The inverse of a Kac-Murdock-Szegő matrix I want to calculate the inverse of a matrix resembling the following form:
\begin{bmatrix}
1 &\rho &\rho^2 &\cdots &\rho^{n-1}\\
\rho& 1& \rho& \cdots &\rho^{n-2}\\
\rho^2& \rho& 1 &\cdots&\rho^{n-3}\\
\vdots&\vdots &\vdots &\ddots &\vdots\\
\rho^{n-1} & \rho^{n-2}&\rho^{n-3} &\cdots&1
\end{bmatrix}
each line of the off-diagonal is a polynomial of $\rho$ and the exponential is increasing towards the boundary.
I wonder how to solve this problem.
 A: Your matrix $\bf A$ (say) has the typical element of the form $a_{ij}=\rho^{|i-j|}$. 
This in fact is a known matrix form called the Kac-Murdock-Szegő matrix.
Let $D_n=\det \bf A$, then it is easily seen that it satisfies the relation $D_n=(1-\rho^2)D_{n-1}\quad, n\ge2$ by applying the operation $R_i'=R_i-\rho R_{i+1}\quad,i=1,2,...,n-1$ on $\bf A$.
As the matrix is transformed to a lower triangular form, it follows that $\det \mathbf A=(1-\rho^2)^{n-1}$.
So $\bf A$ is invertible iff $\rho\ne \pm1$ $\quad(\bf A$ is positive definite when $-1<\rho<1)$  .
Now as is shown in this answer, when $\bf A$ is positive definite, it can be decomposed as $\bf BB^\top$ where 
$\bf B=\begin{bmatrix} 1 &0 & 0& \dots&0\\\rho & \sqrt{1-\rho^2} &0 &\dots &0\\\rho^2 & \rho\sqrt{1-\rho^2} & \sqrt{1-\rho^2} &\dots &0\\\ & \ddots & \ddots & \ddots &\\\rho^{n-1} &\rho^{n-2}\sqrt{1-\rho^2}  & \rho^{n-3}\sqrt{1-\rho^2} & \dots &\sqrt{1-\rho^2} \end{bmatrix}$
In that case you have $\mathbf A^{-1}=(\mathbf B^{-1})^\top\mathbf B^{-1}$
$=\dfrac{1}{1-\rho^2}\begin{bmatrix} 1 &-\rho &0 &\dots  &0\\-\rho & 1+\rho^2 &-\rho &\dots  &0\\\ & \ddots & \ddots &\ddots &\\0 & \dots &-\rho  & 1+\rho^2 &-\rho\\0 & \dots &0  &-\rho  & 1\end{bmatrix}$ 
Check this
link out for more on these matrices.
