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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!

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    $\begingroup$ You can bound it by Gershgorin. out of interest, does this come from some kind of random matrix problem? $\endgroup$
    – Yemon Choi
    Feb 14, 2012 at 16:40
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    $\begingroup$ I ask because this matrix is the expected value of the natural tridagonalization of the unnormalized GOE, although that may not be relevant here $\endgroup$
    – Yemon Choi
    Feb 14, 2012 at 16:55
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    $\begingroup$ you can bound it above $2\sqrt{n}$. $\endgroup$
    – Kate Juschenko
    Feb 14, 2012 at 17:50
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    $\begingroup$ @Yemon: It looks to me like the sum of a creation and an annihilation operator on (a finite-dimensional piece of) a Fock space in quantum field theory. Would that have any connection with the GOE that you mentioned? $\endgroup$
    – Andreas Blass
    Feb 14, 2012 at 19:03
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    $\begingroup$ The dominant eigenvalue is asymptotic to $2 \sqrt n$. This can be seen by comparison with multiples of the $m \times m$ matrix, call it $L_m$, whose $(i,j)$ entry is $1$ if $|i-j|=1$ and $0$ else. The eigenvalues of $L_m$ are known explicitly and the largest one approaches $2$ from below as $m \rightarrow \infty$. Kate Juschenko may have used this to obtain the upper bound $2 \sqrt n$ (the matrix is smaller than $\sqrt n M_{n+1}$). $\endgroup$
    – Noam D. Elkies
    Feb 14, 2012 at 19:52

1 Answer 1

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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).

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  • $\begingroup$ I'd like to add that the connection with random matrices that Yemon Choi suggests above may be the following : The mean characteristic polynomial of a $n\times n$ GUE random matrix X, namely $\mathbb{E}_{GUE}\det(xI_n−X)$, is actually the Hermite polynomial (this formula goes back to Heine), for which the matrix of recurrence coefficients is given by An, as you pointed out. Even so I don't really see GOE there, even it is plausible. The matrix An is also interpreted as a Toeplitz operator associated with the sum of a creation and annihilation operator as pointed out by Andreas Blass. $\endgroup$
    – Adrien Hardy
    Feb 14, 2012 at 23:34
  • $\begingroup$ The connection with such creation/ annihilation operator has been used by Voiculescu once developing free probability theory. I think this is explained in the book of Hiai and Petz, "The Semicircle Law, Free Random Variables and Entropy". I hope this helps ... $\endgroup$
    – Adrien Hardy
    Feb 14, 2012 at 23:34
  • $\begingroup$ @Adrien: see my comments to the main question for some attempts to correct and clarify my initial claims $\endgroup$
    – Yemon Choi
    Feb 15, 2012 at 0:11

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