Suppose V is a vector space and $T\in L(V)$. Let $G(\lambda,T) $ denote the generalized eigenspace of $T$ corresponding to $\lambda $.
In other words, $G(\lambda,T)=\{\;v\in V\;\vert\;{(T-\lambda I)}^{\;k}\;v=0\;\;for\;some\;integer\;k>0\;\}$.
If V is finite-dimensional, we konw that $ G(\lambda,T)=null\;{(T-\lambda I)}^{dim\;V} $. So we have:
(a)$\quad$ $G(\lambda,T) $ is a subspace of $V$;
(b)$\quad$ $G(\lambda,T) $ is invariant under $ T $;
(c)$\quad$ $G(\lambda,T) $ is invariant under $ T-\lambda I $;
(d)$\quad$ $ (T-\lambda I)\;\vert_{G(\lambda,T)} $ is nilpotent.
If V is infinite-dimensional, do the results above still hold? $\ $Thanks!