A $5 \times 5$ real matrix has an eigenvector in $\mathbb{R}^5$.

Let $A : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ be a linear transformation with eigenvalues $\frac { 2 } { 3 }$ and $\frac { 9 } { 5 }$. Then, there exists a non-zero vector $v \in \mathbb { R } ^ { 2 }$ such that
(a) $\| A v \| > 2 \| v \|$;
(b) $\| A v \| < \frac { 1 } { 2 } \| v \|$;
(c) $\| A v \| = \| v \|$;
(d) $A v = 0$;

Let $M _ { n } ( \mathbb { C } )$ denote the set of $n \times n$ matrices over $\mathbb { C }$. Think of $M _ { n } ( \mathbb { C } )$ as the $2 n ^ { 2 }$-dimensional Euclidean space $\mathbb { R } ^ { 2 n ^ { 2 } }$. Show that the set of all diagonalizable $n \times n$ matrices is dense in $M _ { n } ( \mathbb { C } )$.

Let $v$ be a (fixed) unit vector in $\mathbb { R } ^ { 3 }$. (We think of elements of $\mathbb { R } ^ { n }$ as column vectors.) Let $M = I _ { 3 } - 2 v v ^ { t }$. Pick the correct statement(s) from below.
(A) $O$ is an eigenvalue of $M$.
(B) $M ^ { 2 } = I _ { 3 }$.
(C) 1 is an eigenvalue of $M$.
(D) The eigenspace for the eigenvalue $-1$ is 2-dimensional.

Let $A \in \mathrm { GL } ( 3 , \mathbb { Q } )$ with $A ^ { t } A = I _ { 3 }$. Assume that $$A \left[ \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right] = \lambda \left[ \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right]$$ for some $\lambda \in \mathbb { C }$.
(A) Determine the possible values of $\lambda$.
(B) Determine $x + y + z$ where $x , y , z$ is given by $$\left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = A \left[ \begin{array} { c } 1 \\ - 1 \\ 0 \end{array} \right]$$

Consider the real matrix
$$A = \left( \begin{array} { l l } \lambda & 2 \\ 3 & 5 \end{array} \right)$$
Assume that $-1$ is an eigenvalue of $A$. Which of the following are true?
(A) The other eigenvalue is in $\mathbb { C } \backslash \mathbb { R }$.
(B) $A + I _ { 2 }$ is singular.
(C) $\lambda = 1$.
(D) Trace of $A$ is 5.

The system of linear equations in $x$ and $y$ $$\left( \begin{array} { c c } 5 - \log _ { 2 } a & 2 \\ 3 & \log _ { 2 } a \end{array} \right) \binom { x } { y } = \binom { 0 } { 0 }$$ has a solution other than $x = 0 , y = 0$. What is the sum of all values of $a$? [3 points]
(1) 8
(2) 10
(3) 12
(4) 16
(5) 20

In this section, the dimension of $E$ is 12. We assume that there exists in $\mathscr{L}(E)$, a family $(f_{1}, f_{2}, f_{3}, f_{4})$ of antisymmetric orthogonal automorphisms satisfying: $\forall i \neq j, f_{i}f_{j} + f_{j}f_{i} = 0$. By using $f_{4}$, show that $f_{3}$ cannot be equal to $\pm f_{1}f_{2}$.

In this section, the dimension of $E$ is 12. We assume that there exists in $\mathscr{L}(E)$, a family $(f_{1}, f_{2}, f_{3}, f_{4})$ of antisymmetric orthogonal automorphisms satisfying: $\forall i \neq j, f_{i}f_{j} + f_{j}f_{i} = 0$. Show that $f_{1}f_{2}f_{3}$ is an orthogonal automorphism, symmetric and not collinear to $\operatorname{Id}_{E}$.

In this section, the dimension of $E$ is 12. We assume that there exists in $\mathscr{L}(E)$, a family $(f_{1}, f_{2}, f_{3}, f_{4})$ of antisymmetric orthogonal automorphisms satisfying: $\forall i \neq j, f_{i}f_{j} + f_{j}f_{i} = 0$. What is the spectrum of $f_{1}f_{2}f_{3}$? Show that there exists $x \in E$ of norm 1 such that $\langle f_{1}f_{2}f_{3}(x), x \rangle = 0$. We fix such an $x$ for the rest.

