Eigenvalues of a certain tridiagonal matrix Consider $A \in M_n(\mathbb R)$ defined by:
$$A=\begin{bmatrix}
a & -1 & 0 & \cdots & 0 \\
-1 & a & -1 &\cdots& 0\\
0 & -1 & a & \cdots & 0\\
\vdots & \vdots & \vdots &\ddots & -1\\
0 & 0 & 0 & -1 & a
\end{bmatrix}$$
How to find the eigenvalues of $A$?
I know:
$$P_A(\lambda) = \left| \begin{array}{ccc}
\lambda - a & 1 & 0 & \cdots & 0 \\
1 & \lambda - a & 1 &\cdots& 0\\
0 & 1 & \lambda - a & \cdots & 0\\
\vdots & \vdots & \vdots &\ddots & 1\\
0 & 0 & 0 & 1 & \lambda - a
\end{array} \right|$$
or more simply
$$P_A(a + \lambda) = \left| \begin{array}{ccc}
\lambda & 1 & 0 & \cdots & 0 \\
1 & \lambda & 1 &\cdots& 0\\
0 & 1 & \lambda & \cdots & 0\\
\vdots & \vdots & \vdots &\ddots & 1\\
0 & 0 & 0 & 1 & \lambda
\end{array} \right|$$
What to do next?
 A: Hint:
Determinants of tridiagonal matrices can be calculated with a linear recurrence of order $2$. In the present case, denoting $D_n$ this determinant of order $n$, we have:
\begin{align*} D_n&=\lambda D_{n-1}- D_{n-2}\\
\text{with initials conditions:}\quad D_0&=1,\enspace D_1=\lambda
\end{align*}
The solutions are linear combinations of geometric sequences $r^n$, $r$ satisfying the characteristic equation:
$$r^2=\lambda r-1.$$
A: It is enough to calculate the eigenvalues of the matrix
$$ A=\begin{bmatrix}
0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 1 &\cdots& 0\\
0 & 1 & 0 & \cdots & 0\\
\vdots & \vdots & \vdots &\ddots & 1\\
0 & 0 & 0 & 1 & 0
\end{bmatrix}. $$
For that, I suggest that instead of writing a recurrence for the characteristic polynomial (which still leaves you to find the roots of the polynomial) you write a recurrence relation for the eigenvectors (which are guaranteed to exist since $A$ is symmetric). More explicitly, if $v = (x_1, \ldots, x_n)^T$ is an eigenvector of $A$ with eigenvalue $\lambda$, then by writing the equation $Av = \lambda v$ explicitly, we see that the sequence $(x_i)$ must satisfy the linear recurrence
$$ x_{i-1} + x_{i + 1} = \lambda x_i, \,\,\, 1 \leq i \leq n $$
where we let $x_0 = x_{n+1} = 0$. Thus, we obtain linear recurrence 
$$ x_{i+1} = \lambda x_i - x_{i-1} $$
with boundary conditions $x_0 = x_{n+1} = 0$ for which we are looking for a non-trivial solution. The general solution to a recurrence relation is obtained by finding the characteristic roots of the associated equation $u^2 = \lambda u - 1$ whose solutions are $u_{1,2} = \frac{\lambda \pm \sqrt{\lambda^2 - 4}}{2}$. If the equation has a double root then the general solution to the recurrence relation is of the form
$$ x_{i} = Au^i + Biu^i $$
(where $u = u_1 = u_2$) for $A, B \in \mathbb{R}$. However, such a solution cannot satisfy the boundary conditions $x_0 = x_{n+1} = 0$ without being trivial contradicting the fact that $v$ is an eigenvector. Thus, we see that we must have $u_1 \neq u_2$ and thus
$$ x_i = Au_1^i + Bu_2^i $$
for some $A,B \in \mathbb{R}$. In order to have a non-trivial solution to the recurrence relation with $x_0 = x_{n+1} = 0$, the linear equations 
$$ x_0 = A + B = 0, \\
   x_{n+1} = Au_1^{n+1} + Bu_2^{n+1} = 0 $$
must be linearly dependent and so $u_1^{n+1} = u_2^{n+1}$ or (since $u_1,u_2 \neq 0$)
$$ \left( \frac{\lambda + \sqrt{\lambda^2 - 4}}{\lambda - \sqrt{\lambda^2 - 4}} \right)^{n+1} = 1. $$
Solving explicitly for $\lambda$, you can easily deduce that there are $n$ distinct solutions given by
$$ \lambda_j = 2\cos \left( \frac{j \pi}{n+1} \right). $$
