# Largest eigenvalue of a periodic Jacobi matrix

There is a vast literature on Jacobi matrices, I just don't know where to start looking. I'm interested in estimating the largest eigenvalue of the $n\times n$ periodic Jacobi matrix $D+P+P^{-1}$, where $P$ is the matrix of the cyclic permutation of coordinates, and $D$ is the diagonal matrix with diagonal entries $2\cos\frac{2k\pi}{n}$ (where $0\leq k\leq n-1$). I can't help thinking that this has been already considered...

-

## 1 Answer

It's roughly $2 - O(1/n^2)$. For an appropriate choice of $K$ evaluate $(D + P + P^{-1}) u$ for $$u(k) = \begin{cases} 1 - \frac{|k|}{K} & |k| \leq K \\\ 0 & otherwise.\end{cases}$$ An application of the uncertainty principle shows that this is optimal up to constants.

The main point why this works is that $2 \cos( 2 \pi \frac{k}{n})$ is almost constant in $k$. If you would instead take for some $\ell$ coprime to $n$ the potential $2 \cos(2 \pi \frac{\ell k}{n})$. The question is much harder.

-
Don't you mean $4-O(1/n^2)$ ? By the way, let $A_\ell$ be the matrix where, as you suggest, you replace the potential $2\cos\frac{2\pi k}{n}$ by $2\cos\frac{2\pi k\ell}{n}$. Is it true that the maximum of the $\|A_\ell\|$'s is attained for $\ell=1$? – Alain Valette May 21 '11 at 19:37
It's 4 not 2. The maximum is at l = 1. See en.wikipedia.org/wiki/Almost_Mathieu_operator l is roughly the vertical axis in the Hofstadter butterfly. – Helge May 21 '11 at 20:13
Gosh! And I knew it 15 years ago, as I needed it in: On the spectrum of a random walk on the discrete Heisenberg group and the norm of Harper's operator Authors: Beguin C.1; Valette A.; Zuk A. Source: Journal of Geometry and Physics, Volume 21, Number 4, March 1997 , pp. 337-356(20) Thanks anyway for your help! – Alain Valette May 21 '11 at 20:47