinite dimensional? When the such $S$ exists, we can easy to find the such $v$. But I can not do the converse statement. In fact, I need to find a $S\neq 0$ such that $\operatorname{Image}(S)\subseteq \ker(T)$. We know that $\ker(T)$ is a subspace of $V$ and so $S\in L(V,\ker(T))$. Thus if $L(V,\ker(T))\neq \{0\}$ when $\ker(T)\neq 0$?

share|cite|improve this question
up vote 2 down vote accepted

The same proof works. Let $v$ a non-zero vector so that $T(v)=0$. Extend $\{v\}$ to a basis for $V$, say, $\{v\}\cup\{v_i: i\in I\}$. Now define $S$ on the basis as follows $S(v)=v$ and $S(v_i)=0$ for all $i\in I$ then extend $S$ by linearity to all $V$. Now it is easy that $S$ is as required.

share|cite|improve this answer
    
Thanks very much. :) – Sun Apr 17 '12 at 5:58

Right, this is a local property and so only matters in finite dimensions. If $TS=0$ then $T(S(v))=0$ for any vector. If $T(v)=0$ extend $v$ to a basis $\{v\}\cup B$ and define $S(v)=v$ and $S(b)=0$ for $b\in B$. Then clearly $TS=0$.

share|cite|improve this answer
    
Thanks very much. :) – Sun Apr 17 '12 at 5:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.