Re: Fast solution to very small eigenvalue problem

From: Mark Mackey (markm_at_chiark.greenend.org.uk)
Date: 06/25/04 

Date: 25 Jun 2004 11:58:11 +0100 (BST)

In article <cbf3ba$lnn$1@fb04373.mathematik.tu-darmstadt.de>,
Peter Spellucci <nospamspellucci@fb04373.mathematik.tu-darmstadt.de> wrote:
>
>you did not mention it by from "jacobi" I conclude -> symmetric.

Oops, sorry. Yes, it is symmetric.

>hence:
>1) transform to tridiagonal form. don't use LAPACK, write this yourself, just
> 3 givens rotations.

OK.

>2) use bisection on -norm(A),norm(A) for example the infinity norm =
> max row sum of abs-values.

Ok...

> bisection means counting the negative pivots in the lu-decomposition of
> the tridiagonal matrix -mu*I, without pivoting, replacing a zero pivot by eps>0
> this disturbs the eigenvalues by at most eps. if the number of negative pivots is
> <= 3 at mu then mu = new lower bound, otherwise mu = new upper bound,
> until sufficient precision is attained.

I think you lost me there :).

>3) solve now (A-mu I)*x =0 using complete pivoting and setting artificially x(4)=1,
> substituting back then. 

-- 
Mark Mackey 
"The determined Real Programmer can write Fortran programs in any language."
		 - "Real Programmers don't use Pascal"
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