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Given $a, b \in \Bbb R$, consider the following large tridiagonal matrix

$$M := \begin{pmatrix} a^2 & b & 0 & 0 & \cdots \\ b & (a+1)^2 & b & 0 & \cdots & \\ 0 & b & (a+2)^2 & b & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$

What can be said about its eigenvalues? Are analytic expressions known? Or, at least, properties of the eigenvalues?

Rodrigo de Azevedo
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Nigel1
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  • it is known that the eigenvalues of $M_n$ separate the eigenvalues of $M_{n+1};$ it easily follows from the three term recursion relation for the determinants. – abel Mar 06 '16 at 19:59
  • @abel Thanks. What do you mean by seperation? I am completely ignorant of this. Do you have a reference? – Nigel1 Mar 07 '16 at 21:08
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    Posted also on MO: https://mathoverflow.net/questions/233452/eigenvalues-of-large-tridiagonal-matrix I find the advice given in this answer very reasonable. Of course, you might have a look at other discussions regarding (cross-posting), too. – Martin Sleziak May 29 '17 at 11:09

2 Answers2

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Let's call $M_n$ this matrix, and let's consider its characteristic polynomial $P_n := \det(XI_n-M_n)$. Expansion according to the last column yields the recurrence relation $$ P_n = (X-(a+n)^2)P_{n-1}-b^2P_{n-2}. $$ with initial conditions $P_0 = 1$ and $P_1 = X-a^2$.

ManifoldFR
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  • Thanks. Can writing down recurrence relation of above matrix say anything about eigenvalues or their properties? – Nigel1 Feb 26 '16 at 08:34
  • If you can use the recurrence relation to get information about the characteristic polynomial, you can get information about the eigenvalues. I think the polynomials might be orthogonal for some dot product according to their recurrence relation and a theorem I can't recall. – ManifoldFR Feb 26 '16 at 09:20
  • Favard's theorem? – Semiclassical Mar 03 '16 at 22:53
  • That'd be the one, yes. – ManifoldFR Mar 03 '16 at 22:54
  • @Groovy I failed so far to get usable expression for characteristic polynomial by hand calculation due to large size of matrix. As matrix is very large, do you know any infinite matrix theorems which help to get charcteristic polynomial in usable form? At least for lowest eigenvalues. – Nigel1 Mar 07 '16 at 20:41
  • I'm afraid I don't know anything about infinite matrices, so sorry. :( – ManifoldFR Mar 08 '16 at 20:57
  • @Groovy It makes two of us :-( – Nigel1 Mar 12 '16 at 12:34
  • There are some properties of the roots of orthogonal polynomials listed here. Maybe one of them would be helpful to you? http://mathworld.wolfram.com/OrthogonalPolynomials.html – Davis Yoshida Mar 26 '16 at 10:12
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Since $M_n(a,b)$ and $M_n(a,-b)$ have same real spectrum, we may assume that $b\geq 0$. Let $\lambda_n$ be the smallest eigenvalue of $M_n$. Since there exist hidden orthogonal polynomials, the real sequence $(\lambda_n)_n$ is non-increasing.

Assume that $a\geq 0$. Note that $e_1^TM_ne_1=a^2$; then $\lambda_n\leq a^2$. Denote by $B_n$ the matrix $M_n$ with a zero diagonal (only the $b$'s remain). Then $M_n\geq B_n$ and $\lambda_n\geq \inf(\text{spectrum}(B_n))\geq -2b$. Finally the sequence $(\lambda_n)_n$ converges to $\lambda\in [-2b,a^2]$.

Note that , if $\dfrac{b}{a^2}$ is small enough, then $M_n\geq 0$ and $\lambda\approx a^2$. If $a$ is fixed and $b$ tends to $+\infty$, then $\lambda\rightarrow -2b$.

Hans
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  • What is a hidden orthogonal polynomial? Why does the hidden orthogonal polynomial lead to the real sequence $(\lambda_n)_n$ being non-increasing? – Hans Oct 31 '17 at 06:12
  • @Hans , the characteristic polynomial $P_n$ of $M_n$ satisfies a special recurrence with three terms; according to Favard's theorem (cf. wikipedia), there is a dot product for which the $(P_n)$ are orthogonal; then the roots of $P_{n+1}$ and $P_n$ are intertwined and $(\lambda_n)$ is non-increasing. –  Oct 31 '17 at 08:25
  • OK. Thanks. Why is it that if $a$ is fixed and $b$ tends to $+\infty$, then $\lambda\rightarrow -2b$? I suppose this is for fixed $n$. I think it may be better to say $\frac b{a^2}\to\infty$ than fix $a$. Should we not get $\lambda\to -b$ instead of $-2b$. Is this a typo? – Hans Oct 31 '17 at 18:19