[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.
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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
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-----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
Hello!
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
--
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)
http://definetti.uark.edu/~gpetris/
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