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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
  • 0
    Why must $T(v) = 0$? Just because $S\neq 0$ doesn't mean that $S(v) = 0 \implies v=0$.2012-10-26
  • 1
    The implication is false, since if $S=0$ then $ST=0$ whether or not $T$ is onto.2012-10-26

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