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?



0 eigenvalue for a symmetric tridiagonal matrix
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.




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For additional information and details about this material, please see:


1. M. Andelic, and C. M. Da Fonesca. *Sufficient conditions for positive definiteness of tridiagonal matrices revisited*. (2010).


