The complex plane is equipped with a direct orthonormal coordinate system. Consider the equation
$$( E ) : \quad z ^ { 4 } + 2 z ^ { 3 } - z - 2 = 0$$
with unknown complex number $z$.

1. Give an integer solution of ( $E$ ).
2. Prove that, for every complex number $z$, $$z ^ { 4 } + 2 z ^ { 3 } - z - 2 = \left( z ^ { 2 } + z - 2 \right) \left( z ^ { 2 } + z + 1 \right) .$$
3. Solve equation ( $E$ ) in the set of complex numbers.
4. The solutions of equation ( $E$ ) are the affixes of four points $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D }$ in the complex plane such that ABCD is a non-crossed quadrilateral. Is quadrilateral ABCD a rhombus? Justify.

O produto das raízes da equação $2x^2 - 5x + 3 = 0$ é
(A) $\dfrac{2}{3}$ (B) $\dfrac{3}{2}$ (C) $\dfrac{5}{2}$ (D) $2$ (E) $3$

The function $f(x) = x^2 - 4x + 3$ has roots $x_1$ and $x_2$. What is the value of $x_1 + x_2$?
(A) $-4$
(B) $-3$
(C) $3$
(D) $4$
(E) $7$

It is given that the complex number $i - 3$ is a root of the polynomial $3 x ^ { 4 } + 10 x ^ { 3 } + A x ^ { 2 } + B x - 30$, where $A$ and $B$ are unknown real numbers. Find the other roots.

Let $x^{3} + ax^{2} + bx + 8 = 0$ be a cubic equation with integer coefficients. Suppose both $r$ and $-r$ are roots of this equation, where $r > 0$ is a real number. List all possible pairs of values $(a, b)$.

[12 points] Consider polynomials $p(x)$ with the following property, called $(\dagger)$. $(\dagger)$ If $r$ is a root of $p(x)$, then $r^{2} - 4$ is also a root of $p(x)$.
(i) We want to find every quadratic polynomial of the form $p(x) = x^{2} + bx + c$ such that $p(x)$ has two distinct roots, has integer coefficients and has property $(\dagger)$. Prove that there are exactly two such polynomials and list them.
(ii) It is also true that there are exactly two cubic polynomials of the form $p(x) = x^{3} + ax^{2} + bx + c$ with the property $(\dagger)$ such that $p(x)$ shares no root with the polynomials you found in part (i). Explain fully how you will prove this along with the method to find the polynomials, but do not try to explicitly find the polynomials.

Consider the polynomial $$p(x) = x^6 + 10x^5 + 11x^4 + 12x^3 + 13x^2 - 12x - 11.$$ Let $z_1, z_2, z_3, z_4, z_5, z_6$ be the six complex roots of $p(x)$. Evaluate $\sum_{i=1}^{6} z_i^2$. [2 points]

In this section we consider a circle $\mathcal{C}(\Omega, r)$ with center $\Omega$ and radius $r \geqslant 0$ disjoint from the $x$-axis. We denote by $K$ the orthogonal projection of $\Omega$ onto the $x$-axis. Let $A$ be a matrix with proper eigenvalue circle equal to $\mathcal{C}(\Omega, r)$. Let $U$ be one of the points of $\mathcal{C}(\Omega, r)$ at which the tangent line contains $K$.
Express the eigenvalues of $A$, considered as an element of $\mathcal{M}_2(\mathbb{C})$, using the abscissa of $K$ and the distance $KU$ from $K$ to $U$.

We define the matrix $W_n = (w_{i,j})$ in $\mathcal{M}_n(\mathbb{R})$ by $w_{i,j} = \begin{cases} 1 & \text{if } 1 \leqslant i < n \text{ and } j = i+1 \\ 1 & \text{if } i = n \text{ and } j \in \{1,2\} \\ 0 & \text{in all other cases} \end{cases}$
Show that the characteristic polynomial of $W_n$ is $X^n - X - 1$.
Deduce that $W_n^{n^2-2n+1} = \sum_{k=1}^{n-1} \binom{n-2}{k-1} W_n^k$, then that $W_n^{n^2-2n+2} = I_n + W_n + \sum_{k=2}^{n-1} \binom{n-2}{k-2} W_n^k$.

We define the matrix $W_n = (w_{i,j})$ in $\mathcal{M}_n(\mathbb{R})$ by $w_{i,j} = \begin{cases} 1 & \text{if } 1 \leqslant i < n \text{ and } j = i+1 \\ 1 & \text{if } i = n \text{ and } j \in \{1,2\} \\ 0 & \text{in all other cases} \end{cases}$
Specify the shortest circuit passing through index 1 in the matrix $W_n$.
Deduce that the positive matrix $W_n^{n^2-2n+1}$ is not strictly positive.

