Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. We consider an eigenvalue $\lambda \in \mathbb{C}$ of $A$ with modulus 1 and $x$ an eigenvector associated with $\lambda$. Show that $|x| \leqslant A|x|$.

We fix a real vector space $E$ of dimension $n$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set
$$H := \operatorname{Vect}(x)^{\perp}, \quad \mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\} \text{ and } \mathcal{W} := \{v \in \mathcal{V} : v(x) = 0\}$$
We denote by $\pi$ the orthogonal projection of $E$ onto $H$. For $u \in \mathcal{W}$, we denote by $\bar{u}$ the endomorphism of $H$ defined by $\forall z \in H, \bar{u}(z) = \pi(u(z))$. We consider the sets $\overline{\mathcal{V}} := \{\bar{u} \mid u \in \mathcal{W}\}$ and $\mathcal{Z} := \{u \in \mathcal{W} : \bar{u} = 0\}$. Given $a \in E$ and $x \in E$, $(a \otimes x)(z) = (a \mid z) \cdot x$ for all $z \in E$.
Show that there exists a vector subspace $L$ of $E$ such that
$$\mathcal{Z} = \{a \otimes x \mid a \in L\} \quad \text{and} \quad \operatorname{dim} L = \operatorname{dim} \mathcal{Z},$$
and show that then $x \in L^{\perp}$.

Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. We consider an eigenvalue $\lambda \in \mathbb{C}$ of $A$ with modulus 1 and $x$ an eigenvector associated with $\lambda$. We assume that $|x| < A|x|$. Show that there exists $\varepsilon > 0$ such that $A^2|x| - A|x| > \varepsilon A|x|$.

We fix a real vector space $E$ of dimension $n$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set
$$H := \operatorname{Vect}(x)^{\perp}, \quad \mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\} \text{ and } \mathcal{W} := \{v \in \mathcal{V} : v(x) = 0\}$$
We consider the sets $\overline{\mathcal{V}} := \{\bar{u} \mid u \in \mathcal{W}\}$ and $\mathcal{Z} := \{u \in \mathcal{W} : \bar{u} = 0\}$. Given $a \in E$ and $x \in E$, $(a \otimes x)(z) = (a \mid z) \cdot x$ for all $z \in E$. There exists a vector subspace $L$ of $E$ such that $\mathcal{Z} = \{a \otimes x \mid a \in L\}$ and $x \in L^{\perp}$.
By considering $u$ and $a \otimes x$ for $u \in \mathcal{V}$ and $a \in L$, deduce from Lemma A that $\mathcal{V} x \subset L^{\perp}$, and that more generally $u^{k}(x) \in L^{\perp}$ for every $k \in \mathbf{N}$ and every $u \in \mathcal{V}$.

Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. We consider an eigenvalue $\lambda \in \mathbb{C}$ of $A$ with modulus 1 and $x$ an eigenvector associated with $\lambda$. We assume that $|x| < A|x|$ and that there exists $\varepsilon > 0$ such that $A^2|x| - A|x| > \varepsilon A|x|$. We set $B = \frac{1}{1+\varepsilon} A$. Show that for all $k \geqslant 1, B^k A|x| \geqslant A|x|$.

We fix a real vector space $E$ of dimension $n$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set
$$H := \operatorname{Vect}(x)^{\perp}, \quad \mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\}$$
There exists a vector subspace $L$ of $E$ such that $\mathcal{Z} = \{a \otimes x \mid a \in L\}$, $x \in L^{\perp}$, and $\mathcal{V} x \subset L^{\perp}$.
Justify that $\lambda x \notin \mathcal{V} x$ for every $\lambda \in \mathbf{R}^{*}$, and deduce from the two previous questions that
$$\operatorname{dim} \mathcal{V} x + \operatorname{dim} L \leq n-1$$

Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. We set $B = \frac{1}{1+\varepsilon} A$ for some $\varepsilon > 0$. Determine $\lim_{k \rightarrow +\infty} B^k$.

We fix a real vector space $E$ of dimension $n$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set
$$H := \operatorname{Vect}(x)^{\perp}, \quad \mathcal{W} := \{v \in \mathcal{V} : v(x) = 0\}$$
We denote by $\pi$ the orthogonal projection of $E$ onto $H$. For $u \in \mathcal{W}$, we denote by $\bar{u}$ the endomorphism of $H$ defined by $\forall z \in H, \bar{u}(z) = \pi(u(z))$. We consider the set $\overline{\mathcal{V}} := \{\bar{u} \mid u \in \mathcal{W}\}$.
Let $u \in \mathcal{W}$. Show that $(\bar{u})^{k}(z) = \pi(u^{k}(z))$ for every $k \in \mathbf{N}$ and every $z \in H$. Deduce that $\overline{\mathcal{V}}$ is a nilpotent vector subspace of $\mathcal{L}(H)$.

Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. We consider an eigenvalue $\lambda \in \mathbb{C}$ of $A$ with modulus 1 and $x$ an eigenvector associated with $\lambda$. Conclude (that 1 is an eigenvalue of $A$).

We fix a real vector space $E$ of dimension $n \geq 2$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set $H := \operatorname{Vect}(x)^{\perp}$, $\mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\}$, $\mathcal{W} := \{v \in \mathcal{V} : v(x) = 0\}$, $\overline{\mathcal{V}} := \{\bar{u} \mid u \in \mathcal{W}\}$, $\mathcal{Z} := \{u \in \mathcal{W} : \bar{u} = 0\}$, and $L$ the vector subspace such that $\mathcal{Z} = \{a \otimes x \mid a \in L\}$.
We have $\operatorname{dim} \mathcal{V} = \operatorname{dim}(\mathcal{V} x) + \operatorname{dim} \mathcal{Z} + \operatorname{dim} \overline{\mathcal{V}}$, $\operatorname{dim} \mathcal{V} x + \operatorname{dim} L \leq n-1$, $\overline{\mathcal{V}}$ is a nilpotent subspace of $\mathcal{L}(H)$ with $\dim H = n-1$, and by induction hypothesis $\operatorname{dim} \overline{\mathcal{V}} \leq \frac{(n-1)(n-2)}{2}$.
Prove that
$$\operatorname{dim} \mathcal{V} \leq \frac{n(n-1)}{2}$$

Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. Show that $A$ admits a strictly positive eigenvector associated with the eigenvalue 1.

We fix a real vector space $E$ of dimension $n \geq 2$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$ with $\operatorname{dim} \mathcal{V} = \frac{n(n-1)}{2}$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set $H := \operatorname{Vect}(x)^{\perp}$, $\mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\}$, $\mathcal{W} := \{v \in \mathcal{V} : v(x) = 0\}$, $\overline{\mathcal{V}} := \{\bar{u} \mid u \in \mathcal{W}\}$, $\mathcal{Z} := \{u \in \mathcal{W} : \bar{u} = 0\}$, and $L$ the vector subspace such that $\mathcal{Z} = \{a \otimes x \mid a \in L\}$.
Prove that
$$\operatorname{dim} \overline{\mathcal{V}} = \frac{(n-1)(n-2)}{2}, \quad \operatorname{dim}(\operatorname{Vect}(x) \oplus \mathcal{V} x) + \operatorname{dim} L = n$$
and
$$L^{\perp} = \operatorname{Vect}(x) \oplus \mathcal{V} x$$
Deduce that $\operatorname{Vect}(x) \oplus \mathcal{V} x$ contains $v^{k}(x)$ for every $v \in \mathcal{V}$ and every $k \in \mathbf{N}$.

Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. Show that 1 is the only eigenvalue of $A$ with modulus 1.
One may admit without proof that if $z_1, z_2, \ldots, z_k$ are non-zero complex numbers such that $|z_1 + \cdots + z_k| = |z_1| + \cdots + |z_k|$, then $\forall j \in \llbracket 1, k \rrbracket, \exists \lambda_j \in \mathbb{R}^+$ such that $z_j = \lambda_j z_1$.

We fix a real vector space $E$ of dimension $n \geq 2$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$ with $\operatorname{dim} \mathcal{V} = \frac{n(n-1)}{2}$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set $H := \operatorname{Vect}(x)^{\perp}$, $\mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\}$, $\mathcal{W} := \{v \in \mathcal{V} : v(x) = 0\}$, $\overline{\mathcal{V}} := \{\bar{u} \mid u \in \mathcal{W}\}$. We have established that $\operatorname{dim} \overline{\mathcal{V}} = \frac{(n-1)(n-2)}{2}$ and $L^{\perp} = \operatorname{Vect}(x) \oplus \mathcal{V} x$.
By applying the induction hypothesis, show that the generic nilindex of $\mathcal{V}$ is greater than or equal to $n-1$, and that if moreover $\mathcal{V} x = \{0\}$ then there exists a basis of $E$ in which every element of $\mathcal{V}$ is represented by a strictly upper triangular matrix.

Throughout part II, $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix satisfying $\rho(A) = 1$. Show that $\dim\left(\ker\left(A - I_n\right)\right) = 1$.

