3
$\begingroup$

Let $A_n\in\mathbb{R}^{n\times n}$ be defined as $$ A_n=\begin{bmatrix} a & b & 0 & \cdots & \cdots & 0 & 0\\ b & a & b & \cdots & \cdots & 0 & 0\\ 0 & b & a & \cdots & \cdots & 0 & 0\\ \vdots & \vdots & \vdots &\ddots & \ddots & \vdots & \vdots \\ \vdots & \vdots & \vdots &\ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 &\cdots & \cdots & a & b \\ 0 & 0 & 0 &\cdots & \cdots & b & a\end{bmatrix}, $$ where $a,b\in\mathbb{R}$. It is well-known that the eigenvalues of $A_n$ are $$ \text{eig}(A_n) =\left\{ a+2b\cos\left(\frac{\pi}{n+1}k\right), \ k=1,2,\dots,n \right\}. $$

My question. Does there exist a closed-form expression for the eigenvalue density of the sequence $\{A_n\}$ as $n\to \infty$?

$\endgroup$
4
$\begingroup$

For large $n$ we may treat $x\equiv k/n+1$ as a continuous variable with a uniform density in the interval $0 <x< 1$. The corresponding eigenvalue $\lambda(x)=a+2b\cos\pi x$ ranges from $a-2|b|$ to $a+2|b|$. Since $$|d\lambda/dx|=\pi\sqrt{4b^2-(\lambda-a)^2}.$$ The eigenvalue density follows from $$\rho(\lambda)d\lambda=ndx\Rightarrow \rho(\lambda)=\frac{n}{\pi}\frac{1}{\sqrt{4b^2-(\lambda-a)^2}},\;\;a-2|b|<x<a+2|b|.$$ As a check $\int \rho(\lambda)d\lambda=n$.

$\endgroup$
3
$\begingroup$

To put this into context, the limit $$ \lim_{L\to\infty} \frac{\# \textrm{ eigenvalues in }I \textrm{ of the problem on } \{0,\ldots, L\} }{L} $$ (assuming it exists) is one way of defining the density of states measure $\int_I dN(\lambda)$.

For an ergodic (with respect to the shift) system $\mathcal A$ of operators with probability measure $dP$ this will equal the average $$ dN(\lambda)=\int_{\mathcal A} d\mu(\lambda;A)\, dP(A) $$ of the spectral measures $d\mu$.

In your case, for constant coefficients, the system consisting of this single operator is trivially ergodic, so the density of states is the spectral measure $$ d\mu(\lambda) = \chi_{(a-|b|,a+|b|)}(\lambda) \frac{d\lambda}{\sqrt{4b^2-(\lambda -a)^2}} , $$ which Carlo finds by direct computation.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.