Let an $n\times n$ matrix ${\bf A}$, the all ones vector ${\bf w}$, and the $n\times n$ Krylov matrix $${\bf K}_n = \left[ {\bf w}\;\;{\bf A}{\bf w}\;\;\ldots \;\; {\bf A}^{n1}{\bf w}\right].$$ Is there a way to characterize the spectrum of ${\bf K}_n$ in terms of the eigenvalues of ${\bf A}$?

$\begingroup$ Why the "randommatrices" tag? $\endgroup$ – Yemon Choi Nov 6 '11 at 3:39

$\begingroup$ I guess because I consider ${\bf A}$ to be "generic" or random. $\endgroup$ – Anadim Nov 6 '11 at 4:45

$\begingroup$ One trivial situation arises when $w$ is an eigenvector of $A$. $\endgroup$ – Suvrit Nov 6 '11 at 10:22

$\begingroup$ Where do you meet these matrices? What is known about them? I know they used in integrable system theor to construct separated variaables.... $\endgroup$ – Alexander Chervov Nov 6 '11 at 10:28
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 $b$, $K b = \sum_{j=0}^{n1} b_{j+1} A^j w$. So $K$ is singular if and only if $w$ is in the null space of a nontrivial polynomial in $A$ of degree $\le n1$.
I don't see any reason for there to be a nice characterization. For instance if $A$ is diagonal then $K_n$ is a Vandermonde matrix, so its spectrum is fairly complicated...

$\begingroup$ Would you suggest any particular references on random Vandermonde matrices? $\endgroup$ – Anadim Nov 6 '11 at 4:48

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.
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 $H_m$, $m<n$, are fairly readily computed. I think the question of why the (Ritz) eigenvalues of $H_n$ converge to those of $A$ is an open question for general $A$.