# eigenvalues of a symmetric tridiagonal matrix with zero diagonals

I was investigating a problem and came up with the following symmetric tridiagonal matrix (with zero diagonal elements):

$$\left(\begin{array}{cccccc} 0 & a & 0 & \ldots & 0 \\ a & 0 & a^2 & & \vdots \\ 0 & a^2 & 0 & \ddots & 0\\ \vdots & & \ddots & \ddots & a^2\\0 & \ldots & 0 & a^2 & 0 \end{array}\right)$$

The dimension of the matrix is $n \times n$. I am looking for a method to find a formula to calculate its eigenvalues (as a function of $n$) or at least to make a simplification. Any advice is appreciated.

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one possibility is to just compute the characteristic polynomial using some inductive guessing....but not really a research math question I think. For the char poly, have a look at the determinant recurrence on en.wikipedia.org/wiki/Tridiagonal_matrix – Suvrit Sep 5 '13 at 12:02
For a large $n$,it is a perturbation of the matrix obtained when you change the $2$ entries equal to $a$ with $a^2$. Then the eigenvalues are approximately $2a^2\cos(\dfrac{k\pi}{n+1})$, $k=1,\cdots,n$. In a second time, you can use Newton method to have a better approximation. – loup blanc Oct 8 '13 at 14:51