Show that $f$ is cyclic if and only if there exists a basis $\mathcal{B}$ of $E$ in which the matrix of $f$ is of the form $C_Q$, where $Q$ is a monic polynomial of degree $n$.

We assume that $f$ is a nilpotent endomorphism of $E$. We denote by $r$ the smallest natural number such that $f^r = 0$. Show that $f$ is cyclic if and only if $r = n$. Specify the companion matrix.

We assume that $\mathbb{K} = \mathbb{C}$, that $(\mathrm{Id}, f, f^2, \ldots, f^{n-1})$ is free, and we factor the characteristic polynomial of $f$ in the form
$$\chi_f(X) = \prod_{k=1}^{p} \left(X - \lambda_k\right)^{m_k}$$
where the $\lambda_k$ are the $p$ eigenvalues pairwise distinct of $f$ and the $m_k \in \mathbb{N}^*$ their respective multiplicities. For $k \in \llbracket 1, p \rrbracket$, we set $F_k = \ker\left(\left(f - \lambda_k \operatorname{Id}_E\right)^{m_k}\right)$.
Show that the vector subspaces $F_k$ are stable under $f$ and that $E = F_1 \oplus \cdots \oplus F_p$.

We assume that $\mathbb{K} = \mathbb{C}$, that $(\mathrm{Id}, f, f^2, \ldots, f^{n-1})$ is free, and we factor the characteristic polynomial of $f$ in the form
$$\chi_f(X) = \prod_{k=1}^{p} \left(X - \lambda_k\right)^{m_k}$$
where the $\lambda_k$ are the $p$ eigenvalues pairwise distinct of $f$ and the $m_k \in \mathbb{N}^*$ their respective multiplicities. For $k \in \llbracket 1, p \rrbracket$, we set $F_k = \ker\left(\left(f - \lambda_k \operatorname{Id}_E\right)^{m_k}\right)$, and we denote by $\varphi_k$ the endomorphism induced by $f - \lambda_k \operatorname{Id}$ on the vector subspace $F_k$,
$$\varphi_k : \left\lvert\, \begin{aligned} & F_k \rightarrow F_k, \\ & x \mapsto f(x) - \lambda_k x. \end{aligned} \right.$$
Justify that $\varphi_k$ is a nilpotent endomorphism of $F_k$.

We assume that $\mathbb{K} = \mathbb{C}$, that $(\mathrm{Id}, f, f^2, \ldots, f^{n-1})$ is free. For $k \in \llbracket 1, p \rrbracket$, $\varphi_k$ is a nilpotent endomorphism of $F_k$, and $\nu_k$ denotes the smallest natural number such that $\varphi_k^{\nu_k} = 0$. Why do we have $\nu_k \leqslant \operatorname{dim}(F_k)$?

We assume that $\mathbb{K} = \mathbb{C}$, that $(\mathrm{Id}, f, f^2, \ldots, f^{n-1})$ is free, and we factor the characteristic polynomial of $f$ in the form
$$\chi_f(X) = \prod_{k=1}^{p} \left(X - \lambda_k\right)^{m_k}$$
where the $\lambda_k$ are the $p$ eigenvalues pairwise distinct of $f$ and the $m_k \in \mathbb{N}^*$ their respective multiplicities. For $k \in \llbracket 1, p \rrbracket$, $\varphi_k$ is a nilpotent endomorphism of $F_k$, and $\nu_k$ denotes the smallest natural number such that $\varphi_k^{\nu_k} = 0$.
Show, with the proposed hypothesis, that for all $k \in \llbracket 1, p \rrbracket$, we have $\nu_k = m_k$.

