I am working on the inverse of the sum of an identity matrix and a Toepltz matrix, and trying to find the formula for the (1,1) element of the inverse. For example, Assume $c$ is a nonzero constant, and let $$A_{t}=cI_{t}+\Omega_{t},$$ where, for $t=4$,
$$ \Omega _{t}=\left( \begin{array}{cccc} 1 & \rho & \rho ^{2} & \rho ^{3} \\ \rho & 1 & \rho & \rho ^{2} \\ \rho ^{2} & \rho & 1 & \rho \\ \rho ^{3} & \rho ^{2} & \rho & 1% \end{array}% \right), \quad \text{with} \quad \left\vert \rho \right\vert <1. $$
In general, the $(i,j)$-th element of $\Omega _{t}$ is given by $\rho ^{|i-j|}$.
Is there any way to find a general formula for the $(1,1)$-th element of $A_{t}^{-1}$ for different $t,$ say, $t=2,3,4,\ldots ?$
Many thanks!
By Cramer's rule, $U_t=A_t^{-1}(1,1)$ is given by the ratios of two determinants. Let:
$$A_3=\left(\begin{array}{ccc}c+1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right),\qquad B_3=\left(\begin{array}{ccc}1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right)$$ and define $A_t,B_t$ in a similar fashion. By replacing the first column with the first column minus $\rho$ times the second column, then expanding along the first column, we get: $$ \det A_{t+1} = (c+1-\rho^2) \det A_t + c\rho^2 \det B_t, $$ $$ \det B_{t+1} = (1-\rho^2) \det A_t + c\rho^2 \det B_t, $$ and by eliminating $B_t$, $$ \det A_{t+1} = (c+1+c\rho^2-\rho^2) \det A_t -c^2\rho^2\det A_{t-1} $$ then dividing the whole line by $\det A_t$, $$ \frac{1}{U_{t+1}} = (c+1+c\rho^2-\rho^2)-c^2\rho^2 U_{t} $$ or:
where obviously $U_1=\frac{1}{c+1}$. If we set $U_t=\frac{V_t}{c\rho},$
$$ V_{t+1} =\frac{1}{\frac{c+1+c\rho^2-\rho^2}{c\rho}-V_t},$$ so by setting $\beta=\frac{c+1+c\rho^2-\rho^2}{c\rho}$ and assuming $\beta \geq 2$ we have: $$ \lim_{t\to +\infty}U_t = \frac{-\beta+\sqrt{\beta^2-4}}{2c\rho}$$ by the usual convergence properties of the negative continued fractions.