Suppose $V$ is a vector space and $T$ is a linear map on $V$. $\ G(\lambda,T) $ denotes the generalized eigenspace of $T$ corresponding to $\lambda $. In other words, $G(\lambda,T)=\{v\in V\mid {(T-\lambda I)}^{k}v=0\text{ for some integer }k>0\}$.
Prove or give a counterexample: $\ $ $T$ is nilpotent $\iff$ $G(0,T)=V $ .
BTW: It is true when V is finite-dimensional.$\;$This can be seen from 8.18 in Linear Algebra Done Right 3rd Edition.
Any insights are much appreciated.
It does not follow that $T$ is nilpotent in the infinite-dimensional case. Take the vector space with countable basis $$ \{ e_1, e_2, \ldots\} $$ and let $T$ be the left-shift operator that annihilates $e_1$ and sends $e_{i+1}$ to $e_i$. Then the generalized eigenspace (with eigenvalue $0$) for $T$ is all of $V$ since for any basis vector, there is a power of $T$ that eliminates it, but $T$ is not nilpotent since there is no power of $T$ that eliminates all the basis vectors at once.