Let $T$ be linear transformation, defined for some space $V$, where $T: V \to V$.
Let vector $v \in V$ be called generalized eigenvector of rank $m$ with correspondence to eigenvalue $\lambda$, if $(T - \lambda I)^m (v) = 0$, but $(T - \lambda I)^{m-1} (v) \neq 0$.
A set spanned by all generalized vectors for a given $\lambda$ will be called generalized eigen(sub)space $V(\lambda)$
I struggle to understand (and find the information and proofs), why the condition for $V(\lambda)$ to be non-trivial (not zero and not the whole space) is $\lambda$ must be eigenvalue, even if $m = 1$.
I would be happy for any explanation or links. Usually it is just stated in a books with an assumption, that reader understands what they are reading. Unfortunately, I don't. Also, if I made a mistake with these statement or "names" for this mathematical objects, please, point it out (I mean correct me). Thank you.
(upd. also, perhaps, you know any other conditions for $V(\lambda)$ to be non-trivial? If you do, I would be grateful)
If $(T-\lambda)^nv=0$ for some positive integer $n$, with $v\not=0$, let $n_o$ be the smallest exponent such that $(T-\lambda)^{n_o}v=0$. Then $v_1=(T-\lambda)^{n_o-1}v$ is non-zero, and $(T-\lambda)v_1=0$, so $v_1$ is a $\lambda$-eigenvector.