``Graded'' version of the decomposition theorem In this part, we are given: a finite-dimensional vector space $V$; a nilpotent endomorphism $u$ of $V$; a nonzero natural integer $N$ and $\zeta = \exp\frac{2\mathrm{i}\pi}{N}$; an invertible endomorphism $h$ of $V$ such that $h^N = \operatorname{id}_V$ and $h \circ u \circ h^{-1} = \zeta u$. a) Let $n$ be the index of $u$, that is, the integer such that $u^{n-1} \neq 0$ and $u^n = 0$. Prove that there exists a vector $v$ such that $v$ is an eigenvector of $h$ and $u^{n-1}(v) \neq 0$. b) Prove that there exists a basis of $V$ in which the matrices of $u$ and $h$ are block diagonal and the diagonal blocks are respectively of the form $$J_r \quad \text{and} \quad D_{r,a} = \operatorname{diag}(\zeta^a, \zeta^{a+1}, \ldots, \zeta^{a+r-1})$$ for $r \in \mathbb{N}^*$ and $a \in \{0, \ldots, N-1\}$ suitable. We call $(r, a)$ the type of such a pair of matrices $(J_r, D_{r,a})$.
\textbf{``Graded'' version of the decomposition theorem}\\
In this part, we are given: a finite-dimensional vector space $V$; a nilpotent endomorphism $u$ of $V$; a nonzero natural integer $N$ and $\zeta = \exp\frac{2\mathrm{i}\pi}{N}$; an invertible endomorphism $h$ of $V$ such that $h^N = \operatorname{id}_V$ and $h \circ u \circ h^{-1} = \zeta u$.
a) Let $n$ be the index of $u$, that is, the integer such that $u^{n-1} \neq 0$ and $u^n = 0$. Prove that there exists a vector $v$ such that $v$ is an eigenvector of $h$ and $u^{n-1}(v) \neq 0$.
b) Prove that there exists a basis of $V$ in which the matrices of $u$ and $h$ are block diagonal and the diagonal blocks are respectively of the form
$$J_r \quad \text{and} \quad D_{r,a} = \operatorname{diag}(\zeta^a, \zeta^{a+1}, \ldots, \zeta^{a+r-1})$$
for $r \in \mathbb{N}^*$ and $a \in \{0, \ldots, N-1\}$ suitable. We call $(r, a)$ the type of such a pair of matrices $(J_r, D_{r,a})$.