# Eigenvalues of a cyclic symmetric tridiagonal matrix where $M_{k,k+1}=\tfrac12\sqrt{M_{k,k}M_{k+1,k+1}}$

Working on a physics problem, I've encountered some structured cyclic tridiagonal $n\times n$ matrices. They're all of the following form: $$\tiny \begin{bmatrix} \alpha_1 & \frac{\sqrt{\alpha_1\alpha_2}}2 & 0 & \cdots &\cdots &\cdots &\cdots &0 & \frac{\sqrt{\alpha_n\alpha_1}}2 \\ \frac{\sqrt{\alpha_1\alpha_2}}2 & \alpha_2 & \frac{\sqrt{\alpha_2\alpha_3}}2 &0 & \cdots &\cdots &\cdots &\cdots &0 \\ 0 & \ddots&\ddots&\ddots &0 & \cdots &\cdots &\cdots &0 \\ \vdots &0 &\frac{\sqrt{\alpha_{k-2}\alpha_{k-1}}}2 & \alpha_{k-1} & \frac{\sqrt{\alpha_{k-1}\alpha_k}}2 &0 & \cdots& \cdots&\vdots \\ 0& \cdots & 0 &\frac{\sqrt{\alpha_{k-1}\alpha_k}}2 & \alpha_k &\frac{\sqrt{\alpha_k\alpha_{k+1}}}2 & 0 & \cdots & 0\\ \vdots & \cdots & \cdots & 0 &\frac{\sqrt{\alpha_k\alpha_{k+1}}}2 & \alpha_{k+1} &\frac{\sqrt{\alpha_{k+1}\alpha_{k+2}}}2 &0 & \vdots \\ 0 & \cdots& \cdots& \cdots& 0 &\ddots &\ddots &\ddots & 0 \\ 0 & \cdots& \cdots& \cdots& \cdots & 0&\frac{\sqrt{\alpha_{n-2}\alpha_{n-1}}}2 & \alpha_{n-1}&\frac{\sqrt{\alpha_{n-1}\alpha_n}}2\\ \frac{\sqrt{\alpha_n\alpha_1}}2& 0 & \cdots& \cdots& \cdots& \cdots & 0&\frac{\sqrt{\alpha_{n-1}\alpha_n}}2 & \alpha_n \end{bmatrix}$$ i.e. they obey $M_{k,k+1}=M_{k+1,k} = \tfrac12\sqrt{\alpha_k\alpha_{k+1}}$ with $M_{k,k}=\alpha_k$ and $k=n+1$ is remapped to $k=1$.

I am interested in the eigenvalues of such a a matrix, or at least its characteristic polynomial, but I was not able to simplify the problem further than this, even is the geometric means on the second diagonals let me hope there is a solution to this problem.

P.S. One interesting case for me is when $\alpha_k$ is the binomial coefficient $\displaystyle \binom{n}{k}$, but I don't think it simplifies the problem.

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That would be a cyclic tridiagonal (or periodic tridiagonal) matrix, to use the term of art. – J. M. Jul 22 '11 at 16:53
If all the $\alpha_i$ are the same it becomes very easy since then it is not only cyclic tridiagonal but a circulant matrix. Other than that i think you should just follow J.M.'s lead to find some specialzed algorithm to find the eigenvalues... – Peter Sheldrick Jul 22 '11 at 17:00
@ J. M. Thanks for the official name. I'll edit my question accordingly – Frédéric Grosshans Jul 22 '11 at 17:04
Without the "ears", the usual way is to note that the characteristic polynomials of successive minors form a system of orthogonal polynomials (with the tacit assumption that no off-diagonal element vanishes). I'll get back to you after I comb through the literature... – J. M. Jul 22 '11 at 17:19

## 1 Answer

Unless I've messed up something, this can be factorized as

$$M = \frac{1}{2} C^T C$$

where

$$C=\begin{bmatrix} \beta_1 & \beta_2 & 0 & ... & 0\\ 0 & \beta_2 & \beta_3 & ... & 0\\ & & \cdots & & \\ 0& 0 & \cdots & \beta_{N-1} & \beta_{N}\\ \beta_1& 0 & \cdots & 0 & \beta_{N}\\ \end{bmatrix}$$

and $\beta_i = \sqrt{\alpha_i}$

This at least shows that the matrix is semi positive definite: one of its eigenvalues is zero (if N is even), the rest are non-negative.

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Thanks for the decomposition. The Semi-positive definiteness is a good news for me, since the eigenvalues of this matrix are supposed to be proportional to probabilities. – Frédéric Grosshans Jul 26 '11 at 15:03
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