Predicting eigenvalues of bigger matrices Consider the following $(3 \times 3)$ matrix:
$K_3 = \left( \begin{array}{ccc}
a & -1 & 0 \\
-1 & a+1 & -1 \\
0 & -1 & a \end{array} \right)$
The question has a quantum physics context, so we'll assume that $a$ is such that $K$ is hermitian. This matrix has eigenvalues $-1 + a$, $a$, and $a+1$. Now consider growing the matrix to a $(4\times4)$ in the following way:
$K_4 = \left( \begin{array}{ccc}
a & -1 & 0 & 0 \\
-1 & a+1 & -1 & 0 \\
0 & -1 & a+1 & -1 \\
0 & 0 & -1 & a \end{array} \right)$
Its eigenvalues are $-1 + a$, $1+a$, $1-\sqrt{2} + a$, and $1+\sqrt{2} + a$. There seems to be at least some structure in this. I was wondering whether there's a way to predict what the eigenvalues of increasingly bigger matrices of this form will be. Specifically, is it possible to find the eigenvalues of:
$K_n = \left( \begin{array}{ccccccc}
a & -1 & 0 & \dots & 0 & 0 & 0 \\
-1 & a+1 & -1 & \dots & 0 & 0 & 0 \\
0 & -1 & a+1 & \dots & 0 & 0 & 0 \\
0 & 0 & -1 & \dots & -1 & 0 & 0 \\
0 & 0 & 0 & \dots & a+1 & -1 & 0 \\
0 & 0 & 0 & \dots & -1 & a+1 & -1 \\
0 & 0 & 0 & \dots & 0 & -1 & a \end{array} \right)$
I can imagine the expressions become rather monstrous if it's at all possible, so taking something like $a = 1$ is fine for my purposes, if it simplifies matters.

With the help of mookid I managed to find a general form for the characteristic polynomials of $K - Ia$. Defining $P_n$ as the characteristic polynomial of the $(n \times n)$ matrix $K - Ia$, we have:
$P_n = -\lambda Q_{n-1} - Q_{n-2}$
Where $\lambda$ are the requested eigenvalues and $Q_n$ are the characteristic polynomials of the same matrix, where the zero on the $nn$-th entry has been replaced by a one:
$Q_n = (1-\lambda) Q_{n-1} - Q_{n-2}$
With boundary conditions $Q_1 = -\lambda$ and $Q_2 = \lambda^2 - \lambda - 1$.
Quoting wikipedia, with regards to invertibility of tridiagonal matrices: "Closed form solutions can be computed for special cases such as symmetric matrices with all off-diagonal elements equal". I'm hoping that means there also exists an analytic solution for the roots of these polynomials (and hence the eigenvalues for arbitrary $n$). Does anyone know if this is possible, and/or how to go about finding the solution?

You just have to go for the $a=0$ case; in general, the eigenvalues are
$a+\lambda, \lambda$ eigenvalue of $\left( \begin{array}{ccccccc}
0 & -1 & 0 & \dots & 0 & 0 & 0 \\
-1 & 1 & -1 & \dots & 0 & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 & 0 \\
0 & 0 & -1 & \dots & -1 & 0 & 0 \\
0 & 0 & 0 & \dots & 1 & -1 & 0 \\
0 & 0 & 0 & \dots & -1 & 1 & -1 \\
0 & 0 & 0 & \dots & 0 & -1 & 0 \end{array} \right)$
Now refering to wikipedia, the sequences of characteristic polynomials is
$$P_2 = X^2 - 1, P_3 = -X^3+X
\\
P_{n+1} = X\times P_n - (-1)\times(-1)\times P_{n-1}
=-XP_n - P_{n-1}$$
