[R] SVD of a variance matrix

Ravi Varadhan rvaradhan at jhmi.edu
Wed Apr 16 00:03:29 CEST 2008

Yes.  SVD of any symmetric (which is, of course, also square) matrix will
always have U = V.  Also, SVD is the same as spectral decomposition, and the
columns of U and V are the eigenvectors, but the singular values will be the
absolute value of eigenvalues.



Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvaradhan at jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html



-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of Giovanni Petris
Sent: Tuesday, April 15, 2008 5:43 PM
To: r-help at r-project.org
Subject: [R] SVD of a variance matrix


I suppose this is more a matrix theory question than a question on R,
but I will give it a try...

I am using La.svd to compute the singular value decomposition (SVD) of
a variance matrix, i.e., a symmetric nonnegative definite square
matrix. Let S be my variance matrix, and S = U D V' be its SVD. In my
numerical experiments I always got U = V. Is this necessarily the
case? Or I might eventually run into a SVD which has U != V?

Thank you in advance for your insights and pointers. 



Giovanni Petris  <GPetris at uark.edu>
Associate Professor
Department of Mathematical Sciences
University of Arkansas - Fayetteville, AR 72701
Ph: (479) 575-6324, 575-8630 (fax)

R-help at r-project.org mailing list
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

More information about the R-help mailing list