We set $D = (d_{ij})_{(i,j) \in \llbracket 1,n\rrbracket^2} = (\sqrt{m_{ij}})_{(i,j) \in \llbracket 1,n\rrbracket^2} \in \mathcal{M}_n(\mathbb{R})$ and $M_c = \left((d_{ij} + c\xi_i^j)^2\right)$ with $c > 0$, where $\xi_i^j = \begin{cases}0 & \text{if } i = j \\ 1 & \text{otherwise}\end{cases}$. Show that, for all $X \in \mathbb{R}^n$, $${}^t X \Psi(M_c) X = {}^t X \Psi(M) X + 2c\, {}^t X \Psi(D) X + \frac{c^2}{2}\, {}^t X P X$$
Throughout this part, $n$ denotes an integer greater than or equal to 3, and $p = [(n+1)/2]$ is the integer part of $(n+1)/2$. Let $a = (a_0, \ldots, a_{2n-2})$ be an element of $\mathbb{R}^{2n-1}$ and $M = H(a)$. We denote $\operatorname{Spo}(M) = (\lambda_1, \ldots, \lambda_n)$. We define two vectors $v = (v_1, \ldots, v_n)$ and $w = (w_1, \ldots, w_n)$ of $\mathbb{R}^n$ by $$\begin{cases} v_i = \sqrt{2i-1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2i-1}} & \text{if } i \in \{1, \ldots, p\} \\ v_i = \sqrt{2n-2i+1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2n-2i+1}} & \text{if } i \in \{p+1, \ldots, n\} \end{cases}$$ Finally, we set $K_n = n - \|w\|^2$. Show that $\langle v, w \rangle = \sum_{i=1}^{n} \lambda_i$ and $\|v\|^2 \leqslant \sum_{i=1}^{n} \lambda_i^2$.
Throughout this part, $n$ denotes an integer greater than or equal to 3, and $p = [(n+1)/2]$ is the integer part of $(n+1)/2$. Let $a = (a_0, \ldots, a_{2n-2})$ be an element of $\mathbb{R}^{2n-1}$ and $M = H(a)$. We denote $\operatorname{Spo}(M) = (\lambda_1, \ldots, \lambda_n)$. We define two vectors $v = (v_1, \ldots, v_n)$ and $w = (w_1, \ldots, w_n)$ of $\mathbb{R}^n$ by $$\begin{cases} v_i = \sqrt{2i-1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2i-1}} & \text{if } i \in \{1, \ldots, p\} \\ v_i = \sqrt{2n-2i+1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2n-2i+1}} & \text{if } i \in \{p+1, \ldots, n\} \end{cases}$$ Finally, we set $K_n = n - \|w\|^2$. Show that $\displaystyle\sum_{1 \leqslant i < j \leqslant n} (\lambda_i - \lambda_j)^2 = n \sum_{i=1}^{n} \lambda_i^2 - \langle v, w \rangle^2$ and deduce the inequality: $$\sum_{1 \leqslant i < j \leqslant n} (\lambda_i - \lambda_j)^2 \geqslant K_n \sum_{i=1}^{n} \lambda_i^2 \tag{III.1}$$
Throughout this part, $n$ denotes an integer greater than or equal to 3, and $p = [(n+1)/2]$ is the integer part of $(n+1)/2$. Verify that if $n = 3$, condition III.1 is equivalent to: $2(\lambda_1^2 + \lambda_2^2 + \lambda_3^2) \geqslant 3(\lambda_1\lambda_2 + \lambda_1\lambda_3 + \lambda_2\lambda_3)$.
Let $A = \left( a _ { i j } \right) _ { 1 \leqslant i , j \leqslant n } \in \mathcal { M } _ { n } ( \mathbb { R } )$. We define $R ( A ) = \left\{ { } ^ { t } X A X \mid X \in \mathbb { R } ^ { n } , \| X \| = 1 \right\}$. We consider two real numbers $a \in R ( A )$ and $b \in R ( A )$, with $a < b$. Let $X _ { 1 }$ and $X _ { 2 }$ be two vectors of norm 1 such that ${ } ^ { t } X _ { 1 } A X _ { 1 } = a$, ${ } ^ { t } X _ { 2 } A X _ { 2 } = b$. I.C.1) Prove that $X _ { 1 }$ and $X _ { 2 }$ are linearly independent. I.C.2) We set $X _ { \lambda } = \lambda X _ { 1 } + ( 1 - \lambda ) X _ { 2 }$ for $0 \leqslant \lambda \leqslant 1$. Prove that the function $\phi : \lambda \mapsto \frac { { } ^ { t } X _ { \lambda } A X _ { \lambda } } { \left\| X _ { \lambda } \right\| ^ { 2 } }$ is defined and continuous on the interval $[ 0,1 ]$. I.C.3) Deduce that the segment $[ a , b ]$ is included in $R ( A )$.
