We consider $A \in \mathcal{M}_{n}(\mathbb{R})$. We assume $A$ is invertible and, in accordance with the notations of the problem, $(A^{-1})_{s}$ denotes the symmetric part of the inverse of $A$. Show that $(\operatorname{det}(A))^{2}\operatorname{det}\left((A^{-1})_{s}\right) = \operatorname{det}(A_{s})$.
One may consider $A(A^{-1})_{s}A^{\top}$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix in $\mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$.
Show that a matrix in $\mathcal{M}_{n}(\mathbb{R})$ is singular if and only if it is $E_{n}$-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 $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$.
Show that $A$ is $H$-singular if and only if there exist a non-zero vector $X$ in $H$ and a real number $\lambda$ such that $AX = \lambda N$.

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 $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$.
Deduce that $A$ is $H$-singular if and only if the matrix $A_{N} = \begin{pmatrix} A & N \\ N^{\top} & 0 \end{pmatrix} \in \mathcal{M}_{n+1}(\mathbb{R})$ 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 $n \geqslant 2$. 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 $(X \mid Y) = X^{\top}Y$. 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 $E_{n}$.

A matrix $A$ of $\mathcal{M}_{n}(\mathbb{R})$ is said to be positively stable if all its complex eigenvalues have strictly positive real part.
Let $A \in \mathcal{M}_{2}(\mathbb{R})$. Show that $A$ is positively stable if and only if $\operatorname{tr}(A) > 0$ and $\operatorname{det}(A) > 0$.

A matrix $A$ of $\mathcal{M}_{n}(\mathbb{R})$ is said to be positively stable if all its complex eigenvalues have strictly positive real part.
a) Is the sum of two positively stable matrices of $\mathcal{M}_{2}(\mathbb{R})$ necessarily positively stable?
b) Let $A, B$ in $\mathcal{M}_{n}(\mathbb{R})$ be two positively stable matrices that commute. Show that $A + B$ is positively stable.

A matrix $A$ of $\mathcal{M}_{n}(\mathbb{R})$ is said to be positively stable if all its complex eigenvalues have strictly positive real part. Let $A \in \mathcal{M}_{n}(\mathbb{R})$ such that $A_{s}$ is positive definite.
a) Let $X = Y + \mathrm{i}Z$ be a column matrix of $\mathcal{M}_{n,1}(\mathbb{C})$, where $Y$ and $Z$ belong to $\mathcal{M}_{n,1}(\mathbb{R})$. We set $\bar{X} = Y - \mathrm{i}Z$ and we identify the matrix $\bar{X}^{\top}AX \in \mathcal{M}_{1}(\mathbb{C})$ with the complex number equal to its unique entry.
Show that, if $X \neq 0$, then $\operatorname{Re}(\bar{X}^{\top}AX) > 0$, where $\operatorname{Re}(z)$ denotes the real part of $z \in \mathbb{C}$.
b) Show that $A$ is positively stable.

Verify the following property: $\forall ( A , B ) \in \left( \mathcal { M } _ { n } ( \mathbb { C } ) \right) ^ { 2 } \quad \| A B \| _ { 0 } \leqslant n \| A \| _ { 0 } \cdot \| B \| _ { 0 }$

Verify the following property: $\forall A \in \mathcal { M } _ { n } ( \mathbb { C } ) , \forall Y \in \mathbb { C } ^ { n } \quad \| A Y \| _ { \infty } \leqslant n \| A \| _ { 0 } \cdot \| Y \| _ { \infty }$

We fix a natural number $p$ greater than or equal to 2. We recall the following result: for all matrices $A _ { 1 }$ and $A _ { 2 }$ in $\mathcal { M } _ { n } ( \mathbb { C } )$, all matrices $X _ { 1 }$ and $X _ { 2 }$ in $\mathcal { M } _ { n , 1 } ( \mathbb { C } )$ and all complex numbers $\lambda _ { 1 }$ and $\lambda _ { 2 }$: $$\left( \begin{array} { c c } A _ { 1 } & X _ { 1 } \\ 0 _ { 1 , n } & \lambda _ { 1 } \end{array} \right) \left( \begin{array} { c c } A _ { 2 } & X _ { 2 } \\ 0 _ { 1 , n } & \lambda _ { 2 } \end{array} \right) = \left( \begin{array} { c c } A _ { 1 } A _ { 2 } & A _ { 1 } X _ { 2 } + \lambda _ { 2 } X _ { 1 } \\ 0 _ { 1 , n } & \lambda _ { 1 } \lambda _ { 2 } \end{array} \right)$$ Let $A \in \mathcal { M } _ { n } ( \mathbb { C } ) , X \in \mathcal { M } _ { n , 1 } ( \mathbb { C } )$ and $\lambda \in \mathbb { C }$. Prove that, for all integer $k \geqslant 1$ we have: $$\left( \begin{array} { c c } A & X \\ 0 _ { 1 , n } & \lambda \end{array} \right) ^ { k } = \left( \begin{array} { c c } A ^ { k } & X _ { k } \\ 0 _ { 1 , n } & \lambda ^ { k } \end{array} \right)$$ where $X _ { k } = \left( \sum _ { j = 0 } ^ { k - 1 } \lambda ^ { k - 1 - j } A ^ { j } \right) X$.

We fix a natural number $p$ greater than or equal to 2. We denote $\mathcal { V } _ { p } = \left\{ \mathrm { e } ^ { \frac { 2 \mathrm { i } k \pi } { p } } ; k \in \llbracket 1 , p - 1 \rrbracket \right\}$, the set of $p$-th roots of unity different from 1. Let $A = \left( a _ { i , j } \right)$ be a matrix in $\mathcal { M } _ { n } ( \mathbb { C } )$ that is upper triangular and invertible. Let $\lambda$ be a non-zero complex number. Assume that, for all $i \in \llbracket 1 , n \rrbracket , \frac { a _ { i , i } } { \lambda } \notin \mathcal { V } _ { p }$. Prove that the matrix $\sum _ { j = 0 } ^ { p - 1 } \lambda ^ { p - 1 - j } A ^ { j }$ is invertible.

We fix a natural number $p$ greater than or equal to 2. Show that every upper triangular and invertible matrix admits at least one upper triangular $p$-th root.
One may prove by induction on $n \geqslant 1$ the following property: $$\forall B \in \mathcal { T } _ { n } ( \mathbb { C } ) \cap \mathrm { GL } _ { n } ( \mathbb { C } ) , \quad \exists A \in \mathcal { T } _ { n } ( \mathbb { C } ) \quad \text { such that } \left\{ \begin{array} { l } A ^ { p } = B \\ \forall ( i , j ) \in \llbracket 1 , n \rrbracket ^ { 2 } , \frac { a _ { i , i } } { a _ { j , j } } \notin \mathcal { V } _ { p } \end{array} \right.$$