5

I want to find the inverse of the following matrix:

$$ R_{k-1} = \begin{pmatrix} 1 &\rho &\rho^2 & \dots &\rho^{k-2} \\ \rho &1 &\rho & \dots &\rho^{k-3} \\ \rho^2 &\rho &1 & \dots&\rho^{k-4} \\ \vdots &\vdots &\vdots &\ddots &\vdots\\ \rho^{k-2} & \rho^{k-3} & \rho^{k-4} & \dots & 1\end{pmatrix}$$

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?

Rodrigo de Azevedo
  • 23,223
Aeroplane
  • 195

2 Answers2

4

When $\rho$ is close to zero, $R_{k-1}$ is positive definite. Therefore we may perform a Cholesky decomposition and obtain $R_{k-1}=L\,\operatorname{diag}(1,\,1-\rho^2,\ldots,\,1-\rho^2)\,L^T$, where $$ L= \pmatrix{ 1\\ \rho &1\\ \rho^2 &\rho &1\\ \vdots &\ddots &\ddots &\ddots\\ \vdots &\ddots &\ddots &\ddots &1\\ \rho^{k-2} &\rho^{k-3} &\cdots &\rho^2 &\rho &1}. $$ It follows that \begin{align} R_{k-1}^{-1} &=(L^{-1})^T\,\operatorname{diag}\left(1,\frac1{1-\rho^2},\ldots,\frac1{1-\rho^2}\right)\,L^{-1}\\ &=\frac1{1-\rho^2}(L^{-1})^T\,(-\rho^2E_{11}+I)\,L^{-1} \end{align} with $$ L^{-1}= \pmatrix{ 1\\ -\rho&1\\ &\ddots &\ddots\\ &&\ddots &\ddots\\ &&&-\rho &1}. $$ Consequently, $$ \renewcommand{\d}{\phantom{1}-\rho\phantom{^2}} R_{k-1}^{-1}=\frac1{1-\rho^2} \pmatrix{ 1 &\d\\ \d &1+\rho^2 &\d\\ &\d &\ddots&\ddots\\ &&\ddots &\ddots&\ddots\\ &&&\ddots &1+\rho^2 &\d\\ &&&&\d&1}. $$

user1551
  • 149,263
1

Sometimes the easiest way to deduce something is through an experiment, which works here quite well.

As a candidate, try the tridiagonal matrix with the entries $$ \frac{1}{1-\rho^2},\frac{1+\rho^2}{1-\rho^2},\ldots,\frac{1+\rho^2}{1-\rho^2},\frac{1}{1-\rho^2} $$ on the diagonal and the entries $$ -\frac{\rho}{1-\rho^2} $$ above and below the diagonal. Of course, $\rho$ must be different from one.

Algebraic Pavel
  • 23,818
  • Thanks for the answer, but what i really want to do is to prove it without first knowing the exact answer. – Aeroplane Oct 15 '14 at 17:55