Eigenvalues of a certain tridiagonal Toeplitz matrix

Question

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?

Answer 1

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: $$r^2=\lambda r-1.$$

Answer 2

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

$$ \lambda_j = 2\cos \left( \frac{j \pi}{n+1} \right). $$