Let $A$ and $B$ be two arbitrary matrices of $\mathcal{M}_d(\mathbf{R})$. Using the results of questions 8 and 9, deduce that $$\exp(A+B) = \lim_{n \rightarrow +\infty} \left(\exp\left(\frac{A}{n}\right)\exp\left(\frac{B}{n}\right)\right)^n$$

Let $H = l^2(\mathbb{N})$, the vector space of real sequences that are square-summable: $$l^2(\mathbb{N}) = \left\{(u_n)_{n \in \mathbb{N}} \in \mathbb{R}^{\mathbb{N}}, \quad \sum_{n=0}^{+\infty} |u_n|^2 < +\infty\right\}$$ equipped with the norm: $$\|u\|_2 = \sqrt{\sum_{n=0}^{+\infty} |u_n|^2}$$ We denote by $S$, respectively $V$, the left shift application: $(Su)_n = u_{n-1}$ if $n \geq 1$ and $(Su)_0 = 0$, respectively the right shift: $(Vu)_n = u_{n+1}$ if $n \geq 0$ in $H = l^2(\mathbb{N})$.
Show that $S$ and $V$ belong to $\mathcal{L}(H)$.

We now work in the space of bounded real sequences $F = l^{\infty}(\mathbb{N})$ equipped with the norm $$\|u\|_{\infty} = \sup_{n \in \mathbb{N}} |u_n|$$ We denote by $S$, respectively $V$, the left shift application: $(Su)_n = u_{n-1}$ if $n \geq 1$ and $(Su)_0 = 0$, respectively the right shift: $(Vu)_n = u_{n+1}$ if $n \geq 0$.
Show that $S$ and $V$ belong to $\mathcal{L}(F)$.

We recall that a pre-Hilbert space $H$ is a normed vector space whose norm is derived from an inner product denoted $\langle .,. \rangle$. We call Hilbert basis of $H$ any family $B = (b_i)_{i \in \mathbb{N}}$ such that: (i) the family is orthonormal: for all $i$ and $j$ in $\mathbb{N}$, $\langle b_i, b_j \rangle = 1$ if $i = j$ and $0$ otherwise. (ii) every element $x$ of $H$ can be written: $x = \sum_{i=0}^{+\infty} \langle x, b_i \rangle b_i$, that is $$\lim_{N \rightarrow +\infty} \left\| x - \sum_{i=0}^{N} \langle x, b_i \rangle b_i \right\| = 0$$
Show that if $B = (b_i)_{i \in \mathbb{N}}$ is a Hilbert basis of $H$, then $$\forall x \in H, \quad \|x\|^2 = \sum_{i=0}^{+\infty} |\langle x, b_i \rangle|^2$$

Show that $H = l^2(\mathbb{N})$ equipped with the norm $\|.\|_2 = \sqrt{\sum_{n=0}^{+\infty} |u_n|^2}$ is a pre-Hilbert space for the inner product: $$\langle u, v \rangle = \sum_{n=0}^{+\infty} u_n v_n$$ (justify that this is indeed an inner product) then determine a Hilbert basis of $H$.

In $H = l^2(\mathbb{N})$ equipped with the inner product $\langle u, v \rangle = \sum_{n=0}^{+\infty} u_n v_n$, let $T$ be an operator on $H$. We admit the existence of an operator $\tilde{T} \in \mathcal{L}(H)$ such that $$\forall (x,y) \in H^2, \quad \langle T(x), y \rangle = \langle x, \tilde{T}(y) \rangle$$ Let $B = (b_i)_{i \in \mathbb{N}}$ and $C = (c_i)_{i \in \mathbb{N}}$ be two Hilbert bases of $H$ such that $$\sum_{i=0}^{+\infty} \|T(b_i)\|^2 < +\infty$$ Show that $$\sum_{i=0}^{+\infty} \|T(b_i)\|^2 = \sum_{i=0}^{+\infty} \|\tilde{T}(c_i)\|^2$$

In $H = l^2(\mathbb{N})$ equipped with the inner product $\langle u, v \rangle = \sum_{n=0}^{+\infty} u_n v_n$, let $B = (b_i)_{i \in \mathbb{N}}$ be a Hilbert basis of $H$ and $T \in \mathcal{L}(H)$. Show that the quantity (possibly infinite) $$\sum_{i=0}^{+\infty} \|T(b_i)\|^2$$ does not depend on the basis $B$. We denote $$\|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2$$ and we set $$\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 < +\infty\right\}$$

In $H = l^2(\mathbb{N})$ equipped with the inner product $\langle u, v \rangle = \sum_{n=0}^{+\infty} u_n v_n$, with $S$ the left shift $(Su)_n = u_{n-1}$ if $n \geq 1$, $(Su)_0 = 0$, and $V$ the right shift $(Vu)_n = u_{n+1}$, and $$\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2 < +\infty\right\}$$ Show that the operators $S$ and $V$ defined in part 2 are not in $\mathcal{L}^2(H)$. Give an example of a non-zero operator in $\mathcal{L}^2(H)$.

