Theorem: Let T be a linear mapping on a vector space V, and let $\lambda$ be an eigenvalue of T. A vector $v \in V$ is an eigenvector of T corresponding to $\lambda$ if and only if $v \neq 0$ and$ v \in N(T-\lambda I)$
By definition, onsider $0 \neq v$ be an eigen vector corresponding to an eigen value $\lambda.$ Then $Tv = \lambda v$ which implies $(T - \lambda I)v = 0.$
Question: Can I then say $v \in N(T- \lambda I)$? Are we treating $(T-\lambda I)$ as a linear mapping? Why is that? I know the nullspace is defined to be $ker(L)={v \in V |L(v)=0}$ where L is a linear mapping.
Yes.
Well, it is a linear mapping. It takes a vector $v$, and maps it to $Tv - \lambda v$. You can easily prove that this is a linear map.
We are treating it like a linear map (or, more importantly, just a map) so that we can write $N(T - \lambda I)$ and have it make sense. The proof is simple, and shows the connection between eigenvectors for operator $T$ with eigenvalue $\lambda$, and the nullspace of a different linear operator.
In particular, this leads to other results, like $\lambda$ is an eigenvalue if and only if $\operatorname{det}(T - \lambda I) = 0$, and hence arguments with the characteristic polynomial. Note that $\operatorname{det}(T - \lambda I)$ doesn't make sense either without considering $T - \lambda I$ as a linear operator.