In this section, the dimension of $E$ is 12. We assume that there exists in $\mathscr{L}(E)$, a family $(f_{1}, f_{2}, f_{3}, f_{4})$ of antisymmetric orthogonal automorphisms satisfying: $\forall i \neq j, f_{i}f_{j} + f_{j}f_{i} = 0$. We fix $x \in E$ of norm 1 such that $\langle f_{1}f_{2}f_{3}(x), x \rangle = 0$. Show that $F = (x, f_{1}(x), f_{2}(x), f_{3}(x), f_{1}f_{2}(x), f_{1}f_{3}(x), f_{2}f_{3}(x), f_{1}f_{2}f_{3}(x))$ is an orthonormal family.

In this section, the dimension of $E$ is 12. We assume that there exists in $\mathscr{L}(E)$, a family $(f_{1}, f_{2}, f_{3}, f_{4})$ of antisymmetric orthogonal automorphisms satisfying: $\forall i \neq j, f_{i}f_{j} + f_{j}f_{i} = 0$. We set $V = \operatorname{Vect}(F)$ where $F = (x, f_{1}(x), f_{2}(x), f_{3}(x), f_{1}f_{2}(x), f_{1}f_{3}(x), f_{2}f_{3}(x), f_{1}f_{2}f_{3}(x))$. It is therefore a vector subspace of $E$ of dimension 8. a) Show that $V^{\perp}$ is stable under $f_{1}, f_{2}, f_{3}$. b) We denote by $f_{i}^{\prime}$ the endomorphism induced by $f_{i}$ on $V^{\perp}$, $i = 1, 2, 3$. Justify that there exists $\delta^{\prime} \in \{-1, 1\}$ such that $f_{3}^{\prime} = \delta^{\prime} f_{1}^{\prime} f_{2}^{\prime}$. If necessary, by replacing $f_{3}$ by $-f_{3}$, we consider for the rest that $f_{3}^{\prime} = f_{1}^{\prime} f_{2}^{\prime}$. c) Let $e$ be fixed in $V^{\perp}$, of norm 1. By proceeding as in II.B.1.a) (but this is not to be redone), one can show that $(e, f_{1}(e), f_{2}(e), f_{1}f_{2}(e))$ is an orthonormal basis of $V^{\perp}$. By noting that $f_{3}(e) = f_{1}f_{2}(e)$, use this basis to show that: $\forall y \in V^{\perp}, f_{4}(y) \in V$. Thus $W = f_{4}(V^{\perp})$ is a vector subspace of $V$ of dimension 4. d) Show that the sum of $W$ and $V^{\perp}$ is direct and that $W \oplus V^{\perp}$ is stable under $f_{1}, f_{2}, f_{3}, f_{4}$. Then reach a contradiction.

In this section, the dimension of $E$ is 12. Deduce the value of $d_{12}$.

In this section, the dimension of $E$ is 8. Show that, for all $(x_{0}, \ldots, x_{7}) \in \mathbb{R}^{8}$, $$\left(\begin{array}{cccccccc} x_{0} & -x_{1} & -x_{2} & -x_{4} & -x_{3} & -x_{5} & -x_{6} & -x_{7} \\ x_{1} & x_{0} & -x_{4} & x_{2} & -x_{5} & x_{3} & -x_{7} & x_{6} \\ x_{2} & x_{4} & x_{0} & -x_{1} & -x_{6} & x_{7} & x_{3} & -x_{5} \\ x_{4} & -x_{2} & x_{1} & x_{0} & x_{7} & x_{6} & -x_{5} & -x_{3} \\ x_{3} & x_{5} & x_{6} & -x_{7} & x_{0} & -x_{1} & -x_{2} & x_{4} \\ x_{5} & -x_{3} & -x_{7} & -x_{6} & x_{1} & x_{0} & x_{4} & x_{2} \\ x_{6} & x_{7} & -x_{3} & x_{5} & x_{2} & -x_{4} & x_{0} & -x_{1} \\ x_{7} & -x_{6} & x_{5} & x_{3} & -x_{4} & -x_{2} & x_{1} & x_{0} \end{array}\right)$$ is a similarity matrix. What can we deduce from this?

Conjecture the value of $d_{n}$ in the general case.

Throughout this part, $n$ denotes an integer greater than or equal to 3. We say that a matrix $M = (m_{i,j})_{1 \leqslant i,j \leqslant n}$ in $\mathcal{M}_n(\mathbb{R})$ is a Hankel matrix if there exists $a = (a_0, \ldots, a_{2n-2}) \in \mathbb{R}^{2n-1}$ such that for all $i$ and $j$ in $\{1, \ldots, n\}$, $m_{i,j} = a_{i+j-2}$. Such a matrix is denoted $M = H(a)$.
Show that if $M$ is a Hankel matrix of size $n$ then it admits $n$ real eigenvalues $\lambda_1, \ldots, \lambda_n$ (each repeated as many times as its multiplicity) which can be ordered in decreasing order $\lambda_1 \geqslant \lambda_2 \geqslant \ldots \geqslant \lambda_n$.

