Show that $\mathcal{S}_{n}(\mathbb{R})$ and $\mathcal{A}_{n}(\mathbb{R})$ are two supplementary orthogonal vector subspaces in $\mathcal{M}_{n}(\mathbb{R})$ and specify their dimensions. (The inner product on $\mathcal{M}_{n}(\mathbb{R})$ is given by $(M,N) \mapsto \operatorname{tr}(M^{\top}N)$.)
Let $A \in \mathcal{M}_{n}(\mathbb{R})$. Show that for every matrix $S \in \mathcal{S}_{n}(\mathbb{R})$, $\|A - A_{s}\|_{2} \leqslant \|A - S\|_{2}$. Specify under what condition on $S \in \mathcal{S}_{n}(\mathbb{R})$ this inequality is an equality.
If $M \in \mathcal{M}_{n}(\mathbb{R})$ and $X, Y \in \mathcal{M}_{n,1}(\mathbb{R})$, the matrix $X^{\top}MY$ belongs to $\mathcal{M}_{1}(\mathbb{R})$ and we agree to identify it with the real number equal to its unique entry. With this convention, show that $A_{s} \in \mathcal{S}_{n}^{+}(\mathbb{R})$ if and only if $\forall X \in \mathcal{M}_{n,1}(\mathbb{R}), X^{\top}A_{s}X \geqslant 0$ and that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$ if and only if $\forall X \in \mathcal{M}_{n,1}(\mathbb{R}) \setminus \{0\}, X^{\top}A_{s}X > 0$.
We consider $A \in \mathcal{M}_{n}(\mathbb{R})$. For every real eigenvalue $\lambda$ of $A$, show that $\min \operatorname{sp}_{\mathbb{R}}(A_{s}) \leqslant \lambda \leqslant \max \operatorname{sp}_{\mathbb{R}}(A_{s})$. Deduce that if $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$ then $A$ is invertible.
We consider $A \in \mathcal{M}_{n}(\mathbb{R})$ and assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$. a) Show that there exists a unique matrix $B$ in $\mathcal{S}_{n}^{++}(\mathbb{R})$ such that $B^{2} = A_{s}$. b) Show that there exists a matrix $Q$ in $\mathcal{A}_{n}(\mathbb{R})$ such that $\operatorname{det}(A) = \operatorname{det}(A_{s})\operatorname{det}(I_{n} + Q)$. c) Deduce that $\operatorname{det}(A) \geqslant \operatorname{det}(A_{s})$.
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 $A(A^{-1})_{s}A^{\top}$.
Give an example of a symmetric matrix $S$ in $\mathcal{S}_{2}(\mathbb{R})$ such that $\operatorname{sp}_{\mathbb{R}}(S) \subset [-1,1]$ and for which there does not exist a matrix $A \in \mathrm{O}_{2}(\mathbb{R})$ satisfying $A_{s} = S$.
Let $S \in \mathcal{S}_{n}(\mathbb{R})$. a) We assume that $\operatorname{sp}_{\mathbb{R}}(S) \subset [-1,1]$ and that for every eigenvalue $\lambda$ of $S$ in $]-1,1[$, the eigenspace of $S$ associated with $\lambda$ has even dimension. Show that there exists $A \in \mathrm{O}_{n}(\mathbb{R})$ such that $A_{s} = S$. b) Conversely, show that if there exists $A \in \mathrm{O}_{n}(\mathbb{R})$ such that $A_{s} = S$, then $\operatorname{sp}_{\mathbb{R}}(S) \subset [-1,1]$ and for every eigenvalue $\lambda$ of $S$ in $]-1,1[$, the eigenspace of $S$ associated with $\lambda$ has even dimension.
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 $(X \mid Y) = X^{\top}Y$. 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 $$A = A(\mu) = \begin{pmatrix} 2-\mu & -1 & \mu \\ -1 & 2-\mu & \mu-1 \\ 0 & -1 & 1 \end{pmatrix}$$ 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 $E_{n}$.
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.
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. Give an example of a positively stable matrix $A$ such that $A_{s}$ is not positive definite.
