Answer
See below
Work Step by Step
Given $T:V \rightarrow V$
is a linear transformation.
With $v \in Rng(T)=$, there exist $u \in Rng(T)$ such as $T(u)=v$
Given $T^2=0 \rightarrow T(T(u))=0$
We know the fact that $Rng(T) \subseteq Ker (T)$.
Since $Ker (T)$ and $Rng(T)$ are both subspaces of $V$, $Rng(T)$ is also a subspace of $Ker (T)$