With $\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 < +\infty\right\}$ where $\|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2$ for any Hilbert basis $B = (b_i)_{i \in \mathbb{N}}$ of $H = l^2(\mathbb{N})$,
Show that $\mathcal{L}^2(H)$ equipped with $\|.\|_2$ has the structure of a normed vector space.

With $\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 < +\infty\right\}$ where $\|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2$ for any Hilbert basis $B = (b_i)_{i \in \mathbb{N}}$ of $H = l^2(\mathbb{N})$,
Let $L$ and $U$ be in $\mathcal{L}^2(H)$ and $B = (b_i)_{i \in \mathbb{N}}$ a Hilbert basis of $H$. Show that the quantity $$\sum_{i=0}^{+\infty} \langle L(b_i), U(b_i) \rangle$$ is finite, independent of the basis $B$ chosen, and defines an inner product on $\mathcal{L}^2(H)$.

With $\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 < +\infty\right\}$ where $\|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2$ for any Hilbert basis $B = (b_i)_{i \in \mathbb{N}}$ of $H = l^2(\mathbb{N})$,
We consider $L$ and $U$ two operators in $\mathcal{L}(H)$. Show that if $L \in \mathcal{L}^2(H)$, then so is $UL$.

With $\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 < +\infty\right\}$ where $\|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2$ for any Hilbert basis $B = (b_i)_{i \in \mathbb{N}}$ of $H = l^2(\mathbb{N})$,
We consider $L$ and $U$ two operators in $\mathcal{L}(H)$. What happens for $UL$ assuming this time that $U \in \mathcal{L}^2(H)$?

Let $M \in \mathcal{S}_{n}(\mathbb{R})$, we denote by $m = s^{\downarrow}(M)$ its ordered spectrum, and $\left(v_{1}, \ldots, v_{n}\right)$ an orthonormal basis from the spectral resolution of $M$. Calculate $$\sup_{\|x\|=1} \langle x, Mx \rangle$$ as a function of the coordinates of $m$. Is this supremum attained? (One may decompose $x$ and $Mx$ on the orthonormal basis $\left(v_{1}, \ldots, v_{n}\right)$ of question 2a).

In this part, we consider two real symmetric matrices $A, B \in \mathcal{S}_{n}(\mathbb{R})$ and their sum $C = A + B$. We denote by $a = s^{\downarrow}(A)$, $b = s^{\downarrow}(B)$ and $c = s^{\downarrow}(C)$.
Show that $a_{1} + b_{1} \geqslant c_{1}$.

In this part, we consider two real symmetric matrices $A, B \in \mathcal{S}_{n}(\mathbb{R})$ and their sum $C = A + B$. We denote by $a = s^{\downarrow}(A)$, $b = s^{\downarrow}(B)$ and $c = s^{\downarrow}(C)$.
Show that $a_{n} + b_{n} \leqslant c_{n}$.

In this part, we consider two real symmetric matrices $A, B \in \mathcal{S}_{n}(\mathbb{R})$ and their sum $C = A + B$. We denote by $a = s^{\downarrow}(A)$, $b = s^{\downarrow}(B)$ and $c = s^{\downarrow}(C)$.
By using spectral resolutions of $A$, $B$ and $C$, show that if the strictly positive integers $j$ and $k$ satisfy $j + k \leqslant n + 1$, we have $$c_{j+k-1} \leqslant a_{j} + b_{k}.$$ Deduce that for every integer $j$, $1 \leqslant j \leqslant n$, $$a_{j} + b_{n} \leqslant c_{j}.$$

We denote by $a_{ii}$ for $1 \leqslant i \leqslant n$ the diagonal elements of $A \in \mathcal{S}_{n}(\mathbb{R})$ with ordered spectrum $a = s^{\downarrow}(A)$. Prove that $a_{11} \leqslant a_{1}$.

