# How to estimate the lower bound of a given Toeplitz matrix's eigenvalue?

Given the Toeplitz matrix

$$\begin{pmatrix} 1 & a & a^2 & \cdots & a^n \\ a &1 &a & \cdots & a^{n-1} \\ a^2&a & 1 & \cdots& a^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a^n & a^{n-1} & a^{n-2} & \cdots & 1\\ \end{pmatrix}$$

where $$a \in (0,1)$$, can one find the eigenvalues of the matrix? If not, can one find a lower bound?

Any links or reference materials? Thanks.

Let's call the matrix in question $A$. Its inverse looks quite easy, $$A^{-1} = \frac{1}{1-a^2} \begin{bmatrix} 1 & -a & & & & \\ -a & 1 + a^2 & -a & & & \\ & -a & 1+a^2 & -a & & \\ & & \ddots & \ddots & \ddots & \\ & & & -a & 1 + a^2 & -a \\ & & & & -a & 1 \\ \end{bmatrix}$$ A lower bound on $\lambda_{\min}(A)$ can be estimated using the reciprocal of an upper bound on $\lambda_{\max}(A^{-1})$ for which one can use the Gershgorin theorem: $$\lambda_{\max}(A^{-1})\leq\max\left\{\frac{1 + a}{1 - a^2}, \frac{1+2a+a^2}{1 - a^2}\right\}.$$ We clean up a bit the expressions, $$\frac{1+a}{1-a^2}=\frac{1}{1-a},\quad \frac{1+2a+a^2}{1-a^2}=\frac{1+a}{1-a},$$ and see that the latter is generally larger. So $$\lambda_{\max}(A^{-1})\leq\frac{1+a}{1-a}\quad\Rightarrow\quad\lambda_{\min}(A)\geq\frac{1-a}{1+a}.$$
Also, the same approach can be used to bound the maximal eigenvalues of $A$: $$\lambda_{\max}(A)\leq 1+a+\cdots+a^n=1+\frac{a(1-a^n)}{1-a}\leq \frac{1}{1-a}.$$
• Can something (more) be said about the eigenvalues if $n\to\infty$? What if $a$ is a complex number? – pisoir Dec 5 '14 at 9:52