# [R] SVD of a variance matrix

Giovanni Petris GPetris at uark.edu
Wed Apr 16 02:06:45 CEST 2008

```Hi Ravi,

Thank you for your useful reply. Does the result also hold for
variance-covariance matrices that have one or more zero eigenvalues?
Do you have a reference to suggest?

Thank you again!

Giovanni

> Date: Tue, 15 Apr 2008 18:14:11 -0400
> From: Ravi Varadhan <rvaradhan at jhmi.edu>
> Thread-index: AcifQeEz9B1geo3TQyesYlQGMCSuNgAAWF1QAACa9sA=
>
> Let me correct my reply a bit.
>
> U and V will differ by a factor of (-1) corresponding to negative
> eigenvalues (if any) of a general symmetric A.  However, for symmetric
> positive-definite matrices (e.g. variance-covariance matrix), they will be
> identical.
>
> Ravi.
>
> ----------------------------------------------------------------------------
> -------
>
> 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 Ravi Varadhan
> Sent: Tuesday, April 15, 2008 6:03 PM
> To: 'Giovanni Petris'; r-help at r-project.org
> Subject: Re: [R] SVD of a variance matrix
>
> 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.
>
> ----------------------------------------------------------------------------
> -------
>
> 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
>
>
> 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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>

```

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