1
$\begingroup$

During my research work related to the convergence of estimates, I needed to calculate the eigenvalues of the following symmetric matrix.

$$\Sigma_{n\times n} := \begin{pmatrix} 1 & \frac{a}{2} & \frac{a^2}{2} & \frac{a^4}{2} &\cdots & \frac{a^{n-2}}{2} & \frac{a^{n-1}}{2} \\ \frac{a}{2} & 1 & \frac{a}{2} & \frac{a^2}{2} & \cdots & \frac{a^{n-3}}{2} & \frac{a^{n-2}}{2} \\ \frac{a^2}{2} & \frac{a}{2} & 1 & \frac{a}{2} & \cdots & \frac{a^{n-4}}{2} & \frac{a^{n-3}}{2} \\ \vdots & \vdots & \ddots &\ddots & \cdots & \vdots & \vdots \\ \frac{a^{n-1}}{2} & \cdots & \cdots & \cdots &\cdots & \frac{a}{2}& 1 \end{pmatrix} $$

I tried to guess the answer using analysis on small matrices, but even with a $3 \times 3$ matrix it is difficult to cope. For $2 \times 2$ matrices, everything is obvious. The recurrence has not been compiled.

$\endgroup$
6
  • 3
    $\begingroup$ This is, up to a multiplication and translation, (i.e. taking $\Sigma' := 2 \Sigma - I$) a Kac-Murdoch-Szegö matrix. See here $\endgroup$
    – Jean Marie
    May 14 at 6:06
  • $\begingroup$ Related $\endgroup$ May 14 at 6:36
  • $\begingroup$ @JeanMarie *Murdock $\endgroup$ May 14 at 6:37
  • 1
    $\begingroup$ Oleg, are you sure the (1,4)-th entry is $\frac{a^4}{2}$? Shouldn't it be $\frac{a^3}{2}$? $\endgroup$ May 14 at 6:38
  • $\begingroup$ Compiled? I don't understand that part. $\endgroup$ May 14 at 6:41

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.