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.
 A: 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 $A$:
$$
\lambda_{\max}(A)\leq 1+a+\cdots+a^n=1+\frac{a(1-a^n)}{1-a}\leq \frac{1}{1-a}.
$$
