For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, the spectral radius $\rho(A)$ is a dominant eigenvalue of $A$ and the associated eigenspace is a line that possesses a strictly positive direction vector $x > 0$. We denote $r$ the spectral radius of $A$. We denote $x$ (respectively $y$) a strictly positive direction vector of the line $D = \operatorname{Ker}(A - rI_n)$ (respectively of the line $\Delta = \operatorname{Ker}(A^\top - rI_n)$). We denote $H = \operatorname{Im}(A - rI_n)$. If necessary by multiplying $y$ by an appropriate strictly positive coefficient, we assume $(y \mid x) = y^\top x = 1$. We denote $L = xy^\top$.
Show that $AL = LA = rL$. Deduce: $\forall m \in \mathbb{N}^*, (A - rL)^m = A^m - r^m L$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$.
Let $\lambda$ be a nonzero eigenvalue of $B$ and let $z$ be an associated eigenvector.
Show that $Lz = 0$, then $Az = \lambda z$. Deduce $\rho(B) \leqslant r$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$.
Let $\lambda$ be a nonzero eigenvalue of $B$ and let $z$ be an associated eigenvector. By contradiction, we assume $\rho(B) = r$. We can therefore choose $\lambda$ such that $|\lambda| = r$. Show that then $\lambda = r$ then $Lz = z$ and reach a contradiction. Conclude.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$, and we have shown $\rho(B) < r$ and $\forall m \in \mathbb{N}^*, (A - rL)^m = A^m - r^m L$.
Deduce from the above (and from subsection IV.A) that $\lim_{m \rightarrow +\infty} \left(\frac{1}{r}A\right)^m = L$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$. Let $\mu$ be the multiplicity of $r$ as an eigenvalue of $A$ and let $T = PAP^{-1}$ be a triangular reduction of $A$.
By examining the diagonal of $\left(\frac{1}{r}T\right)^m$ when $m \rightarrow +\infty$, show that $\mu = 1$.

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 $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
Show that there exists a matrix $B = \begin{pmatrix} B_{1} & B_{2} \\ B_{3} & B_{4} \end{pmatrix}$ with $B_{1} \in \mathcal{M}_{n}(\mathbb{R}), B_{2} \in \mathcal{M}_{n,1}(\mathbb{R}), B_{3} \in \mathcal{M}_{1,n}(\mathbb{R})$, $B_{4} \in \mathcal{M}_{1}(\mathbb{R})$ such that: $$A_{N}B = \begin{pmatrix} I_{n} & 0 \\ N^{\top}A^{-1} & -N^{\top}A^{-1}N \end{pmatrix}$$

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$, and $A_{N} = \begin{pmatrix} A & N \\ N^{\top} & 0 \end{pmatrix}$.
Deduce that $\operatorname{det}(A_{N}) = -N^{\top}A^{-1}N\operatorname{det}(A)$.

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 $$A = A(\mu) = \begin{pmatrix} 2-\mu & -1 & \mu \\ -1 & 2-\mu & \mu-1 \\ 0 & -1 & 1 \end{pmatrix}$$
Show that $A(\mu)$ is invertible for every real $\mu$.

We consider the matrix $$A = A(\mu) = \begin{pmatrix} 2-\mu & -1 & \mu \\ -1 & 2-\mu & \mu-1 \\ 0 & -1 & 1 \end{pmatrix}$$
Calculate $A(\mu)_{s}$ and show that $A(\mu)_{s}$ is singular for $\mu = 1, 1-\sqrt{3}, 1+\sqrt{3}$.

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 $n \geqslant 3$. 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 $\mathcal{M}_{n}(\mathbb{R})$.
Show that there exists a matrix $B = \begin{pmatrix} B_{1} & B_{2} \\ B_{3} & B_{4} \end{pmatrix}$ with $B_{1} \in \mathcal{M}_{n}(\mathbb{R}), B_{2} \in \mathcal{M}_{n,2}(\mathbb{R}), B_{3} \in \mathcal{M}_{2,n}(\mathbb{R})$ and $B_{4} \in \mathcal{M}_{2}(\mathbb{R})$ such that $$A_{N}B = \begin{pmatrix} I_{n} & 0 \\ N^{\top}A^{-1} & -N^{\top}A^{-1}N \end{pmatrix}$$

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 $\mathcal{M}_{n}(\mathbb{R})$, and $A_{N} = \begin{pmatrix} A & N \\ N^{\top} & 0_{2} \end{pmatrix}$.
Deduce that $\operatorname{det}(A_{N}) = \operatorname{det}(N^{\top}A^{-1}N)\operatorname{det}(A)$.

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 $n \geqslant 3$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$. Let $N' = \begin{pmatrix} N_{1}' & N_{2}' \end{pmatrix}$.
Show that $$\operatorname{det}(N'^{\top}AN') = \left(N_{1}'^{\top}A_{s}N_{1}'\right)\left(N_{2}'^{\top}A_{s}N_{2}'\right) - \left(N_{1}'^{\top}A_{s}N_{2}'\right)^{2} + \left(N_{1}'^{\top}A_{a}N_{2}'\right)^{2}$$

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 $E_{n}$.