Throughout this part $A$ and $B$ denote real symmetric matrices of $\mathcal { M } _ { 2 } ( \mathbb { R } )$. We denote by $\lambda _ { 1 } \leqslant \lambda _ { 2 }$ (resp. $\mu _ { 1 } \leqslant \mu _ { 2 }$) the eigenvalues of $A$ (resp. $B$). Prove that $\operatorname { Tr } ( A B ) \leqslant \lambda _ { 1 } \mu _ { 1 } + \lambda _ { 2 } \mu _ { 2 }$. One may use an orthogonal matrix $P$ such that ${ } ^ { t } P B P$ is a diagonal matrix, to obtain ${ } ^ { t } P A P = A ^ { \prime } = \left( a _ { i j } ^ { \prime } \right)$ with $\operatorname { Tr } ( A ) = \lambda _ { 1 } + \lambda _ { 2 } = a _ { 11 } ^ { \prime } + a _ { 22 } ^ { \prime }$.
Throughout this part $A$ and $B$ denote real symmetric matrices of $\mathcal { M } _ { 2 } ( \mathbb { R } )$. We set $$A = \left( \begin{array} { l l } a _ { 1 } & b _ { 1 } \\ b _ { 1 } & d _ { 1 } \end{array} \right) \quad B = \left( \begin{array} { l l } a _ { 2 } & b _ { 2 } \\ b _ { 2 } & d _ { 2 } \end{array} \right)$$ We assume in this section that $A \geqslant 0$ and $B \geqslant 0$. II.E.1) By applying the Cauchy-Schwarz inequality to the vectors $( b _ { 1 } , \sqrt { \operatorname { det } A }$ ) and $( b _ { 2 } , \sqrt { \operatorname { det } B }$ ), prove that $$b _ { 1 } b _ { 2 } \leqslant \sqrt { a _ { 1 } a _ { 2 } d _ { 1 } d _ { 2 } } - \sqrt { \operatorname { det } A \operatorname { det } B }$$ II.E.2) By computing $\operatorname { det } ( A + B ) - \operatorname { det } A - \operatorname { det } B$, deduce that $$\operatorname { det } ( A + B ) \geqslant \operatorname { det } ( A ) + \operatorname { det } ( B ) + 2 \sqrt { \operatorname { det } ( A ) \operatorname { det } ( B ) }$$
We consider the following relation on the set of real symmetric matrices of format $(2,2)$: we say that $S \leqslant S ^ { \prime }$ if and only if the symmetric matrix $S ^ { \prime } - S$ satisfies $S ^ { \prime } - S \geqslant 0$ (i.e., all eigenvalues of $S' - S$ are $\geqslant 0$). Prove that the relation $\leqslant$ above is indeed an order relation on real symmetric matrices of format $(2,2)$.
We consider a sequence $\left( A _ { n } \right) _ { n \geqslant 0 }$ $$A _ { n } = \left( \begin{array} { l l } a _ { n } & b _ { n } \\ b _ { n } & d _ { n } \end{array} \right)$$ of symmetric matrices of $\mathcal { M } _ { 2 } ( \mathbb { R } )$. We say that $S \leqslant S'$ if and only if $S' - S \geqslant 0$. We assume that the sequence $\left( A _ { n } \right) _ { n \geqslant 0 }$ is increasing and bounded for this order relation. II.H.1) Prove that for every vector $X$, the sequence $\left( { } ^ { t } X A _ { n } X \right) _ { n \geqslant 0 }$ is increasing and bounded. II.H.2) Prove that the sequences $\left( a _ { n } \right) _ { n \geqslant 0 }$ and $\left( d _ { n } \right) _ { n \geqslant 0 }$ are increasing and bounded. II.H.3) By considering the vector $X = ( 1,1 )$, prove that the sequence of matrices $\left( A _ { n } \right) _ { n \geqslant 0 }$ is convergent in $\mathcal { M } _ { 2 } ( \mathbb { R } )$, that is, the sequences $\left( a _ { n } \right) _ { n \geqslant 0 }$, $\left( b _ { n } \right) _ { n \geqslant 0 }$ and $\left( d _ { n } \right) _ { n \geqslant 0 }$ are convergent in $\mathbb { R }$.