We assume that $\mathbb{K} = \mathbb{C}$, that $(\mathrm{Id}, f, f^2, \ldots, f^{n-1})$ is free, and we factor the characteristic polynomial of $f$ in the form
$$\chi_f(X) = \prod_{k=1}^{p} \left(X - \lambda_k\right)^{m_k}$$
where the $\lambda_k$ are the $p$ eigenvalues pairwise distinct of $f$ and the $m_k \in \mathbb{N}^*$ their respective multiplicities. For $k \in \llbracket 1, p \rrbracket$, $F_k = \ker\left(\left(f - \lambda_k \operatorname{Id}_E\right)^{m_k}\right)$ and $\nu_k = m_k$.
Specify the dimension of $F_k$ for $k \in \llbracket 1, p \rrbracket$, then deduce the existence of a basis $\mathcal{B} = (u_1, \ldots, u_n)$ of $E$ in which $f$ has a block diagonal matrix, these blocks belonging to $\mathcal{M}_{m_k}(\mathbb{C})$ and being of the form
$$\left(\begin{array}{cccccc} \lambda_k & 0 & \cdots & \cdots & \cdots & 0 \\ 1 & \lambda_k & \ddots & & & \vdots \\ 0 & 1 & \lambda_k & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & & \ddots & \ddots & \lambda_k & 0 \\ 0 & \cdots & \cdots & 0 & 1 & \lambda_k \end{array}\right)$$

We assume that $\mathbb{K} = \mathbb{C}$, that $(\mathrm{Id}, f, f^2, \ldots, f^{n-1})$ is free. There exists a basis $\mathcal{B} = (u_1, \ldots, u_n)$ of $E$ in which $f$ has a block diagonal matrix with Jordan blocks of sizes $m_k$ associated to eigenvalues $\lambda_k$. We set $x_0 = u_1 + u_{m_1+1} + \cdots + u_{m_1 + \cdots + m_{p-1}+1}$.
Determine the polynomials $Q \in \mathbb{C}[X]$ such that $Q(f)(x_0) = 0$.

We assume that $\mathbb{K} = \mathbb{C}$, that $(\mathrm{Id}, f, f^2, \ldots, f^{n-1})$ is free. There exists a basis $\mathcal{B} = (u_1, \ldots, u_n)$ of $E$ in which $f$ has a block diagonal matrix with Jordan blocks of sizes $m_k$ associated to eigenvalues $\lambda_k$. We set $x_0 = u_1 + u_{m_1+1} + \cdots + u_{m_1 + \cdots + m_{p-1}+1}$.
Justify that $f$ is cyclic.

We assume that $f$ is cyclic and we choose a vector $x_0$ in $E$ such that $(x_0, f(x_0), \ldots, f^{n-1}(x_0))$ is a basis of $E$. Let $g \in \mathcal{C}(f)$, an endomorphism that commutes with $f$.
Justify the existence of $\lambda_0, \lambda_1, \ldots, \lambda_{n-1}$ of $\mathbb{K}$ such that
$$g(x_0) = \sum_{k=0}^{n-1} \lambda_k f^k(x_0)$$

We assume that $f$ is cyclic and we choose a vector $x_0$ in $E$ such that $(x_0, f(x_0), \ldots, f^{n-1}(x_0))$ is a basis of $E$. Let $g \in \mathcal{C}(f)$, an endomorphism that commutes with $f$, and suppose $g(x_0) = \sum_{k=0}^{n-1} \lambda_k f^k(x_0)$.
Show then that $g \in \mathbb{K}[f]$.

We assume that $f$ is cyclic. Establish that $g \in \mathcal{C}(f)$ if and only if there exists a polynomial $R \in \mathbb{K}_{n-1}[X]$ such that $g = R(f)$.

We denote by $d$ the degree of $\pi_f$. Justify the existence of a vector $x_1$ of $E$ such that $\left(x_1, f(x_1), \ldots, f^{d-1}(x_1)\right)$ is free.

We denote by $d$ the degree of $\pi_f$, and $x_1$ is a vector of $E$ such that $\left(x_1, f(x_1), \ldots, f^{d-1}(x_1)\right)$ is free. We set $e_1 = x_1, e_2 = f(x_1), \ldots, e_d = f^{d-1}(x_1)$ and $E_1 = \operatorname{Vect}(e_1, e_2, \ldots, e_d)$.
Show that $E_1$ is stable under $f$ and that $E_1 = \{P(f)(x_1) \mid P \in \mathbb{K}[X]\}$.

We denote by $d$ the degree of $\pi_f$, $E_1 = \operatorname{Vect}(x_1, f(x_1), \ldots, f^{d-1}(x_1))$, and $\psi_1$ is the endomorphism induced by $f$ on the vector subspace $E_1$,
$$\psi_1 : \left\lvert\, \begin{aligned} & E_1 \rightarrow E_1, \\ & x \mapsto f(x). \end{aligned} \right.$$
Justify that $\psi_1$ is cyclic.

