Let $V$ be a finite dimensional vector space and $T \in \mathrm{End}(V)$. Let $\alpha,\beta$ also be two distint eigenvalues for $T$ and $V_\alpha, V_\beta$ are the generalized eigenspaces relative to them. I want to prove that $V_\alpha \cap V_\beta=\{0\}$.
It is given a hint:
Take $v \in V_\alpha \cap V_\beta$ and consider the subspace $U$ generated by $\{v,Tv,\dots,T^kv\}$ where $k$ is the maximum number s.t. they are linear indipendent and compute the matrix of $T|_U$ in that basis.
The matrix is
$$ A= \begin{pmatrix} 0 & 0 & 0 & \dots & 0 & a_0\\ 1 & 0 & 0 & \dots & 0 & a_1\\ 0 & 1 & 0 & \dots & 0 & a_2\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \dots & 0 & a_{k-1}\\ 0 & 0 & 0 & \dots & 1 & a_k \end{pmatrix} $$
where $(a_i)$ are the coordinates of $T^{k+1}v$.
I noticed that $U \subseteq V_\alpha \cap V_\beta$ so exists $n>0$ such that we have $(A-\alpha I)^n=0=(A-\beta I)^n$ (from this condition maybe we can get $\alpha=\beta$ but it gets messy very soon)
I don't see the utility of computing the matrix of $T|_U$, as is suggested.
I would recommend the following approach. Show that because $(T|_U - \alpha I)^n = 0$, there can be no vector $u \in U$ for which $u \neq 0$ and $(T|_U - \beta I)u = 0$. However, $(T|_U - \beta I)^n = 0$. Why does this allow us to conclude that $U = \{0\}$?
We are given that $(T|_U - \alpha I)^n = (T|_U - \beta I)^n = 0$. Suppose for the sake of contradiction that $U \neq \{0\}$. Let $u \in U$ be a non-zero vector. There exists a largest $k$ for which $(T|_U - \alpha I)^k u \neq 0$; let $v = (T|_U - \alpha I)^k u$. We see that $v \neq 0$ with $Tv = \alpha v$. It follows that $(T|_U - \beta I)^n v = (\alpha - \beta)^n v \neq 0$, which contradicts our supposition that $(T|_U - \beta I)^n = 0$.
Conclude that $U = \{0\}$.