I'm having trouble with a French test, which seems to have errors.
The final aim is to find the eigenvalues of the matrix $A_n$(and its eigenvectors): $$A_7 = \begin{bmatrix} 0& 1& 0& 0& 0& 0& 0 \\ 1& 0& 1& 0& 0& 0& 0 \\ 0& 1& 0& 1& 0& 0& 0 \\ 0& 0& 1& 0& 1& 0& 0 \\ 0& 0& 0& 1& 0& 1& 0 \\ 0& 0& 0& 0& 1& 0& 1 \\ 0& 0& 0& 0& 0& 1& 0 \\ \end{bmatrix} $$ To do so, however, the test forces us to answer some questions. It sets up a sequence called $u_n$: $$ u_n = \det(2\cos(α) * I_n - A_n) $$
The polynomial expression of the matrix evaluated at $2\cos(α)$; with $α$ in $(0, \pi)$: $$ 2\cos(α) * I_n - A_n = \begin{bmatrix} 2\cos(α)& -1& 0& \dots& 0& 0& 0 \\ -1& 2\cos(α)& -1& \dots& 0& 0& 0 \\ 0& -1& 2\cos(α)& ...& 0& 0& 0 \\ \vdots& \vdots& & \ddots& & & \\ 0& 0& 0& ...& -1& 2\cos(α)& -1 \\ 0& 0& 0& ...& 0& -1& 2\cos(α) \\ \end{bmatrix} $$ Also seen here
The first question asks to set up a relation between $u_n$, $u_{n-1}$, $u_{n-2}$. I found $$u_n = 2\cos(\alpha) * u_{n-1} - u_{n-2}$$
and checked that it does work for $n = 4$.
The second question asks us to deduce that
$$u_n = \frac{\sin(n+1)\alpha}{\sin(\alpha)}$$
This is where I think the mistake is, as $$ u_2 = 4\cos(\alpha)^2 - 1 \neq \frac{\sin(3)\alpha}{\sin(α)} \text{ for } \alpha \in (0, \pi) $$ I've found that this relation does hold true for $α = 1$, and possibly for $α = 2.2$, but not for other angles.
That is the biggest issue I've encountered, and is the main problem of this question.
The third question asks us to deduce the eigenvalue from the previous results, and the fourth asks for all of the eigenvectors associated with those values.
You can find the original test, in French, here (see page 3, section I-B):
https://www.concours-centrale-supelec.fr/CentraleSupelec/2015/TSI/sujets/2014-021.pdf
The symbol $\sin(n+1)\alpha$ means $\sin\left((n+1)\alpha\right)$, not $\left(\sin(n+1)\right)\alpha$.