# 0 eigenvalue for a symmetric tridiagonal matrix

Let $T\in \mathbb{R}^{n\times n}$ be a symmetric tridiagonal matrix having the off--diagonal entries equal to -1. The diagonal entries are all positive, $a_i>0$, $i=\overline{1,n}$, and there exist $j$ and $k$, $j\neq k$ such that $a_j=a_k\leq 1$. $a_j$ and $a_k$ are the smallest diagonal entries.

I'm interested under what supplemental conditions can such a matrix have the smallest eigenvalue equal to 0?

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could you give some motivation/background? – Vladimir Dotsenko Feb 25 '12 at 13:26
The matrix $T$ is the linearization of a tridiagonal cooperative dynamical system around an equilibrium point. – Andreea Feb 25 '12 at 14:07

To simplify things a little, I describe conditions under which the smallest eigenvalue is strictly positive. These can be adjusted to get equality to zero.

Necessary and sufficient conditions for positive definiteness of the tridiagonal matrix in question are described below.

Definition (Chain Sequence). A sequence $\lbrace x_k \rbrace_{k > 0}$ is a chain sequence if there exists another sequence $\lbrace y_k \rbrace_{k\ge 0}$ such that \begin{equation*} x_k = y_k(1-y_{k-1}), \end{equation*} where $y_0 \in [0,1)$ and $y_k \in (0,1)$ for $k > 0$.

By the Wall-Wetzel Theorem, your tridiagonal matrix is positive definite if and only if

\begin{equation*} \left\lbrace \frac{1}{a_ka_{k+1}} \right\rbrace_{k=1}^{n-1} \end{equation*}

is a chain sequence.

Example. In particular, if the entries of the matrix satisfy,

\begin{equation*} 0 < \frac{1}{a_ka_{k+1}} < \frac{1}{4\cos^2\left(\frac{\pi}{n+1}\right)},\quad k=1,\ldots,n-1, \end{equation*} then it is positive definite.