Throughout this part, $n$ denotes an integer greater than or equal to 3. We say that a matrix $M = (m_{i,j})_{1 \leqslant i,j \leqslant n}$ in $\mathcal{M}_n(\mathbb{R})$ is a Hankel matrix if there exists $a = (a_0, \ldots, a_{2n-2}) \in \mathbb{R}^{2n-1}$ such that for all $i$ and $j$ in $\{1, \ldots, n\}$, $m_{i,j} = a_{i+j-2}$. Such a matrix is denoted $M = H(a)$.
Show that if $\lambda \in \mathbb{R}^*$ then the $n$-tuple $(\lambda, \ldots, \lambda)$ is not the ordered $n$-tuple of eigenvalues of a Hankel matrix of size $n$.

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 $$\sum_{i=1}^{n} \lambda_i = \sum_{k=0}^{n-1} a_{2k} \quad \text{and} \quad \sum_{i=1}^{n} \lambda_i^2 = \sum_{k=0}^{n-1} (k+1) a_k^2 + \sum_{k=n}^{2n-2} (2n-k-1) a_k^2$$

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 $\langle v, w \rangle = \sum_{i=1}^{n} \lambda_i$ 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: $$\sum_{1 \leqslant i < j \leqslant n} (\lambda_i - \lambda_j)^2 \geqslant K_n \sum_{i=1}^{n} \lambda_i^2 \tag{III.1}$$

Let $A = \left( a _ { i j } \right) _ { 1 \leqslant i , j \leqslant n } \in \mathcal { M } _ { n } ( \mathbb { R } )$. We define $R ( A ) = \left\{ { } ^ { t } X A X \mid X \in \mathbb { R } ^ { n } , \| X \| = 1 \right\}$.
Let $Q$ be a real orthogonal matrix. Prove that $R ( A ) = R \left( { } ^ { t } Q A Q \right)$.

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 the following conditions:
(C1) $\quad \operatorname { Tr } ( A ) \in R ( A )$
(C2) There exists a real orthogonal matrix $Q$ such that the diagonal of the matrix ${ } ^ { t } Q A Q$ is of the form $( \operatorname { Tr } ( A ) , 0 , \ldots , 0 )$
I.F.1) Prove that condition (C2) implies condition (C1).
I.F.2) We assume that $x \in R ( A )$.
Prove that there exists an orthogonal matrix $Q _ { 1 }$ such that $${ } ^ { t } Q _ { 1 } A Q _ { 1 } = \left( \begin{array} { c c } x & L \\ C & B \end{array} \right)$$ where $B$ is a matrix of format $( n - 1 , n - 1 )$ $\left( B \in \mathcal { M } _ { n - 1 } ( \mathbb { R } ) \right)$, $C$ a column vector with $n - 1$ elements $\left( C \in \mathcal { M } _ { n - 1,1 } ( \mathbb { R } ) \right)$ and $L$ a row vector with $n - 1$ elements $\left( L \in \mathcal { M } _ { 1 , n - 1 } ( \mathbb { R } ) \right)$.
I.F.3) Prove that if the matrix $A$ is symmetric then so is the matrix $B$ above.
I.F.4) Prove that $\operatorname { Tr } ( A ) = \operatorname { Tr } \left( { } ^ { t } Q _ { 1 } A Q _ { 1 } \right)$.
I.F.5) Deduce that if $A$ is symmetric, condition (C1) implies condition (C2).
One may reason by induction on $n$.

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. $B$). We define $R ( A ) = \left\{ { } ^ { t } X A X \mid X \in \mathbb { R } ^ { n } , \| X \| = 1 \right\}$.
Prove that $R ( A ) = \left[ \lambda _ { 1 } , \lambda _ { 2 } \right]$.

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 $$A = \left( \begin{array} { l l } a & b \\ b & d \end{array} \right)$$ and assume $A \geqslant 0$ (i.e., all eigenvalues of $A$ are $\geqslant 0$).
II.D.1) Prove that $\operatorname { det } ( A ) \geqslant 0$.
II.D.2) Prove that ${ } ^ { t } X A X \geqslant 0$ for every vector $X$.
II.D.3) Prove that $a \geqslant 0$ and $d \geqslant 0$.
II.D.4) Let $S \in \mathcal { M } _ { 2 } ( \mathbb { R } )$ be symmetric. Prove that: $$S \geqslant 0 \quad \text { if and only if } \quad ( \operatorname { Tr } ( S ) \geqslant 0 \text { and } \operatorname { det } ( S ) \geqslant 0 )$$