similar to: SVD least squares sub-space projection

Dear R list users, the singluar value decomposition of a symmetric matrix M is UDV^(T), where U = V. La.svd(M) gives as output three elements: the diagonal of D and the two orthogonal matrices u and vt (which is already the transpose of v). I noticed that the transpose of vt is not exactly u. Why is that? thank you for your attention and your help Stefano AVVISO IMPORTANTE: Questo messaggio di

Hi. I have been looking at the PCAproj function in package pcaPP (R 2.4.1) for robust principal components, and I'm trying to interpret the results. I started with a data matrix of dimensions RxC (R is the number of rows / observations, C the number of columns / variables). PCAproj returns a list of class princomp, similar to the output of the function princomp. In a case where I can

Hi: Does R have a function as gsorth is SAS, that perform a the Gram-Schmidt orthonormal factorization of the m ?n matrix A, where m is greater than or equal to n? That is, the GSORTH subroutine in SAS computes the column-orthonormal m ?n matrix P and the upper triangular n ?n matrix T such that A = P*T. or any other version of Gram-Schmidt orthonormal factorization? I search the help, but I

Hi All, I performed an svd on a matrix X and saved the first three column of the left singular matrix U. ( I assume that they correspond to the projection of the matrix on the first three eigen vectors that corresponds to the first three largest eigenvalues). I would like to know how much variance is explained by the first eigenvectors? how can I find that. Thanks for your help -- View this

Hi, Why the qr() produces a negative Q compared with Gram-Schmidt? (note example below, except Q[2,3]) Here is an example, I calculate the Q by Gram-Schmidt process and compare the output with qr.Q() a <- c(1,0,1) b <- c(1,0,0) c <- c(2,1,0) x <- matrix(c(a,b,c),3,3) ########################## # Gram-Schmidt ########################## A <- matrix(a,3,1) q1 <-

I have a number of space separated files of weather data, with some equivalent column names, and differing number of fields in each file. Some of the files have 40 or more vars, but I only want a subset of the fields. I can use colClasses with read.table to drop some of the fields, but only if I know where those columns are in the first place, and they're not always in the same place. So I

Per the discussion about the problems with prcomp() when n << p, which boils down to a problem with svd() when n << p, here is a patch to prcomp() which substitutes La.svd() instead of svd(). -Greg (This is really a feature enhancement, but submitted to R-bugs to make sure it doesn't get lost. ) *** R-1.6.0/src/library/mva/R/prcomp.R Mon Aug 13 17:41:50 2001 ---

(I tried to send this earlier, but it doesnt seem to have come through, due to problems on my system) Hola: Both cannot be correct: > m <- matrix(1:4, 2) > svd(m) $d [1] 5.4649857 0.3659662 $u [,1] [,2] [1,] -0.5760484 -0.8174156 [2,] -0.8174156 0.5760484 $v [,1] [,2] [1,] -0.4045536 0.9145143 [2,] -0.9145143 -0.4045536 > La.svd(m) $d [1]

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

Hi, I'm trying to do PCA on a n by p wide matrix (n < p), and I'd like to get more principal components than there are rows. However, svd() only returns a V matrix of with n columns (instead of p) unless the argument nv=p is set (prcomp calls svd without setting it). Moreover, the eigenvalues returned are always min(n, p) instead of p, even if nv is set: > x <-

Hello List, i need help for eigen and svd functions. I have a non-symmetric square matrix. These matrix is not positive (some eigenvalues are negative). I want to diagonalise these matrix. So, I use svd and eigen and i compare the results. eigen give me the "good" eigenvalues (positive and negative). I compare with another software and the results are the same. BUT, when i use svd,

Dear R-helpers, could anybody explain me briefly what is the difference between eigenvectors returned by 'eigen' and 'svd' functions and how they are related? Thanks in advance Ondrej Mikula

Hi, I met a probem recently and need your help. I would really appreciate it. I kept receiving the following error message when running a program: 'Error in svd(X) : infinite or missing values in x'. However, I did not use any svd function in this program though I did include the function pseudoinverse. Is the problem caused by doing pseudoinverse? Best regards, Tongtong

--=====================_24736660==_ Content-Type: text/plain; charset="iso-8859-1"; format=flowed Content-Transfer-Encoding: quoted-printable SVD-Error on R 1.1.0 Windows 98 I get the following error applying svd on a positive definite matrix : > sk2 [,1] [,2] [,3] [,4] [,5] [1,] 1.0460139783 0.084356992 -2.810553e-04

Dear All, Is it possible to manage a svd analysis within a matrix containing NA values. If not how do I could overcome this problem. Thanks in advance Antonio

Generator Microsoft Word 11 (filtered medium) Hi, I'm the mannova package maintainer. We used La.svd(X, method="dgesvd") in maanova package before. After R-2.3.0, the old La.svd() method was deprecated for option method="dgesvd". I changed maanova code correspondingly, which will call method="dgesdd" instead. But after that, we keep getting below error message

Dear R-users, I computed determinant of a square matrix "var.r" using the SVD output: detr _ 1 d _ svd(var.r)$d for (i in 1:length(d)) { detr _ detr*d[i] } print(detr) 30.20886 BUT when I tried : det(var.r) I got : -30.20886 Is this because SVD output will only give absolute of the eigenvalues ?, If this is the case how can I get the original eigenvalues? Thanks, Agus

Ordinary functions should not alter their inputs but in R-2.15.2 svd(LINPACK=TRUE,X) does. (It worked in 2.15.0 but not in 2.15.1 or 2.15.2 and became deprecated in 2.15.2.) > X <- matrix(c(1,2,3, 5,7,11, 13,17,19), 3, 3) > X [,1] [,2] [,3] [1,] 1 5 13 [2,] 2 7 17 [3,] 3 11 19 > svd(X, LINPACK=TRUE)$d [1] 31.9718214 2.3882717 0.3143114 Warning message:

Is there a way to determine the SVD of a complex matrix using R? (I'm using v1.0.1 and svd() won't do the trick). I know LAPACK has a function to do this. Thanks -- Ben Stapley Biomolecular Modelling Lab Imperial Cancer Research -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

Dear list, I searched the libraries but could not find means to compute the svd of a coupled field. Is it possible in R Thanks nuncio -- Nuncio.M Research Scientist National Center for Antarctic and Ocean research Head land Sada Vasco da Gamma Goa-403804 [[alternative HTML version deleted]]