Eigenvalue distribution of matrix (specific Töplitz matrix)

Question

Given the $N$-by-$N$ matrix $J$ with entries $J_{kl}=0$ for $k=l$ and \begin{align*} J_{kl}=\frac{i \left(3-4 e^{\frac{i \pi (k-l)}{2}}+e^{2 i \pi (k-l)}\right)}{6 \pi (k-l)}\,, \end{align*} I would like to show that in the limit $N\to\infty$, 1/4 of the eigenvalues approaches 1 and 3/4 of the eigenvalues approaches -1/3. This is really only true in the limit, because there are always a few eigenvalues that are a little off. For $N=40$, the eigenvalues are for instance: \begin{align*} \{1.,1.,1.,1.,0.999997,0.99993,0.998958,0.988395,0.910196,0.594114,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333333,-0.333327,-0.333239,-0.332132,-0.321125,-0.242825,0.0710588\}\,. \end{align*}

I'm not sure what the best approach is to show this behavior analytically from the closed form of the matrix entries. Note that it is a Töplitz matrix, meaning that the matrix entry only depends on $(k-l)$.

Answer 1

I made some progress on this problem that I wanted to share. It is not the full solution, but close. I used Szegö's theorem to compute the Fourier transform of the matrix entries, namely \begin{align*} f(\lambda)=\sum_{k=-\infty}^\infty J_d e^{id\lambda}=-\frac{2 i \left(\log \left(1-e^{i \lambda }\right)-\log \left(e^{-i \lambda } \left(-1+e^{i \lambda }\right)\right)+\log \left(e^{-i \lambda } \left(e^{i \lambda }+i\right)\right)-\log \left(1-i e^{i \lambda }\right)\right)}{3 \pi }\,, \end{align*} where $J_d=J_{kl}$ with $d=k-l$. This function looks a little complicated, but it turns out to be exactly the eigenvalue distribution that I was looking for. Plotting it in Mathematica, it looks like this:

I'm not exactly sure why this is the case, but I believe that this may be a special case of Szegö's theorem applied to my problem. Any ideas?