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 $c > 0$. The hyperplane $\mathcal{H}$ has normal vector $Z$ and equation $x_1 + \cdots + x_n = 0$.
Show that if $\lambda_{\min}$ and $\mu_{\min}$ denote the respective minimal eigenvalues of $\Psi(M)$ and $\Psi(D)$, then $$\forall X \in \mathcal{H}, \quad {}^t X \Psi(M) X \geqslant \lambda_{\min}\, {}^t XX \quad \text{and} \quad {}^t X \Psi(D) X \geqslant \mu_{\min}\, {}^t XX$$

For every real eigenvalue $\lambda$ of $A$, show that $\min \operatorname{sp}_{\mathbb{R}}\left(A_{s}\right) \leqslant \lambda \leqslant \max \operatorname{sp}_{\mathbb{R}}\left(A_{s}\right)$.
Deduce that if $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$ then $A$ is invertible.

Let $A \in \mathrm{O}_{n}(\mathbb{R})$. Show that the eigenvalues of $A_{s}$ are in $[-1,1]$.

Let $E$ be a Euclidean space of dimension $N$. Let $u$ be a self-adjoint endomorphism of $E$ such that for all $x \in E$, $q_u(x) = (u(x) \mid x) \geq 0$. We assume that 0 is a simple eigenvalue of $u$ and we denote by $\lambda_2$ the smallest nonzero eigenvalue of $u$. We denote by $p : E \rightarrow E$ the orthogonal projection onto the vector line $\ker(u)$. Show that for all $x \in E$, $q_u(x - p(x)) \geq \lambda_2 \|x - p(x)\|^2$.

Let $f$ be a symmetric endomorphism of $\mathbf{R}^n$ with eigenvalues $\lambda_1 \leqslant \ldots \leqslant \lambda_n$ and associated orthonormal eigenbasis $(e_1, \ldots, e_n)$. Let $k \in \llbracket 1, n \rrbracket$, $S_k$ a vector subspace of $\mathbf{R}^n$ of dimension $k$, and $T_k = \operatorname{Vect}(e_k, \ldots, e_n)$.
By considering $x \in S_k \cap T_k$, justify that: $$\max_{x \in S_k, \|x\|=1} (x, f(x)) \geq \lambda_k.$$

Let $f$ be a symmetric endomorphism of $\mathbf{R}^n$ with eigenvalues $\lambda_1 \leqslant \ldots \leqslant \lambda_n$ and associated orthonormal eigenbasis $(e_1, \ldots, e_n)$. For $k \in \llbracket 1, n \rrbracket$, let $\pi_k$ denote the set of vector subspaces of $\mathbf{R}^n$ of dimension $k$.
Let $k \in \llbracket 1, n \rrbracket$. Using $S = \operatorname{Vect}(e_1, \ldots, e_k) \in \pi_k$, show the equality: $$\lambda_k = \min_{S \in \pi_k} \left( \max_{x \in S, \|x\|=1} (x, f(x)) \right)$$ This is the Courant-Fischer theorem.

As in the third part, we suppose that $B = A + \mathbf{u}\mathbf{u}^T$ with $A \in \mathcal{S}_n(\mathbb{R})$ a symmetric matrix, and $\mathbf{u} \in \mathbb{R}^n$ a vector such that $\|\mathbf{u}\| = 1$. We denote by $\lambda_1 \leqslant \lambda_2 \leqslant \cdots \leqslant \lambda_n$ the eigenvalues of $A$ and $\mu_1 \leqslant \mu_2 \leqslant \cdots \leqslant \mu_n$ those of $B$. We admit that $$\lambda_1 \leqslant \mu_1 \leqslant \lambda_2 \leqslant \mu_2 \leqslant \cdots \leqslant \lambda_n \leqslant \mu_n.$$ We further suppose that there exists an integer $m \in \{1, 2, \ldots, n-1\}$ such that the eigenvalues of $A$ satisfy $$0 = \lambda_1 = \lambda_2 = \cdots = \lambda_m < \lambda_{m+1} \leqslant \cdots \leqslant \lambda_n.$$ Let $\varepsilon \in ]0, \lambda_{m+1}[$. We suppose that $\left\langle \mathbf{u}, \left(A - \varepsilon \mathbb{I}_n\right)^{-1}\mathbf{u}\right\rangle < -1$.
Show that $\mu_m > \varepsilon$.

As in the third part, we assume that $B = A + \mathbf{u u}^T$ with $A \in \mathcal{S}_n(\mathbb{R})$ a symmetric matrix, and $\mathbf{u} \in \mathbb{R}^n$ a vector such that $\|\mathbf{u}\| = 1$. We denote by $\lambda_1 \leqslant \lambda_2 \leqslant \cdots \leqslant \lambda_n$ the eigenvalues of $A$ and $\mu_1 \leqslant \mu_2 \leqslant \cdots \leqslant \mu_n$ those of $B$. We admit that $$\lambda_1 \leqslant \mu_1 \leqslant \lambda_2 \leqslant \mu_2 \leqslant \cdots \leqslant \lambda_n \leqslant \mu_n.$$ We further assume that there exists an integer $m \in \{1,2,\ldots,n-1\}$ such that the eigenvalues of $A$ satisfy $$0 = \lambda_1 = \lambda_2 = \cdots = \lambda_m < \lambda_{m+1} \leqslant \cdots \leqslant \lambda_n.$$ Let $\varepsilon \in ]0, \lambda_{m+1}[$. We suppose that $\left\langle \mathbf{u}, \left(A - \varepsilon \mathbb{I}_n\right)^{-1} \mathbf{u} \right\rangle < -1$. Show that $\mu_m > \varepsilon$.