Let $A = (a_{i,j})$ in $\mathcal{M}_n(\mathbb{R})$, with $A \geqslant 0$. We say that $A$ is irreducible if, for all $i$ and $j$ in $\llbracket 1, n \rrbracket$, there exists $m \geqslant 0$ (depending a priori on $i$ and $j$) such that $a_{i,j}^{(m)} > 0$.
Express the irreducibility of $A$ in terms of paths in $A$.

Let $A = (a_{i,j})$ in $\mathcal{M}_n(\mathbb{R})$, with $A \geqslant 0$, irreducible.
Show that if $A$ is irreducible, then for all $i$ and $j$ in $\llbracket 1, n \rrbracket$, there exists $m \in \llbracket 0, n-1 \rrbracket$ (depending a priori on $i$ and $j$) such that $a_{i,j}^{(m)} > 0$.

Give a simple example of a square irreducible matrix that is not primitive.

Let $A = (a_{i,j})$ in $\mathcal{M}_n(\mathbb{R})$, with $A \geqslant 0$.
Show that if $A$ is not irreducible, then $A^2$ is not irreducible.
On the other hand, give a simple example of an irreducible matrix $A$ such that $A^2$ is not irreducible.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 2$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
Show that if $\operatorname{det}\left((A^{-1})_{s}\right) = 0$, then there exists a hyperplane $H$ of $E_{n}$ such that $A$ is $H$-singular.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 2$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
Deduce that if $\operatorname{det}(A_{s}) = 0$, then there exists a hyperplane $H$ of $E_{n}$ such that $A$ is $H$-singular.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 2$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
We assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$. Show that $A$ is $H$-regular for every hyperplane $H$ of $E_{n}$.

We consider the matrix $$A = A(\mu) = \begin{pmatrix} 2-\mu & -1 & \mu \\ -1 & 2-\mu & \mu-1 \\ 0 & -1 & 1 \end{pmatrix}$$
A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$.
Determine a hyperplane $H$ such that $A(1)$ is $H$-singular.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 3$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-2$. We consider $(N_{1}, N_{2})$ a basis of $F^{\perp}$ and we set $N = \begin{pmatrix} N_{1} & N_{2} \end{pmatrix} \in \mathcal{M}_{n,2}(\mathbb{R})$.
Show that $A$ is $F$-singular if and only if there exist a non-zero element $X$ of $F$ and two real numbers $\lambda_{1}$, $\lambda_{2}$ such that $AX = \lambda_{1}N_{1} + \lambda_{2}N_{2}$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 3$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-2$. We consider $(N_{1}, N_{2})$ a basis of $F^{\perp}$ and we set $N = \begin{pmatrix} N_{1} & N_{2} \end{pmatrix} \in \mathcal{M}_{n,2}(\mathbb{R})$.
Deduce that $A$ is $F$-singular if and only if the matrix $$A_{N} = \begin{pmatrix} A & N_{1} & N_{2} \\ N_{1}^{\top} & 0 & 0 \\ N_{2}^{\top} & 0 & 0 \end{pmatrix} = \begin{pmatrix} A & N \\ N^{\top} & 0_{2} \end{pmatrix} \in \mathcal{M}_{n+2}(\mathbb{R})$$ is singular.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 3$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$.
Show that there exists $P \in \mathcal{G}_{n,2}(\mathbb{R})$ such that $\operatorname{det}(P^{\top}A^{-1}P) = 0$ if and only if there exists $P' \in \mathcal{G}_{n,2}(\mathbb{R})$ such that $\operatorname{det}(P'^{\top}AP') = 0$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 3$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-2$. We consider $(N_{1}, N_{2})$ a basis of $F^{\perp}$ and we set $N = \begin{pmatrix} N_{1} & N_{2} \end{pmatrix} \in \mathcal{M}_{n,2}(\mathbb{R})$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
Deduce that if $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$, then $\operatorname{det}(N^{\top}A^{-1}N) > 0$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 3$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
Conclude that if $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$, then $A$ is $F$-regular for every vector subspace $F$ of dimension $n-2$ of $E_{n}$.

We return to the example of subsection II.B with $\mu = 1$, i.e. $$A(1) = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}$$ If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$.
How should we choose $N' = \begin{pmatrix} N_{1}' & N_{2}' \end{pmatrix}$ so that $\operatorname{det}(N'^{\top}A(1)N') = 0$?