Let $\lambda \in \mathbb{C}$ such that $\operatorname{Re}(\lambda) > 0$. Let $u$ be a function with complex values of class $\mathcal{C}^{1}$ on $\mathbb{R}^{+}$. Suppose that the function $v = u' + \lambda u$ is bounded on $\mathbb{R}^{+}$. Show that $u$ is bounded on $\mathbb{R}^{+}$. One may consider the differential equation $y' + \lambda y = v$.
Let $T \in \mathcal{M}_{n}(\mathbb{C})$ be an upper triangular matrix with complex entries. Suppose that the diagonal entries of $T$ are complex numbers with strictly positive real part. Let $u_{1}, \ldots, u_{n}$ be functions with complex values, defined and of class $\mathcal{C}^{1}$ on $\mathbb{R}^{+}$ and let, for all $t \in \mathbb{R}^{+}$, $$U(t) = \begin{pmatrix} u_{1}(t) \\ \vdots \\ u_{n}(t) \end{pmatrix}$$ Suppose that, for all $t \in \mathbb{R}^{+}$, $U'(t) + TU(t) = 0$. Show that the functions $u_{j}$, where $1 \leqslant j \leqslant n$, are bounded on $\mathbb{R}^{+}$.
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. Recall that for any matrix $M \in \mathcal{M}_{n}(\mathbb{C})$, $\exp(M) = \sum_{k=0}^{\infty} \frac{M^{k}}{k!}$. Let $A \in \mathcal{M}_{n}(\mathbb{R})$ be a positively stable matrix with complex eigenvalues $\lambda_{1}, \ldots, \lambda_{n}$ and let $\alpha$ be a real number such that $0 < \alpha < \min_{1 \leqslant j \leqslant n} \operatorname{Re}(\lambda_{j})$. Show that the function $t \mapsto \mathrm{e}^{\alpha t}\exp(-tA)$ is bounded on $\mathbb{R}^{+}$. One may apply question III.B.2 to an upper triangular matrix $T$ similar to $A - \alpha I_{n}$.
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 $A \in \mathcal{M}_{n}(\mathbb{R})$ be a positively stable matrix. We consider the endomorphism $\Phi$ of $\mathcal{M}_{n}(\mathbb{R})$ such that $$\forall M \in \mathcal{M}_{n}(\mathbb{R}), \quad \Phi(M) = A^{\top}M + MA$$ Show that $\Phi$ is positively stable, that is, its matrix in any basis of $\mathcal{M}_{n}(\mathbb{R})$ 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 $A \in \mathcal{M}_{n}(\mathbb{R})$ be a positively stable matrix. We consider the endomorphism $\Phi$ of $\mathcal{M}_{n}(\mathbb{R})$ such that $\forall M \in \mathcal{M}_{n}(\mathbb{R}), \Phi(M) = A^{\top}M + MA$. a) Show that there exists a unique matrix $B \in \mathcal{M}_{n}(\mathbb{R})$ such that $A^{\top}B + BA = I_{n}$. b) Show that $B$ is symmetric and that $\operatorname{det}(B) > 0$.
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. Let $A \in \mathcal{M}_{n}(\mathbb{R})$ be a positively stable matrix. Recall that $\exp(M) = \sum_{k=0}^{\infty} \frac{M^{k}}{k!}$ for any $M \in \mathcal{M}_{n}(\mathbb{C})$. For all real $t$, we set $V(t) = \exp(-tA^{\top})\exp(-tA)$ and $W(t) = \int_{0}^{t} V(s)\,\mathrm{d}s$. a) Show that, for all real $t$, $V(t) \in \mathcal{S}_{n}^{++}(\mathbb{R})$ and that, if $t > 0$, $W(t) \in \mathcal{S}_{n}^{++}(\mathbb{R})$. b) Show that, for all real $t$, $A^{\top}W(t) + W(t)A = I_{n} - V(t)$. c) What do we obtain by letting $t$ tend to $+\infty$ in the previous equality? Deduce that the matrix $B$ of question III.C.2 is positive definite.