We denote by $d$ the degree of $\pi_f$, $E_1 = \operatorname{Vect}(e_1, e_2, \ldots, e_d)$ where $e_i = f^{i-1}(x_1)$. We complete $(e_1, e_2, \ldots, e_d)$ to a basis $(e_1, e_2, \ldots, e_n)$ of $E$. Let $\Phi$ be the $d$-th coordinate form which associates to any vector $x$ of $E$ its coordinate along $e_d$. We denote by $F = \{x \in E \mid \forall i \in \mathbb{N}, \Phi(f^i(x)) = 0\}$.
Show that $F$ is stable under $f$ and that $E_1$ and $F$ are in direct sum.

We denote by $d$ the degree of $\pi_f$, $E_1 = \operatorname{Vect}(e_1, e_2, \ldots, e_d)$ where $e_i = f^{i-1}(x_1)$, and $\Phi$ is the $d$-th coordinate form. Let $\Psi$ be the linear map from $E$ to $\mathbb{K}^d$ defined, for all $x \in E$, by
$$\Psi(x) = \left(\Phi\left(f^i(x)\right)\right)_{0 \leqslant i \leqslant d-1} = \left(\Phi(x), \Phi(f(x)) \ldots, \Phi\left(f^{d-1}(x)\right)\right)$$
Show that $\Psi$ induces an isomorphism between $E_1$ and $\mathbb{K}^d$.

We denote by $d$ the degree of $\pi_f$, $E_1 = \operatorname{Vect}(e_1, e_2, \ldots, e_d)$ where $e_i = f^{i-1}(x_1)$, $F = \{x \in E \mid \forall i \in \mathbb{N}, \Phi(f^i(x)) = 0\}$, and $\Psi$ is the linear map from $E$ to $\mathbb{K}^d$ defined by $\Psi(x) = \left(\Phi(f^i(x))\right)_{0 \leqslant i \leqslant d-1}$.
Show that $E = E_1 \oplus F$.

Deduce that there exist $r$ vector subspaces of $E$, denoted $E_1, \ldots, E_r$, all stable under $f$ such that:

• $E = E_1 \oplus \cdots \oplus E_r$;
• for all $1 \leqslant i \leqslant r$, the endomorphism $\psi_i$ induced by $f$ on the vector subspace $E_i$ is cyclic;
• if we denote by $P_i$ the minimal polynomial of $\psi_i$, then $P_{i+1}$ divides $P_i$ for all integer $i$ such that $1 \leqslant i \leqslant r-1$.

Show that the dimension of $\mathcal{C}(f)$ is greater than or equal to $n$.

We assume that $f$ is an endomorphism such that the algebra $\mathcal{C}(f)$ is equal to $\mathbb{K}[f]$. Show that $f$ is cyclic.

In this part, we assume that $\mathbb{K} = \mathbb{R}$ and that $E$ is a Euclidean space. The inner product of two vectors $x, y$ of $E$ is denoted $(x \mid y)$. We say that an endomorphism $f$ of $E$ is orthocyclic if there exists an orthonormal basis of $E$ in which the matrix of $f$ is of the form $C_Q$ (companion matrix).
Let $f$ be a nilpotent endomorphism of $E$. Deduce that $f$ is orthocyclic if and only if
$$f \text{ has rank } n-1 \quad \text{and} \quad \forall x, y \in (\ker f)^{\perp}, \quad (f(x) \mid f(y)) = (x \mid y).$$

What can be said about a nilpotent endomorphism of index 1?

We assume that $n = 2$. Let $u$ be an endomorphism of $E$ nilpotent of index $p \geqslant 2$.
Show that there exists a vector $x$ in $E$ such that $u^{p-1}(x) \neq 0$.

We assume that $n = 2$. Let $u$ be an endomorphism of $E$ nilpotent of index $p \geqslant 2$.
Verify that the family $\left(u^{k}(x)\right)_{0 \leqslant k \leqslant p-1}$ is free. Deduce that $p = 2$.