'Condition number' for Rayleigh-Ritz quotient - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:12:03Z http://mathoverflow.net/feeds/question/104009 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient 'Condition number' for Rayleigh-Ritz quotient Felix Goldberg 2012-08-05T10:44:01Z 2012-12-07T22:10:22Z <p>Suppose that $A$ is a Hermitian matrix and that $u,v$ are two vectors. Is there some known function $\kappa(A)$ so that $||u-v|| \leq \kappa(A) |\frac{u^{*}Au}{u^{*}u}-\frac{v^{*}Av}{v^{*}v}|$?</p> <p>UPDATE: Andrew T. Baker has shown that the answer is "no" in general (to take an even simpler counter-example, take $A=I$) - so let's add the <strong>assumption</strong> that $u$ is a simple eigenvector of $A$.</p> http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient/104199#104199 Answer by Andrew T. Barker for 'Condition number' for Rayleigh-Ritz quotient Andrew T. Barker 2012-08-07T14:02:05Z 2012-08-07T16:38:23Z <p>I think the answer for a general Hermitian matrix $A$ is no.</p> <p>Let $u, v$ be distinct eigenvectors of $A$ with the same eigenvalue $\lambda$ and normalized so that $u^\ast u = v^\ast v = 1$. Then $\| u - v \| > 0$ and \begin{equation*} | u^\ast Au - v^\ast A v | = | \lambda u^\ast u - \lambda v^\ast v | = | \lambda - \lambda | = 0. \end{equation*}</p> <p>A concrete counterexample is \begin{equation*} A = \left( \begin{array}{ccc} 2&amp; &amp; \\ &amp; 2&amp; \\ &amp; &amp; 1 \end{array} \right), u = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right), v = \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) \end{equation*}</p> <p>ADDED: if we assume all eigenvalues are distinct, then the above argument leads to some kind of bound. Again, $u,v$ are eigenvectors of $A$ with corresponding eigenvalues $\lambda_1, \lambda_2$ and normalized with $u^\ast u = v^\ast v = 1$. Then \begin{equation*} \| u - v \| \leq 2 \end{equation*} and \begin{equation*} | u^\ast Au - v^\ast A v | = | \lambda_1 - \lambda_2 | \end{equation*}</p> <p>so that if we choose \begin{equation*} \kappa(A) \geq \min_{\lambda_i, \lambda_j} \frac{2}{|\lambda_i - \lambda_j|} \end{equation*} we can guarantee your inequality will hold.</p> http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient/104210#104210 Answer by Sharkos for 'Condition number' for Rayleigh-Ritz quotient Sharkos 2012-08-07T16:31:49Z 2012-08-07T16:31:49Z <p>Another trivial problem, I'm afraid:</p> <p>The two sides scale differently under rescaling of $u, v$. That is, if it holds for some $\kappa, u \neq v$ then simply multiply $u, v$ by some huge number $Z$ - then the left-hand side is now $Z$ times bigger whilst the right-hand side is constant.</p> <p>How many further qualifications are necessary?!</p> http://mathoverflow.net/questions/104009/condition-number-for-rayleigh-ritz-quotient/115736#115736 Answer by RLC for 'Condition number' for Rayleigh-Ritz quotient RLC 2012-12-07T19:10:42Z 2012-12-07T22:10:22Z <p>If u is an eigenvector for an eigenvalue that is in the interior of the spectrum, then there is no such bound becomes there are numerous vectors that generate Rayleigh quotients for points in the convex hull of the spectrum. And as for Sharkos comment, you would want something like the sine of the angle between u and v.</p> <p>An absolute condition number for a Rayleigh quotient would look more like $|x^*Ax-y'Ay|&lt;=\kappa$(x'Ax) ||x-y||\$, where K would be the condition number for x'Ax, where x and y are unit vectors. It shows how changes in the Rayleigh quotient can be bounded by changes in the vector.</p> <p>Your suggested condition number is more like the condition number for the generating vector. </p>