We define the matrix $W_n = (w_{i,j})$ in $\mathcal{M}_n(\mathbb{R})$ by $w_{i,j} = \begin{cases} 1 & \text{if } 1 \leqslant i < n \text{ and } j = i+1 \\ 1 & \text{if } i = n \text{ and } j \in \{1,2\} \\ 0 & \text{in all other cases} \end{cases}$
Show that for all $i, j$ in $\llbracket 1, n \rrbracket$, with $i \neq j$, there exists in $W_n$ at least one path with origin $i$, endpoint $j$, and length less than or equal to $n-1$.
You may treat successively the two cases $1 \notin \{i,j\}$ and $1 \in \{i,j\}$.
Deduce that the matrix $W_n^{n^2-2n+2}$ is strictly positive and conclude.

Throughout this subsection, $A$ is a given primitive matrix in $\mathcal{M}_n(\mathbb{R})$. We denote by $\ell \in \llbracket 1, n \rrbracket$ the smallest length of an elementary circuit of $A$.
By contradiction, we suppose $\ell = n$.
Show that then all circuits of $A$ have length a multiple of $n$. Deduce that the matrices $A^{kn+1}$ (with $k \in \mathbb{N}$) have zero diagonal and reach a contradiction.

Throughout this subsection, $A$ is a given primitive matrix in $\mathcal{M}_n(\mathbb{R})$. According to what precedes, there exists in $A$ an elementary circuit $\mathcal{C}$ of length $\ell \leqslant n-1$. To simplify the exposition, and because it does not affect the generality of the problem, we assume that it is the circuit $1 \rightarrow 2 \rightarrow \ldots \rightarrow \ell-1 \rightarrow \ell \rightarrow 1$ (the remaining $n-\ell$ indices $\ell+1, \ell+2, \ldots, n$ being thus located ``outside'' the circuit $\mathcal{C}$).
We will show that $A^{n+\ell(n-2)}$ is strictly positive. For this, we are given $i$ and $j$ in $\llbracket 1, n \rrbracket$. Everything comes down to establishing that there exists in $A$ a path with origin $i$, endpoint $j$ and length $n + \ell(n-2)$.
a) Show that in $A$, we can form a path with origin $i$, of length $n-\ell$, whose endpoint is in $\{1, 2, \ldots, \ell\}$ (we will denote by $k$ this endpoint). You may treat the case $1 \leqslant i \leqslant \ell$, then the case $\ell+1 \leqslant i \leqslant n$.
b) State the reason why the first $\ell$ diagonal coefficients of $A^\ell$ (and in particular the $k$-th) are strictly positive. Show then that there exists a path of length $n-1$ in $A^\ell$ (that is, a path of length $\ell(n-1)$ in $A$) with origin $k$ and endpoint $j$.
c) Finally deduce $A^{n+\ell(n-2)} > 0$, then $A^{n^2-2n+2} > 0$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, the spectral radius $\rho(A)$ is a dominant eigenvalue of $A$ and the associated eigenspace is a line that possesses a strictly positive direction vector $x > 0$. We denote $r$ the spectral radius of $A$. We denote $x$ (respectively $y$) a strictly positive direction vector of the line $D = \operatorname{Ker}(A - rI_n)$ (respectively of the line $\Delta = \operatorname{Ker}(A^\top - rI_n)$). We denote $H = \operatorname{Im}(A - rI_n)$.
Show that $H$ is the hyperplane orthogonal to the line $\Delta$ (that is $H = \Delta^\perp$).

