In this part, we assume $n \geqslant 2$. For all $(a_{0}, \ldots, a_{n-1}) \in \mathbb{R}^{n}$, we set $$J(a_{0}, \ldots, a_{n-1}) = \left( \begin{array}{cccc} a_{0} & a_{n-1} & \cdots & a_{1} \\ a_{1} & a_{0} & \cdots & a_{2} \\ \vdots & \vdots & & \vdots \\ a_{n-1} & a_{n-2} & \cdots & a_{0} \end{array} \right)$$ Let $\mathcal{A}$ be the set of matrices of $\mathcal{M}_{n}(\mathbb{R})$ of the form $J(a_{0}, \ldots, a_{n-1})$ where $(a_{0}, \ldots, a_{n-1}) \in \mathbb{R}^{n}$. Let $J \in \mathcal{M}_{n}(\mathbb{R})$ be the matrix canonically associated with the endomorphism $\varphi \in \mathcal{L}(\mathbb{R}^{n})$ defined by $\varphi: e_{j} \mapsto e_{j+1}$ if $j \in \{1, \ldots, n-1\}$ and $\varphi(e_{n}) = e_{1}$, where $(e_{1}, \ldots, e_{n})$ is the canonical basis of $\mathbb{R}^{n}$.
Show that $(I_{n}, J, J^{2}, \ldots, J^{n-1})$ is a basis of $\mathcal{A}$.

We introduce the map $$\begin{aligned} J : \mathbb { R } ^ { N } & \rightarrow \mathbb { R } \\ x & \mapsto \frac { 1 } { 2 } \langle x , A x \rangle - \langle b , x \rangle \end{aligned}$$ where $A \in \mathcal { S } _ { N } ^ { + } ( \mathbb { R } )$, $b \in \mathbb { R } ^ { N }$. We set $H _ { 0 } = \{ 0 \}$ and for $k \geq 1$, $$H _ { k } = \left\{ P ( A ) r _ { 0 } \mid P \in \mathbb { R } [ X ] , \operatorname { deg } ( P ) \leq k - 1 \right\}$$ and $x _ { 0 } + H _ { k }$ denotes the subset of points of $\mathbb { R } ^ { N }$ of the form $x _ { 0 } + x$ where $x$ ranges over $H _ { k }$.
Show that $J$ admits a unique minimizer on the subset $x _ { 0 } + H _ { k }$, for any $k \in \mathbb { N }$.

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, 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 $F_k$.

Let $A$ denote a matrix in $\mathcal{M}_n(\mathbb{C})$.
Show that a matrix is nilpotent if, and only if, its characteristic polynomial is equal to $X^n$.

In this part, we assume $n \geqslant 2$. For all $(a_{0}, \ldots, a_{n-1}) \in \mathbb{R}^{n}$, we set $$J(a_{0}, \ldots, a_{n-1}) = \left( \begin{array}{cccc} a_{0} & a_{n-1} & \cdots & a_{1} \\ a_{1} & a_{0} & \cdots & a_{2} \\ \vdots & \vdots & & \vdots \\ a_{n-1} & a_{n-2} & \cdots & a_{0} \end{array} \right)$$ Let $\mathcal{A}$ be the set of matrices of $\mathcal{M}_{n}(\mathbb{R})$ of the form $J(a_{0}, \ldots, a_{n-1})$ where $(a_{0}, \ldots, a_{n-1}) \in \mathbb{R}^{n}$. Let $J \in \mathcal{M}_{n}(\mathbb{R})$ be the matrix canonically associated with the endomorphism $\varphi \in \mathcal{L}(\mathbb{R}^{n})$ defined by $\varphi: e_{j} \mapsto e_{j+1}$ if $j \in \{1, \ldots, n-1\}$ and $\varphi(e_{n}) = e_{1}$, where $(e_{1}, \ldots, e_{n})$ is the canonical basis of $\mathbb{R}^{n}$.
Let $M \in \mathcal{M}_{n}(\mathbb{R})$. Show that $M$ commutes with $J$ if and only if $M$ commutes with every element of $\mathcal{A}$.

We introduce the map $$\begin{aligned} J : \mathbb { R } ^ { N } & \rightarrow \mathbb { R } \\ x & \mapsto \frac { 1 } { 2 } \langle x , A x \rangle - \langle b , x \rangle \end{aligned}$$ where $A \in \mathcal { S } _ { N } ^ { + } ( \mathbb { R } )$, $b \in \mathbb { R } ^ { N }$, and $\tilde { x }$ is the unique vector satisfying $A \tilde { x } = b$. We denote by $x _ { k }$ the minimizer of $J$ on $x _ { 0 } + H _ { k }$.
Show that $x _ { k }$ identifies with the projection onto $x _ { 0 } + H _ { k }$ for the norm $\| \cdot \| _ { A }$ associated with the matrix $A$, that is, $$\left\| x _ { k } - \tilde { x } \right\| _ { A } = \min _ { x \in x _ { 0 } + H _ { k } } \| x - \tilde { x } \| _ { A }$$

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 integer 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. 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 $\nu_k \leqslant \operatorname{dim}(F_k)$?

Let $A$ denote a matrix in $\mathcal{M}_n(\mathbb{C})$.
Show the converse of question 12: if 0 is the unique eigenvalue of $A$, then $A$ is nilpotent.