We return to the example of subsection II.B with $\mu = 1$, i.e. $$A(1) = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}$$ A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$.
Determine a vector subspace $F$ of $E_{3}$ such that $\dim F = 1$ and such that $A(1)$ is $F$-singular.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$.
Show that $A$ is $F$-singular if $\operatorname{det}(N'^{\top}AN') = 0$ for a matrix $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ that one will define.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We now assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Show that if $X \in \mathcal{M}_{p,1}(\mathbb{R})$ is non-zero then $X^{\top}N'^{\top}AN'X > 0$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Deduce that the real eigenvalues of $N'^{\top}AN'$ are strictly positive.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Deduce that $\operatorname{det}(N'^{\top}AN') > 0$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. We assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Deduce that $A$ is $F$-regular for every non-zero vector subspace $F$ of $E_{n}$.

In all that follows, $\mathbf{K = C}$. We set $E = \mathbf{C}^n$. We are given $A \in \mathcal{M}_n(\mathbf{C})$. For every eigenvalue $\lambda$ of the matrix $A$, we denote by $m_\lambda$ its multiplicity, and we introduce the vector subspace $$F_\lambda = \operatorname{Ker}\left((A - \lambda I_n)^{m_\lambda}\right) = \operatorname{Ker}\left((u - \lambda \operatorname{Id}_E)^{m_\lambda}\right).$$
$\mathbf{20}$ ▷ Show that $\mathbf{C}^n = \bigoplus_{\lambda \in \operatorname{Sp}(A)} F_\lambda$.

In all that follows, $\mathbf{K = C}$. We set $E = \mathbf{C}^n$. We are given $A \in \mathcal{M}_n(\mathbf{C})$. For every eigenvalue $\lambda$ of the matrix $A$, we denote by $m_\lambda$ its multiplicity, and we introduce the vector subspace $$F_\lambda = \operatorname{Ker}\left((A - \lambda I_n)^{m_\lambda}\right).$$
$\mathbf{21}$ ▷ Deduce from question 20) the existence of three matrices $P, D$ and $N$ in $\mathcal{M}_n(\mathbf{C})$ such that $A = P(D + N)P^{-1}$, where $D$ is diagonal, $N$ is nilpotent, $DN = ND$, and $P$ is invertible.

We consider $M \in \mathscr { D } _ { \rho } \left( S _ { n } ( \mathbb { R } ) \right)$. We fix a real eigenvalue $\lambda$ of $M _ { \mid t = 0 }$ and denote by $d$ its multiplicity as a root of $\chi _ { \mid t = 0 }$ where $\chi = \det(XI_n - M)$. We set $A = F ( M )$, $B = G ( M )$, $Q = ( B U \mid A V ) \in \mathscr { D } _ { \rho _ { 2 } } \left( \mathscr { M } _ { n } ( \mathbb { R } ) \right)$.
Show that, for all $a \in U _ { \rho _ { 2 } }$, the direct sum $\operatorname { im } \left( B _ { a } U \right) \oplus \operatorname { im } \left( A _ { a } V \right) = \mathbb { R } ^ { n }$ of question 22a is orthogonal for the standard inner product on $\mathbb { R } ^ { n }$.

We consider $M \in \mathscr { D } _ { \rho } \left( S _ { n } ( \mathbb { R } ) \right)$. We fix a real eigenvalue $\lambda$ of $M _ { \mid t = 0 }$ and denote by $d$ its multiplicity as a root of $\chi _ { \mid t = 0 }$ where $\chi = \det(XI_n - M)$. We set $Q = ( B U \mid A V ) \in \mathscr { D } _ { \rho _ { 2 } } \left( \mathscr { M } _ { n } ( \mathbb { R } ) \right)$ and $Q^{-1} \cdot M \cdot Q = \operatorname{Diag}(M_1, M_2)$.
Show that there exists $\rho _ { 3 } \in \mathbb { R } _ { + } ^ { * }$ such that $\rho _ { 3 } \leqslant \rho _ { 2 }$ and matrices $R _ { 1 } \in \mathrm { GL } _ { d } \left( \mathscr { D } _ { \rho _ { 3 } } ( \mathbb { R } ) \right) , R _ { 2 } \in \mathrm { GL } _ { n - d } \left( \mathscr { D } _ { \rho _ { 3 } } ( \mathbb { R } ) \right)$ such that the matrix $Q \cdot \operatorname { Diag } \left( R _ { 1 } , R _ { 2 } \right)$ is orthogonal. (One may use the result of question 17.)

Prove Theorem 2: Let $M \in \mathscr { D } _ { \rho } \left( S _ { n } ( \mathbb { R } ) \right)$. Then there exists $r \in \mathbb { R } _ { + } ^ { * }$ such that $r \leqslant \rho$ and an orthogonal matrix $P \in \mathscr { D } _ { r } \left( \mathscr { M } _ { n } ( \mathbb { R } ) \right)$ such that $P ^ { \mathrm { T } } \cdot M \cdot P$ is diagonal.