Let $f$ be a linear form on $\mathcal{M}_n(\mathbb{R})$, and let $A$ be the unique matrix in $\mathcal{M}_n(\mathbb{R})$ such that $\forall M \in \mathcal{M}_n(\mathbb{R}), f(M) = \operatorname{Tr}(AM)$. Justify the existence of $M_n = \sup(\{f(O), O \in \mathrm{O}(n)\})$.
Let $B \in M_3(\mathbb{R})$ be antisymmetric. Assume that $$B = P \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -\beta \\ 0 & \beta & 0 \end{pmatrix} P^{-1}$$ for some $P \in O_3(\mathbb{R})$ and $\beta \in \mathbb{R}$. Show that $|\beta| = \frac{\|B\|_2}{\sqrt{2}}$.
We are given a matrix $S \in \mathbf{S}_n$, an angle $\theta \in \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[$ and integers $1 \leqslant p < q \leqslant n$ such that $s_{pq} \neq 0$. We define $S' = {}^t R_{p,q}(\theta) S R_{p,q}(\theta)$ and denote its coefficients by $s_{ij}'$. Suppose in this question that $s_{pq}$ is the coefficient of largest absolute value in $E$. (a) Show that $\|E'\| \leqslant \rho \|E\|$ where $\rho < 1$ is a constant that we will make explicit. (b) If we choose the root $t_0$, show furthermore that $\|D' - D\| \leqslant \|E\|$.
We are given a matrix $S \in \mathbf{S}_n$, an angle $\theta \in \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[$ and integers $1 \leqslant p < q \leqslant n$ such that $s_{pq} \neq 0$. We define $S' = {}^t R_{p,q}(\theta) S R_{p,q}(\theta)$ and denote its coefficients by $s_{ij}'$. From now on we choose the root $t_0$ of (1). We define $$R = \sum_{i,j=1}^{n} \left|s_{jj} - s_{ii}\right| \quad \text{and} \quad R' = \sum_{i,j=1}^{n} \left|s_{jj}' - s_{ii}'\right|$$ Show that $$R' - R \geqslant 2\left(\left|s_{qq}' - s_{qq}\right| + \left|s_{pp}' - s_{pp}\right|\right) = 2\sum_{i=1}^{n} \left|s_{ii}' - s_{ii}\right|$$
In the Jacobi algorithm, we start with a matrix $\Sigma \in \mathbf{S}_n$ and construct a sequence of symmetric matrices $\Sigma^{(m)}$, whose coefficients are denoted $\sigma_{ij}^{(m)}$, as follows:
We set $\Sigma^{(0)} = \Sigma$.
When $\Sigma^{(m)}$ is known, we choose a pair $(p_m, q_m)$ with $p_m < q_m$.
We then apply the calculations from Part 2 to the matrix $S = \Sigma^{(m)}$ and the pair $(p,q) = (p_m, q_m)$: we form the matrix $S'$ studied in this part, and call it $\Sigma^{(m+1)}$.
We define $$R_m = \sum_{i,j=1}^{n} \left|\sigma_{jj}^{(m)} - \sigma_{ii}^{(m)}\right|, \quad \varepsilon_m = \sum_{i=1}^{n} \left|\sigma_{ii}^{(m+1)} - \sigma_{ii}^{(m)}\right|$$ Verify that $R_{m+1} - R_m \geqslant 2\varepsilon_m$. Deduce that the series $\sum_{m=1}^{\infty} \varepsilon_m$ is convergent.