Let $j$ and $k$ be non-negative integers such that $1 \leqslant j < k$ and $s_{1} \geqslant s_{2} \geqslant \cdots \geqslant s_{k}$ be real numbers. We define $\mathcal{D}_{j,k} = \left\{\left(t_{1}, \ldots, t_{k}\right) \in [0,1]^{k} \mid t_{1} + \cdots + t_{k} = j\right\}$ and $f$ the function from $\mathcal{D}_{j,k}$ to $\mathbb{R}$ defined by $$f\left(t_{1}, \ldots, t_{k}\right) = \sum_{i=1}^{k} s_{i} t_{i}.$$ Prove that for every $\left(t_{1}, \ldots, t_{k}\right) \in \mathcal{D}_{j,k}$, $$\sum_{i=1}^{j} s_{i} - f\left(t_{1}, \ldots, t_{k}\right) \geqslant \sum_{i=1}^{j} \left(s_{i} - s_{j}\right)\left(1 - t_{i}\right).$$ Deduce that $$\sup_{\mathcal{D}_{j,k}} f = \sum_{i=1}^{j} s_{i}.$$

We denote by $a_{ii}$ for $1 \leqslant i \leqslant n$ the diagonal elements of $A \in \mathcal{S}_{n}(\mathbb{R})$ with ordered spectrum $a = s^{\downarrow}(A)$. Show that, more generally than in 8a, we have for every integer $1 \leqslant j \leqslant n$ $$\sum_{i=1}^{j} a_{ii} \leqslant \sum_{i=1}^{j} a_{i}.$$

We denote by $a_{ii}$ for $1 \leqslant i \leqslant n$ the diagonal elements of $A \in \mathcal{S}_{n}(\mathbb{R})$ with ordered spectrum $a = s^{\downarrow}(A)$. Deduce that for every integer $1 \leqslant j \leqslant n$ $$\sum_{i=1}^{j} a_{i} = \sup_{\left(x_{1}, \ldots, x_{j}\right) \in \mathcal{R}_{j}} \sum_{i=1}^{j} \langle x_{i}, A x_{i} \rangle,$$ where $\mathcal{R}_{j}$ is the set of orthonormal families of cardinality $j$ in $\mathbb{R}^{n}$.

In this part, we consider two real symmetric matrices $A, B \in \mathcal{S}_{n}(\mathbb{R})$ and their sum $C = A + B$. We denote by $a = s^{\downarrow}(A)$, $b = s^{\downarrow}(B)$ and $c = s^{\downarrow}(C)$. We denote by $\mathcal{R}_{j}$ the set of orthonormal families of cardinality $j$ in $\mathbb{R}^{n}$.
Conclude that for every integer $1 \leqslant j \leqslant n$ $$\sum_{i=1}^{j} c_{i} \leqslant \sum_{i=1}^{j} a_{i} + \sum_{i=1}^{j} b_{i}.$$

For $m \in \mathbb{R}$, we denote by $\mathcal{M}$ the matrix $$\mathcal{M} = \left(\begin{array}{ccc} 0 & -m & 0 \\ m & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$
Give a necessary and sufficient condition on $m$ for the sequence $\left((I_{3} + \mathcal{M})^{n}\right)_{n \in \mathbb{N}}$ to converge in $\mathcal{M}_{3}(\mathbb{R})$.

For all $(M, N) \in (\mathcal{M}_n(\mathbb{R}))^2$, we denote $$(M \mid N) = \operatorname{tr}({}^t M N)$$ Prove that this defines an inner product on $\mathcal{M}_n(\mathbb{R})$. Explicitly express $(M \mid N)$ in terms of the coefficients of $M$ and $N$.

We denote by $\|M\|$ the Euclidean norm associated with the inner product $(M \mid N) = \operatorname{tr}({}^t M N)$. Fix $A \in \mathcal{M}_n(\mathbb{R})$, prove that there exists a matrix $M \in \mathcal{Y}_n$ such that: $$\forall N \in \mathcal{Y}_n \quad \|A - M\| \leqslant \|A - N\|$$

We denote by $\|M\|$ the Euclidean norm associated with the inner product $(M \mid N) = \operatorname{tr}({}^t M N)$. Justify the uniqueness of the matrix $M \in \mathcal{Y}_n$ minimizing $\|A - M\|$ over $\mathcal{Y}_n$ and explicitly express its coefficients in terms of those of $A$.