Re: [R] SVD of a variance matrix

ssage- From: Giovanni Petris [mailto:[EMAIL PROTECTED] Sent: Tuesday, April 15, 2008 8:07 PM To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED]; r-help@r-project.org Subject: Re: [R] SVD of a variance matrix Hi Ravi, Thank you for your useful reply. Does the result also hold for variance-covariance matrice

Re: [R] SVD of a variance matrix

-- > > > > -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On > Behalf Of Ravi Varadhan > Sent: Tuesday, April 15, 2008 6:03 PM > To: 'Giovanni Petris'; r-help@r-project.o

Re: [R] SVD of a variance matrix

etris'; r-help@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 th

Re: [R] SVD of a variance matrix

-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

[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