most recent 30 from http://mathoverflow.net 2013-06-18T06:41:29Z http://mathoverflow.net/feeds/question/80184 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80184/eigenvalues-of-krylov-matrices Anadim 2011-11-06T02:54:49Z 2011-11-06T16:56:00Z

<p>Let an $n\times n$ matrix ${\bf A}$, the all ones vector ${\bf w}$, and the $n\times n$ Krylov matrix $${\bf K}_n = \left[ {\bf w}\;\;{\bf A}{\bf w}\;\;\ldots \;\; {\bf A}^{n-1}{\bf w}\right].$$ Is there a way to characterize the spectrum of ${\bf K}_n$ in terms of the eigenvalues of ${\bf A}$?</p>

http://mathoverflow.net/questions/80184/eigenvalues-of-krylov-matrices/80187#80187 Robert Israel 2011-11-06T04:24:59Z 2011-11-06T04:24:59Z

<p>Certainly not in terms of the eigenvalues of $A$, because this won't be invariant under similarity transformations on $A$. One thing I can say is that for any vector $b$, $K b = \sum_{j=0}^{n-1} b_{j+1} A^j w$. So $K$ is singular if and only if $w$ is in the null space of a nontrivial polynomial in $A$ of degree $\le n-1$. </p>

http://mathoverflow.net/questions/80184/eigenvalues-of-krylov-matrices/80188#80188 Faisal 2011-11-06T04:25:30Z 2011-11-06T04:25:30Z

<p>I don't see any reason for there to be a nice characterization. For instance if <code>$A$</code> is diagonal then <code>$K_n$</code> is a Vandermonde matrix, so its spectrum is fairly complicated...</p>

http://mathoverflow.net/questions/80184/eigenvalues-of-krylov-matrices/80224#80224 Nilima Nigam 2011-11-06T16:56:00Z 2011-11-06T16:56:00Z

<p>The short answer is: no. You can see the difficulty if $w$ is an eigenvector of $A$:the Krylov matrix becomes singular, while $A$ may not be. </p> <p>The Krylov matrix is generated, as you probably know, during the Arnoldi iteration for locating eigenvalues of A. As part of the (stabilized version) of the process, A is partially reduced through orthogonal projections onto $\cal{K}_n$ to Hessenberg form, $H_n$. The eigenvalues of $H_m$, $m