In this part, we assume $n \geqslant 2$. For all $(a_{0}, \ldots, a_{n-1}) \in \mathbb{R}^{n}$, we set $$J(a_{0}, \ldots, a_{n-1}) = \left( \begin{array}{cccc} a_{0} & a_{n-1} & \cdots & a_{1} \\ a_{1} & a_{0} & \cdots & a_{2} \\ \vdots & \vdots & & \vdots \\ a_{n-1} & a_{n-2} & \cdots & a_{0} \end{array} \right)$$ Let $\mathcal{A}$ be the set of matrices of $\mathcal{M}_{n}(\mathbb{R})$ of the form $J(a_{0}, \ldots, a_{n-1})$ where $(a_{0}, \ldots, a_{n-1}) \in \mathbb{R}^{n}$. Let $J \in \mathcal{M}_{n}(\mathbb{R})$ be the matrix canonically associated with the endomorphism $\varphi \in \mathcal{L}(\mathbb{R}^{n})$ defined by $\varphi: e_{j} \mapsto e_{j+1}$ if $j \in \{1, \ldots, n-1\}$ and $\varphi(e_{n}) = e_{1}$, where $(e_{1}, \ldots, e_{n})$ is the canonical basis of $\mathbb{R}^{n}$.
Show that $\mathcal{A}$ is a commutative subalgebra of $\mathcal{M}_{n}(\mathbb{R})$.

We keep the notations from Parts II and III. We denote $r _ { k } = b - A x _ { k }$, $e _ { k } = x _ { k } - \tilde { x }$, and note that $r _ { k } = - A e _ { k }$. We denote by $m$ the smallest integer $k$ such that $H _ { k + 1 } = H _ { k }$.
Show that $e _ { k } \neq 0$ for $k \in \{ 0 , \ldots , m - 1 \}$, and that $e _ { k } = 0$ for $k \geq m$.

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 integer 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 $k \in \llbracket 1, p \rrbracket$, $\varphi_k$ 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 $\nu_k = m_k$.

Let $A$ denote a matrix in $\mathcal{M}_n(\mathbb{C})$.
Show that an upper triangular matrix in $\mathcal{M}_n(\mathbb{C})$ with zero diagonal is nilpotent and that a nilpotent matrix is similar to an upper triangular matrix with zero diagonal.

In this part, we assume $n \geqslant 2$. Let $J \in \mathcal{M}_{n}(\mathbb{R})$ be the matrix canonically associated with the endomorphism $\varphi \in \mathcal{L}(\mathbb{R}^{n})$ defined by $\varphi: e_{j} \mapsto e_{j+1}$ if $j \in \{1, \ldots, n-1\}$ and $\varphi(e_{n}) = e_{1}$, where $(e_{1}, \ldots, e_{n})$ is the canonical basis of $\mathbb{R}^{n}$.
Determine the characteristic polynomial of $J$.

We keep the notations from Parts II and III. We denote $e _ { k } = x _ { k } - \tilde { x }$ and $e _ { 0 } = x _ { 0 } - \tilde { x }$. We recall that $I _ { N }$ is the identity matrix of order $N$.
Show that $$\left\| e _ { k } \right\| _ { A } = \min \left\{ \left\| \left( I _ { N } + A Q ( A ) \right) e _ { 0 } \right\| _ { A } \mid Q \in \mathbb { R } [ X ] , \operatorname { deg } ( Q ) \leq k - 1 \right\}$$

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 $F_k$ with $\nu_k = m_k$.
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, 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
$$\left(\begin{array}{cccccc} \lambda_k & 0 & \cdots & \cdots & \cdots & 0 \\ 1 & \lambda_k & \ddots & & & \vdots \\ 0 & 1 & \lambda_k & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & & \ddots & \ddots & \lambda_k & 0 \\ 0 & \cdots & \cdots & 0 & 1 & \lambda_k \end{array}\right)$$

Let $A$ denote a matrix in $\mathcal{M}_n(\mathbb{C})$.
Prove that, if $A$ is a nilpotent matrix of index $p$, then every polynomial in $\mathbb{C}[X]$ that is a multiple of $X^p$ is an annihilating polynomial of $A$.

In this part, we assume $n \geqslant 2$. Let $J \in \mathcal{M}_{n}(\mathbb{R})$ be the matrix canonically associated with the endomorphism $\varphi \in \mathcal{L}(\mathbb{R}^{n})$ defined by $\varphi: e_{j} \mapsto e_{j+1}$ if $j \in \{1, \ldots, n-1\}$ and $\varphi(e_{n}) = e_{1}$, where $(e_{1}, \ldots, e_{n})$ is the canonical basis of $\mathbb{R}^{n}$.
Show that $J$ is diagonalisable in $\mathcal{M}_{n}(\mathbb{C})$.

We keep the notations from Parts II and III. We denote $e _ { k } = x _ { k } - \tilde { x }$ and $e _ { 0 } = x _ { 0 } - \tilde { x }$. We recall that $I _ { N }$ is the identity matrix of order $N$, and $\| \cdot \|$ denotes the matrix norm defined in question 2.
Show that $$\left\| e _ { k } \right\| _ { A } \leq \left\| e _ { 0 } \right\| _ { A } \min \left\{ \left\| I _ { N } + A Q ( A ) \right\| \mid Q \in \mathbb { R } [ X ] , \operatorname { deg } ( Q ) \leq k - 1 \right\}$$ (One may use the properties of $A ^ { 1 / 2 }$ demonstrated in question 6.)

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

Let $A$ denote a matrix in $\mathcal{M}_n(\mathbb{C})$. Assume that $P$ is an annihilating polynomial of $A$ nilpotent.
Prove that 0 is a root of $P$.