We fix a real vector space $E$ of dimension $n \geq 2$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$ with $\operatorname{dim} \mathcal{V} = \frac{n(n-1)}{2}$, equipped with an inner product $(-\mid-)$. We choose $x$ in $\mathcal{V}^{\bullet} \backslash \{0\}$ (where $\mathcal{V}^{\bullet}$ is the subset of $E$ formed by vectors belonging to at least one of the sets $\operatorname{Im} u^{p-1}$ for $u$ in $\mathcal{V}$). We denote by $p$ the generic nilindex of $\mathcal{V}$ (with $p \geq n-1$ by question 21), and we fix $u \in \mathcal{V}$ such that $x \in \operatorname{Im} u^{p-1}$. We have $\mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\}$ and $L^{\perp} = \operatorname{Vect}(x) \oplus \mathcal{V} x$.
Let $v \in \mathcal{V}$ such that $v(x) \neq 0$. Show that $\operatorname{Im} v^{p-1} \subset \operatorname{Vect}(x) \oplus \mathcal{V} x$. One may use the results of questions 5 and 20.

By combining the results of sub-parts II.B and II.C, justify that we have proved Proposition 1: If $A \in \mathcal{M}_n(\mathbb{R})$ is a strictly positive matrix, then $\rho(A)$ is a dominant eigenvalue of $A$. The associated eigenspace $\ker\left(A - \rho(A) I_n\right)$ is one-dimensional and is spanned by a strictly positive eigenvector.

We assume that $A$ is strictly positive and diagonalizable over $\mathbb{C}$. For all $Y \in \mathcal{M}_{n,1}(\mathbb{R})$, for all $p \in \mathbb{N}^*$, we denote $Y_p = \left(\frac{A}{\rho(A)}\right)^p Y$. Let $\lambda \in S = \operatorname{sp}(A) \setminus \{\rho(A)\}$. Let $Y \in \ker\left(A - \lambda I_n\right)$. Show that the sequence $\left(Y_p\right)_{p \in \mathbb{N}^*}$ converges to 0.

We assume that $A$ is strictly positive and diagonalizable over $\mathbb{C}$. For all $Y \in \mathcal{M}_{n,1}(\mathbb{R})$, for all $p \in \mathbb{N}^*$, we denote $Y_p = \left(\frac{A}{\rho(A)}\right)^p Y$. Let $Y \in \mathcal{M}_{n,1}(\mathbb{R})$ be a positive vector. Show that the sequence $\left(Y_p\right)_{p \in \mathbb{N}^*}$ converges to the projection of $Y$ onto $E_{\rho(A)}(A)$ parallel to $\bigoplus_{\lambda \in S} E_\lambda(A)$. Verify that, if it is non-zero, this latter vector (the projection of $Y$) is strictly positive.

Justify that for all integer $k \geqslant 1$, $A^k$ is similar in $\mathcal{M}_n(\mathbb{C})$ to a triangular matrix, whose diagonal coefficients we will specify.

Let $E$ be a $\mathbb{C}$-vector space of dimension $n \geq 1$. We say that an endomorphism $u$ of $E$ is a permutation endomorphism if there exists a basis $(e_1, \ldots, e_n)$ of $E$ and a permutation $\sigma \in \mathfrak{S}_n$ such that $u(e_j) = e_{\sigma(j)}$ for all $j \in \llbracket 1, n \rrbracket$.
Show that $u$ is a permutation endomorphism if and only if there exists a basis in which its matrix is a permutation matrix.

Let $$B = \frac{1}{4} \left(\begin{array}{cccc} 0 & -5 & 0 & -3 \\ 5 & 0 & 3 & 0 \\ 0 & -3 & 0 & -5 \\ 3 & 0 & 5 & 0 \end{array}\right).$$ Compute $B^{2} \left(\begin{array}{l} 1 \\ 1 \\ 1 \\ 1 \end{array}\right)$.

Let $$B = \frac{1}{4} \left(\begin{array}{cccc} 0 & -5 & 0 & -3 \\ 5 & 0 & 3 & 0 \\ 0 & -3 & 0 & -5 \\ 3 & 0 & 5 & 0 \end{array}\right).$$ Determine a real number $a$ and a matrix $P$ such that $$P \in \mathcal{O}_{4}(\mathbb{R}) \cap \mathrm{Sp}_{4}(\mathbb{R}) \quad \text{and} \quad P^{\top} B P = \left(\begin{array}{cccc} 0 & a & 0 & 0 \\ -a & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/a \\ 0 & 0 & -1/a & 0 \end{array}\right).$$

Show that, for every $M$ in $\mathcal{M}_{n}(\mathbb{R})$ and for all $P$ and $Q$ in $\mathcal{O}_{n}(\mathbb{R})$, we have $\|PMQ\|_{F} = \|M\|_{F}$.

We denote $D_{A} = \operatorname{diag}\left(\lambda_{1}(A), \ldots, \lambda_{n}(A)\right)$ and $D_{B} = \operatorname{diag}\left(\lambda_{1}(B), \ldots, \lambda_{n}(B)\right)$. Show that there exists an orthogonal matrix $P = \left(p_{i,j}\right)_{1 \leqslant i,j \leqslant n}$ such that $\|A - B\|_{F}^{2} = \left\|D_{A}P - PD_{B}\right\|_{F}^{2}$.