0 eigenvalue for a symmetric tridiagonal matrix - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:31:55Z http://mathoverflow.net/feeds/question/89483 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89483/0-eigenvalue-for-a-symmetric-tridiagonal-matrix 0 eigenvalue for a symmetric tridiagonal matrix Andreea 2012-02-25T12:47:00Z 2012-02-25T18:19:33Z <p>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.</p> <p>I'm interested under what supplemental conditions can such a matrix have the smallest eigenvalue equal to 0?</p> http://mathoverflow.net/questions/89483/0-eigenvalue-for-a-symmetric-tridiagonal-matrix/89500#89500 Answer by suVRit for 0 eigenvalue for a symmetric tridiagonal matrix suVRit 2012-02-25T17:13:28Z 2012-02-25T18:19:33Z <p>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.</p> <p><strong>Necessary and sufficient</strong> conditions for positive definiteness of the tridiagonal matrix in question are described below.</p> <p><strong>Definition (Chain Sequence).</strong> 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 $k > 0$.</p> <p>By the Wall-Wetzel Theorem, your tridiagonal matrix is positive definite <strong>if and only if</strong></p> <p>\begin{equation*} \left\lbrace \frac{1}{a_ka_{k+1}} \right\rbrace_{k=1}^{n-1} \end{equation*}</p> <p>is a chain sequence.</p> <p><strong>Example.</strong> In particular, if the entries of the matrix satisfy,</p> <p>\begin{equation*} 0 &lt; \frac{1}{a_ka_{k+1}} &lt; \frac{1}{4\cos^2\left(\frac{\pi}{n+1}\right)},\quad k=1,\ldots,n-1, \end{equation*} then it is positive definite.</p> <hr> <p>For additional information and details about this material, please see:</p> <ol> <li>M. Andelic, and C. M. Da Fonesca. <em>Sufficient conditions for positive definiteness of tridiagonal matrices revisited</em>. (2010).</li> </ol>