I want to find the inverse of the following matrix: $$ R_{k-1}=\begin{pmatrix} 1 &\rho &\rho^2 &\cdots &\rho^{k-2} \\ \rho &1 &\rho &\cdots &\rho^{k-3} \\ \rho^2 &\rho &1 & &\rho^{k-4} \\ \vdots &\vdots &\vdots &\ddots &\vdots\\ \rho^{k-2} &\rho^{k-3} &\cdots & &1\\ \ \end{pmatrix} $$

Let $A_{i,j}$ be the $i,j$ minor of $R_{k-1}$. By considering the pattern of the above matrix and its symmetrical properties, we can conclude that:

  1. $det(A_{11})= det(A_{k-1,k-1})= |R_{k-2}|$
  2. $det(A_{i,j})=det((A_{j,i})^T)$
  3. $det(A_{i,j})=0$ for $|i-j|\le2$

which means that the inverse of $R_{k-1}$ is a tridiagonal symmetric matrix. I've tried to find the inverse using the fact I've described above and using $A^{-1}A=A A^{-1}=I$. But I couldn't find it, since there are more variables than equations. Did i miss something? or may be is there any other easier way to find the inverse?

  • $\begingroup$ The actual answer is very simple in form. You might be able to guess the form from looking at the 2x2 and 3x3 case, which might then help you to see how to derive the general case. $\endgroup$ – Glen_b -Reinstate Monica Oct 16 '14 at 23:04

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