Bounding largest eigenvalue - MathOverflow

Bounding largest eigenvalue

2012-02-14T16:31:37Z

Hi all,

do you know how to compute (as a function of n) the largest eigenvalue of this matrix (or at least to bound it)?

$$ \left(\begin{array}{cccccc} 0 & 1 & & & & \cr 1 & 0 & \sqrt 2 & & & \cr & \sqrt 2 & 0 & & & \cr & & & \ddots & & & \cr & & & & 0 & \sqrt n & \cr & & & & \sqrt n & 0 & \end{array}\right) $$

Thanks!

Answer by Pietro Majer for Bounding largest eigenvalue

2012-02-14T22:26:02Z

If you denote $A_n$ your tri-diagonal matrix of order $n$, and $H_n(x):= \det(x+A_n)$, the sequence $H_n$ satisfies the two-term linear recurrence $H_{n+1}=xH_n - nH_{n-1}$ with initial conditions $H_0=1$ and $H_1=x$. Thus, they are the Hermite polynomials (here in the "probabilist's version"), and their zeros are the eigenvalues of $-A_n$ (on which you can find everything in the literature).