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


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*Is this correct so far?

*I need help with the other direction

*I think I can just take the reverse steps if this is correct


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.
