1
$\begingroup$

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 ${\bf A}$?

$\endgroup$
4
  • $\begingroup$ Why the "random-matrices" tag? $\endgroup$
    – Yemon Choi
    Nov 6, 2011 at 3:39
  • $\begingroup$ I guess because I consider ${\bf A}$ to be "generic" or random. $\endgroup$
    – Anadim
    Nov 6, 2011 at 4:45
  • $\begingroup$ One trivial situation arises when $w$ is an eigenvector of $A$. $\endgroup$
    – Suvrit
    Nov 6, 2011 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.... $\endgroup$ Nov 6, 2011 at 10:28

3 Answers 3

2
$\begingroup$

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 $\le n-1$.

$\endgroup$
1
$\begingroup$

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...

$\endgroup$
2
  • $\begingroup$ Would you suggest any particular references on random Vandermonde matrices? $\endgroup$
    – Anadim
    Nov 6, 2011 at 4:48
  • $\begingroup$ Sorry, but I don't know of any. $\endgroup$
    – Faisal
    Nov 6, 2011 at 19:41
0
$\begingroup$

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 $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.