# What's the distribution of eigenvalues of a real and symmetric Toeplitz matrix?

Can you please help me find the distribution of eigenvalues of a Toeplitz matrix $\mathbf{K}$ that is constructed as follows: $$\mathbf{K}=\left[ \begin{array}{cccc} 1 & \rho & \ldots & \, \, \rho^{N-1} \\ \rho & 1 & \ldots & \, \,\rho^{N-2}\\ \vdots & \vdots & \ddots & \vdots \\ \rho^{N-1} & \rho^{N-2} & \ldots & 1 \\ \end{array} \right].$$ where $0 \leq \rho < 1$.

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## migrated from stats.stackexchange.comAug 15 '11 at 13:40

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what do you mean by 'distribution' ? Are all the elements numbers (as opposed to random variables) ? Do you need this in analytical form or is a numerical solution sufficient ? –  Andre Holzner Aug 15 '11 at 11:33
@ Andre, Thanks a lot for your comment, you are right. Then, there is any analytical equation to evaluate eigenvalues? –  Farzad Aug 15 '11 at 12:23
Because this question is purely about the eigenvalues of these matrices, it is more suited for posting on math.SE. –  whuber Aug 15 '11 at 13:40

For large $N$, the eigenvalues of $K$ are approximately distributed as $2\pi f(\lambda)$ evaluated at frequencies $-\pi + 2\pi j/N$; $f(\lambda)$ is the spectral density associated to the covariance sequence in your Toeplitz matrix. See for instance Hannan, E.J. Time Series Analysis, Chap. 1 towards the end.
@Tusell, thanks a lot for your answer. I have some questions: 1) I think in frequency range $-\pi+2\pi j/N$ there should be $\lambda$ parameter aomewhere, 2) is $f(\lambda) = |\frac{-2\ln \rho}{(\ln \rho)^2+ (2\pi/N\lambda)}|^2$? –  Farzad Aug 17 '11 at 10:42
@Farzad: sorry I couldn't answer earlier. Was on a holiday and not connected. $\lambda$ takes values $\lambda_j = -\pi+2\pi j/N$ for integer values of $0 \le j \le N$. The spectral density is the Fourier transform of the covariance sequence, hence for your case would be something like $\frac{1}{2\pi}(1 + \sum_{j}\rho^{|j|}\cos(\lambda_j))$. –  F. Tusell Sep 1 '11 at 10:44
@Farzad The range $\lambda_j = -\pi + 2\pi j /N$ for $0\leq j \leq N$ doesn't seem right. With that range, I end up with N + 1 values, but the matrix $\mathbf{K}$ will have only ${N}$ eigenvalues. I think it should be $0 \leq j < N$. –  sauravrt Mar 22 '12 at 17:39