1
$\begingroup$

Prove: $T \in L(V, V)$ then $ \exists S \in L(V,V)$ such that $ST = 0 \iff T$ is not onto

Proof: $\rightarrow$

Let $S \in L(V,V)$ s.t $S \neq 0$ and $ST = 0$. Consider $S(T(v))$ for some $v\in V$ Then $T=0$ and we have $S(T(v)) = S(0) = 0. \iff$ is not one to one$\iff T$ is not onto

  1. Is this correct so far?
  2. I need help with the other direction
  3. I think I can just take the reverse steps if this is correct
  • 1
    The implication is false, since if $S=0$ then $ST=0$ whether or not $T$ is onto.2012-10-26

1 Answers 1

0

As noted in the comments your proof is not convincing. To answer the other questions, let me give you some hints:

Hint 1: Prove that if $T$ is onto, then there does not exist such an $S$. I think this is easier since now you can apply the definition of onto meaning that for each $w$ there exists $v$ with $Tv=w$. Now you have $0=STv=Sw$. What can you conclude?

Hint 2: If $T$ is not onto, think about block matrices, or taking a complement of a certain choice of a subspace of $V$ and then define $S$ on each of the direct summands with the easiest possible choice.