In the Jacobi algorithm, we start with a matrix $\Sigma \in \mathbf{S}_n$ and construct a sequence of symmetric matrices $\Sigma^{(m)}$, whose coefficients are denoted $\sigma_{ij}^{(m)}$. We decompose $\Sigma^{(m)}$ in the form $D^{(m)} + E^{(m)}$ where $D^{(m)}$ is diagonal and $E^{(m)}$ has zero diagonal. Show that the sequence $\left(D^{(m)}\right)_{m \in \mathbb{N}}$ is convergent. We denote its limit by $D$.
In the optimal version of the Jacobi algorithm, we choose for each $m$ a pair $(p_m, q_m)$ such that the absolute value of the coefficient $\sigma_{ij}^{(m)}$ is maximal precisely when $(i,j) = (p_m, q_m)$. In other words, $$\forall i < j, \quad \left|\sigma_{ij}^{(m)}\right| \leqslant \left|\sigma_{p_m q_m}^{(m)}\right|$$ Show that as $m \rightarrow +\infty$, the sequence $\left(\Sigma^{(m)}\right)_{m \in \mathbb{N}}$ converges to the diagonal matrix $D$.
In the optimal version of the Jacobi algorithm, we choose for each $m$ a pair $(p_m, q_m)$ such that the absolute value of the coefficient $\sigma_{ij}^{(m)}$ is maximal precisely when $(i,j) = (p_m, q_m)$. In other words, $$\forall i < j, \quad \left|\sigma_{ij}^{(m)}\right| \leqslant \left|\sigma_{p_m q_m}^{(m)}\right|$$ Let $m \in \mathbb{N}$. Show that $$\left\|D - D^{(m)}\right\| \leqslant \frac{\rho^m}{1 - \rho} \left\|E^{(0)}\right\|$$
Show that, for every polynomial $P \in \mathbb{C}[X]$, the map $f_P : A \mapsto P(A)$ is a continuous function from $\mathcal{M}_d(\mathbb{R})$ to $\mathcal{M}_d(\mathbb{C})$.
In the rest of the problem, we denote by $\|\cdot\|$ the norm associated with the inner product $(A, B) \mapsto \operatorname{Tr}({}^t A \times B)$. For all integers $i, j$ between 1 and $d$ and every matrix $A \in \mathcal{M}_d(\mathbb{R})$, compare $\left|A_{i,j}\right|$ and $\|A\|$.
In the rest of the problem, we denote by $\|\cdot\|$ the norm associated with the inner product $(A, B) \mapsto \operatorname{Tr}({}^t A \times B)$. Show that: $\forall (A, B) \in \mathcal{M}_d(\mathbb{R})^2, \|A \times B\| \leqslant \|A\| \cdot \|B\|$.
In the rest of the problem, we denote by $\|\cdot\|$ the norm associated with the inner product $(A, B) \mapsto \operatorname{Tr}({}^t A \times B)$. For $n \in \mathbb{N}^*$ and $A \in \mathcal{M}_d(\mathbb{R})$, compare $\left\|A^n\right\|$ and $\|A\|^n$.
Let $(D_n)_{n \in \mathbf{N}}$ be a sequence of $\mathcal{M}_d(\mathbf{R})$ that converges to $D \in \mathcal{M}_d(\mathbf{R})$. It is therefore bounded: let $\lambda > 0$ be such that for all integers $n \in \mathbf{N}$, $\|D_n\| \leq \lambda$. (a) Let $k \in \mathbf{N}$. Justify that $\frac{n!}{(n-k)! n^k} \rightarrow 1$ when $n \rightarrow +\infty$ and that if $n \geq k$ (and $n \geq 1$), $$0 \leq 1 - \frac{n!}{(n-k)! n^k} \leq 1$$ Deduce that $$\left(I_d + \frac{D_n}{n}\right)^n - \sum_{k=0}^{n} \frac{1}{k!}(D_n)^k \rightarrow 0 \quad \text{when } n \rightarrow +\infty$$ (b) Show that for all integers $k \geq 1$ and $n \geq 0$, $$\left\|(D_n)^k - D^k\right\| \leq k\lambda^{k-1}\|D_n - D\|$$ (c) Conclude that $\left(I_d + \frac{D_n}{n}\right)^n \rightarrow \exp(D)$ when $n \rightarrow +\infty$.