Bounding largest eigenvalue - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:44:01Z http://mathoverflow.net/feeds/question/88440 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88440/bounding-largest-eigenvalue Bounding largest eigenvalue QuantumLeak 2012-02-14T16:31:37Z 2012-02-15T00:48:10Z <p>Hi all, </p> <p>do you know how to compute (as a function of n) the largest eigenvalue of this matrix (or at least to bound it)?</p> <p>$$\left(\begin{array}{cccccc} 0 &amp; 1 &amp; &amp; &amp; &amp; \cr 1 &amp; 0 &amp; \sqrt 2 &amp; &amp; &amp; \cr &amp; \sqrt 2 &amp; 0 &amp; &amp; &amp; \cr &amp; &amp; &amp; \ddots &amp; &amp; &amp; \cr &amp; &amp; &amp; &amp; 0 &amp; \sqrt n &amp; \cr &amp; &amp; &amp; &amp; \sqrt n &amp; 0 &amp; \end{array}\right)$$</p> <p>Thanks!</p> http://mathoverflow.net/questions/88440/bounding-largest-eigenvalue/88464#88464 Answer by Pietro Majer for Bounding largest eigenvalue Pietro Majer 2012-02-14T22:26:02Z 2012-02-14T22:31:33Z <p>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 <a href="http://en.wikipedia.org/wiki/Hermite_polynomials" rel="nofollow">Hermite polynomials</a> (here in the "probabilist's version"), and their zeros are the eigenvalues of $-A_n$ (on which you can find everything in the literature). </p>