A linear operator S has nullity $=0$ iff 0 is not an eigen value of S.
proof:if we consider a n dimensional vector space over the n*n matrix.if 0 is the not an eigen value of the matrix then the matrix is non singular (in particular sometime) and rank of the matrix is n.
We know by Rank Nullity theorem
rank(S)+nullity(S)=n
nullity(S)=n-rank(S)=n-n=0 this is true?
Consider the following arguments (construct the proof based on these):-
Let $T: V \rightarrow V$ be a linear transformation.
If nullity of a linear transformation is $0$, then no vector other than the zero vector itself is mapped to the zero vector in the co - domain. Hence, for a non - zero vector $v$, $T \left( v \right) \neq 0 \cdot v = 0$. Hence, $0$ cannot be an eigenvalue of $T$.
Similarly, if $0$ is an eigenvalue of $T$, then $\exists v \neq 0 \in V$ such that $T \left( v \right) = 0 \cdot v = 0$. Hence, $\dim \ker T > 0$, i.e., $\text{nullity} \neq 0$.