TITLE: Is this matrix obviously positive definite?
QUESTION [18 upvotes]: Consider the matrix $A$ whose elements are $A_{ij} = a^{|i-j|}$ for $-1<a<1$ and $i,j=1,\dots,n$. E.g., for $n=4$, the matrix is
$$A = \left[
\begin{matrix}
1 & a & a^2 & a^3 \\
a & 1 & a & a^2 \\
a^2 & a & 1 & a \\
a^3 & a^2 & a & 1
\end{matrix}
\right]$$
Is this matrix always positive definite? If so, what is the simplest way to see that?
I strongly suspect that the matrix is positive definite for all $n$ and $a$, but am having trouble coming up with a proof.
Extra credit: The eigenvalues seem to lie within an interval $[\lambda_{\rm min}, \lambda_{\rm max}]$ which is a function of $a$ but not of $n$. For example, for $a=1/2$ all eigenvalues lie in $[1/3, 3]$.
Is it true for for general $a$, the eigenvalues lie in $[\lambda(a)^{-1}, \lambda(a)]$? If so, what is $\lambda(a)$?
Also, the eigenvectors seem to have a particularly regular form. In particular, they look like that could be expressed as simple combinations of trigonometric functions. Is this the case?

REPLY [7 votes]: (Too long for a comment.)
What you have is in fact a family of well studied matrices: the Kac-Murdock-Szegő (KMS) matrices. They are generated for instance by the MATLAB command gallery('kms',n,a).
In Grenander and Szegő, and in William Trench's note, there are explicit expressions for the eigenvalues of the KMS matrix $\mathrm{KMS}_{i,j}(a)=a^{|i-j|}$:
$$\lambda_k=\frac{1-a^2}{1+a^2-2a\chi_k}$$
where the $\chi_k$ are the $n$ roots of the polynomial
$$U_n(x)-2a U_{n-1}(x)+a^2 U_{n-2}(x)$$
and $U_n(x)$ is the Chebyshev polynomial of the second kind. (I have observed that Chebyshev polynomials do tend to figure prominently in the theory of Toeplitz matrices.) It is also noted in those references that the inverse of a KMS matrix is tridiagonal.
The $\chi_k$ are shown to satisfy the inequality $-1 < \chi_k < 1$ (they're cosines of angles); it can then be shown from this that the $\lambda_k$ are always positive.