# Eigenvalues of large tridiagonal matrix

Consider large tridiagonal matrix (where $a$ and $b$ are real numbers):

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

What can be said about eigenvalues? Are analytic expressions known?

Or at least properties of eigenvalues?

(Note: a cross-post from mathstack)

• How large is it? Is it actually infinite in size or do you mean that only $N\times N$ with some large $N$? – Igor Khavkine Mar 12 '16 at 14:30
• This is an enormous subject, on which many books are written, it goes under various names, like "Orthogonal polynomials", "Moment problem", etc. Eigenvalues are real and simple, of course, what other properties are you asking about? – Alexandre Eremenko Mar 12 '16 at 14:33
• It's easy to show that $\lambda_n=(n+a)^2 + O(1)$ (use min-max), and I doubt that you can get a whole lot more. – Christian Remling Mar 12 '16 at 21:40
• @Igor Khavkine It is $N \times N$ with $N$ large. – Nigel1 Mar 13 '16 at 16:11
• @Alexandre Eremenko I wonder if expressions for the smallest eigenvalues (in the limit of large $N \times N$ matrix) are possible. – Nigel1 Mar 13 '16 at 16:14

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 othgonal 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 $\lambda_n\geq \inf(spectrum(B_n))\geq -2b$. 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 $\lambda\rightarrow -2b$.