Re: [R] highest eigenvalues of a matrix

From: Simon Wood <s.wood_at_bath.ac.uk>
Date: Thu, 19 Jun 2008 10:25:23 +0100

>
> I happily use eigen() to compute the eigenvalues and eigenvectors of
> a fairly large matrix (200x200, say), but it seems over-killed as its
> rank is limited to typically 2 or 3. I sort of remember being taught
> that numerical techniques can find iteratively decreasing eigenvalues
> and corresponding orthogonal eigenvectors, which would provide a nice
> alternative (once I have the first 3, say, I stop the search).

Lanczos iteration will do this efficiently (see e.g. Golub & van Loan "Matrix Computations"), but I don't think that there are any such routines built into R or LAPACK (although I haven't checked the latest LAPACK release). When I looked it seemed that the LAPACK options that allow you to select eigen-values/vectors still depend on an initial O(n^3) decomposition of the matrix, rather than the O(n^2) that a Lanczos based method would require.

My `mgcv' package (see cran) uses Lanczos iteration for setting up low rank bases for smoothing. The source code is in mgcv/src/matrix.c:lanczos_spd, but I'm afraid that there is no direct R interface, although it would not be too hard to write a suitable wrapper. It requires the matrix to be symmetric.

> Looking at the R source code for eigen and some posts on this list,
> it seems that the function uses a LAPACK routine, but obviously all
> the options are not available through the R wrapper. I'm not
> experienced enough to try and make my own interface with Fortran
> code, so here are two questions:
>
> - is this option (choosing a desired number of eigenvectors) already
> implemented in some function / package that I missed?
--- In the symmetric case you can use `svd' which lets you select (although you'd need to fix up the signs of the singular values to get eigen-values if the matrix is not +ve definite). But this answer is pretty useless as it will be slower than using `eigen' and getting the full decomposition.

Of course if you know that your matrix is low rank because it's a product of non-square matrices then there's usually some way of getting at the eigen-decomposition efficiently. E.g. if A=B'B where B is 3 by 1000, then the cost can easily be kept down to O(1000^2) in R...

best,
Simon

> - is the "range of indices" option in DSYEVR.f < http://
> www.netlib.org/lapack/double/dsyevr.f > what I think, the indices of
> the desired eigenvalues ordered from the highest to lowest?
>
> Many thanks in advance for any piece of advice,
>
> Sincerely,
>

> Baptiste
>
> dummy example if needed:
>
> test <- matrix(c(1, 2, 0, 4, 5, 6, 1.00001, 2, 0), ncol=3)
> eigen(test)
>
>
>
>
> _____________________________
>
> Baptiste Auguié
>
> Physics Department
> University of Exeter
> Stocker Road,
> Exeter, Devon,
> EX4 4QL, UK
>
> Phone: +44 1392 264187
>
> http://newton.ex.ac.uk/research/emag
> http://projects.ex.ac.uk/atto
>
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-- 

> Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK
> +44 1225 386603 www.maths.bath.ac.uk/~sw283
______________________________________________ R-help_at_r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Received on Thu 19 Jun 2008 - 09:24:29 GMT

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