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, the spectral radius $\rho(A)$ is a dominant eigenvalue of $A$ and the associated eigenspace is a line that possesses a strictly positive direction vector $x > 0$. We denote $r$ the spectral radius of $A$. We denote $x$ (respectively $y$) a strictly positive direction vector of the line $D = \operatorname{Ker}(A - rI_n)$ (respectively of the line $\Delta = \operatorname{Ker}(A^\top - rI_n)$). We denote $H = \operatorname{Im}(A - rI_n)$. If necessary by multiplying $y$ by an appropriate strictly positive coefficient, we assume $(y \mid x) = y^\top x = 1$. We denote $L = xy^\top$.
Prove that $L$ is the matrix, in the canonical basis, of the projection of $\mathbb{R}^n$ onto the line $D$, parallel to the hyperplane $H$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, the spectral radius $\rho(A)$ is a dominant eigenvalue of $A$ and the associated eigenspace is a line that possesses a strictly positive direction vector $x > 0$. We denote $r$ the spectral radius of $A$. We denote $x$ (respectively $y$) a strictly positive direction vector of the line $D = \operatorname{Ker}(A - rI_n)$ (respectively of the line $\Delta = \operatorname{Ker}(A^\top - rI_n)$). We denote $H = \operatorname{Im}(A - rI_n)$. If necessary by multiplying $y$ by an appropriate strictly positive coefficient, we assume $(y \mid x) = y^\top x = 1$. We denote $L = xy^\top$.
Verify that $L$ has rank 1, that it is strictly positive, and that $L^\top y = y$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, the spectral radius $\rho(A)$ is a dominant eigenvalue of $A$ and the associated eigenspace is a line that possesses a strictly positive direction vector $x > 0$. We denote $r$ the spectral radius of $A$. We denote $x$ (respectively $y$) a strictly positive direction vector of the line $D = \operatorname{Ker}(A - rI_n)$ (respectively of the line $\Delta = \operatorname{Ker}(A^\top - rI_n)$). We denote $H = \operatorname{Im}(A - rI_n)$. If necessary by multiplying $y$ by an appropriate strictly positive coefficient, we assume $(y \mid x) = y^\top x = 1$. We denote $L = xy^\top$.
Show that $AL = LA = rL$. Deduce: $\forall m \in \mathbb{N}^*, (A - rL)^m = A^m - r^m L$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$.
Let $\lambda$ be a nonzero eigenvalue of $B$ and let $z$ be an associated eigenvector.
Show that $Lz = 0$, then $Az = \lambda z$. Deduce $\rho(B) \leqslant r$.

Let $(a_0, \ldots, a_{n-1}) \in \mathbb{C}^n$. Let $\lambda$ be a complex number. By discussing in $\mathbb{C}^n$ the system $C(a_0, \ldots, a_{n-1})X = \lambda X$, show that $\lambda$ is an eigenvalue of $C(a_0, \ldots, a_{n-1})$ if and only if $\lambda$ is a root of a polynomial of $\mathbb{C}[X]$ to be specified.

Let $(a_0, \ldots, a_{n-1}) \in \mathbb{C}^n$. If $\lambda$ is a root of the polynomial identified in Q24, determine the eigenspace of $C(a_0, \ldots, a_{n-1})$ associated with the eigenvalue $\lambda$ and specify its dimension.

Let $N = \left(\begin{array}{ccccc} 0 & 0 & \cdots & \cdots & 0 \\ 1 & 0 & & & \vdots \\ 0 & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & 0 & 1 & 0 \end{array}\right)$.
Give the eigenvalues of $N$ and the associated eigenspaces. Is it diagonalizable?

Let $\left(a_0, a_1, \ldots, a_{n-1}\right) \in \mathbb{K}^n$ and $Q(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_0$. We consider the matrix
$$C_Q = \left(\begin{array}{cccccc} 0 & \cdots & \cdots & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & \cdots & 0 & -a_1 \\ 0 & 1 & \ddots & & \vdots & -a_2 \\ \vdots & \ddots & \ddots & \ddots & \vdots & \vdots \\ \vdots & & \ddots & 1 & 0 & -a_{n-2} \\ 0 & \cdots & \cdots & 0 & 1 & -a_{n-1} \end{array}\right).$$
Determine as a function of $Q$ the characteristic polynomial of $C_Q$.

Let $\left(a_0, a_1, \ldots, a_{n-1}\right) \in \mathbb{K}^n$ and $Q(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_0$. We consider the companion matrix
$$C_Q = \left(\begin{array}{cccccc} 0 & \cdots & \cdots & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & \cdots & 0 & -a_1 \\ 0 & 1 & \ddots & & \vdots & -a_2 \\ \vdots & \ddots & \ddots & \ddots & \vdots & \vdots \\ \vdots & & \ddots & 1 & 0 & -a_{n-2} \\ 0 & \cdots & \cdots & 0 & 1 & -a_{n-1} \end{array}\right).$$
Let $\lambda$ be an eigenvalue of $C_Q^{\top}$. Determine the dimension and a basis of the associated eigenspace.

Show that: $X^p + \alpha_{p-1}X^{p-1} + \cdots + \alpha_0$ divides the polynomial $\chi_f$.

Prove that $\chi_f(f)$ is the zero endomorphism.