Generalised eigenvectors of a finite dimensional vector space

Question

I'm currently reading Knapp's 'Lie Groups, Beyond an Introduction', and I can't seem to understand this point.

Let $V$, be a finite dimensional $\mathbb{C}$-vector space, $\pi: \mathfrak{h} \rightarrow \text{End}_{\mathbb{C}}(V)$ a representation, and $\alpha \in \mathfrak{h}^*$. Define:$$V_\alpha = \{v \in V: (\pi(H) - \alpha(H)1)^n = 0 \text{ for all } H \in \mathfrak{h} \text{ and some }n = n(H,v)\}$$ Since $V$ is finite-dimensional, $\pi(H) - \alpha(H)1$ has $0$ as its only generalised eigenvalue on $V_\alpha$ and is nilpotent on this space, as a consequence of the theory of Jordan normal form. Therefore $$n(H,v) = \text{dim}(V)$$

There are a few things which I'm not quite understanding here.

1. Why does $V$ being finite dimension mean there are no non-zero generalised eigenvalues?

2. Am I right in thinking by being nilpotent, he just particularly means that $$(\pi(H) - \alpha(H))^2 v = 0$$ since "$0$ is the only generalised eigenvalue".

3. Why is it that $n(H,v) = \text{dim}(V)$, how does this follow from the "theory of Jordan normal form"?

Answer 1

1. First verify that $V_\alpha$ is a $\pi$-invariant subspace. It's saying that it can't have nonzero (generalized) eigenvalue on $V_\alpha$, not on the whole $V$.

2. Well, the exponent can be higher than $2$ to obtain $0$, otherwise yes.

3. In the definition of $V_\alpha$, the nilpotency level $n$ may depend on $H$ and $v$, which is true for the minimum of such $n$. However, here it states only existence, and the choice $n=\dim V_\alpha$ (or $n=\dim V$) works uniformly for all $H$ and $v$.