0 eigenvalue for a symmetric tridiagonal matrix

2012-02-25T12:47:00Z

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?

Answer by suVRit for 0 eigenvalue for a symmetric tridiagonal matrix

2012-02-25T17:13:28Z

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.

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

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

0 eigenvalue for a symmetric tridiagonal matrix - MathOverflow

0 eigenvalue for a symmetric tridiagonal matrix

Answer by suVRit for 0 eigenvalue for a symmetric tridiagonal matrix