I really want to find the eigenvalues of this matrix:
$$ \left[\begin{array}{ccccccc} a_{0} & 1 & 0 & & 0 & 0 & 1\\ 1 & a_{1} & 1 & \cdots & 0 & 0 & 0\\ 0 & 1 & a_{2} & & 0 & 0 & 0\\ & \vdots & & \ddots & & \vdots\\ 0 & 0 & 0 & & a_{L-3} & 1 & 0\\ 0 & 0 & 0 & \cdots & 1 & a_{L-2} & 1\\ 1 & 0 & 0 & & 0 & 1 & a_{L-1} \end{array}\right] $$ where $$a_{n}=2\cos\left(2\pi\frac{n}{L}\right).$$ Note that $a_{n+L}=a_n$.
I've done quite a bit of research on the internet about tridiagonal matrices but I cannot find any source of information suitable to my problem. I don't even know if these specific matrices have a name.
Usually I'd accept that this is as hard a problem as finding the eigenvalues of a tridiagonal matrix with any old $a_n$. However, this has so much symmetry that it makes me wonder if there's a way. If there is such a way, maybe I could use it to solve other problems that arise in a similar context.
If you're interested about the context, I'm studying the Landau levels of a simple tight-binding Hamiltonian on a square lattice with nearest neighbor interactions, periodic boundary conditions and a constant magnetic field using the Landau Gauge.