similar to: the definition of eigenvector in R

Dear all I am so glad the R can provide the efficient calculate about eigenvector and eigenvalue. However, i have some puzzle about the procedure of eigen. Fristly, what kind of procedue does the R utilize such that the eigen are obtained? For example, A=matrix(c(1,2,4,3),2,2) we can define the eigenvalue lamda, such as det | 1-lamda 4 | =0 | 2 3-lamda | then

Hi, All: What tools exist for diagnosing singular gradient problems with 'nls'? Consider the following toy example: DF1 <- data.frame(y=1:9, one=rep(1,9)) nlsToyProblem <- nls(y~(a+2*b)*one, DF1, start=list(a=1, b=1), control=nls.control(warnOnly=TRUE)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial

Hi, If I use the eigen() function to find the eigenvalues of a matrix, how can I find the eigenvector corresponding to the largest eigen value? Thanks! [[alternative HTML version deleted]]

Hi, with small matrices eigen works as expected: > eigen(cbind(c(1,4),c(4,7)), only.values = TRUE) $values [1] 9 -1 $vectors NULL > eigen(cbind(c(1,4),c(4,7))) $values [1] 9 -1 $vectors [,1] [,2] [1,] 0.4472136 -0.8944272 [2,] 0.8944272 0.4472136 > eigen(cbind(c(1,-1),c(1,-1))) $values [1] -3.25177e-17+1.570092e-16i -3.25177e-17-1.570092e-16i $vectors

Hello all, I was wondering if there is a function in R that only computes the eigenvector corresponding to the largest/smallest eigenvalue of an arbitrary real matrix. Thanks Minh -- Living on Earth may be expensive, but it includes an annual free trip around the Sun.

hi all i have a question that about the eigen analysis found in R and in eviews. i used the same data set in the two packages and found different answers. which is incorrect? the data is: aa ( a correlation matrix) 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 now > svd(aa) $d [1] 4.9204

Dear R users & Experts, This is just a curiousity, I was wondering why the dominant eigenvetor and eigenvalue of the following matrix is given as the third. I guess this could complicate automatic selection procedures. 0 0 0 0 0 5 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 Please

I am trying to check the results from an Eigen decomposition and I need to force a scalar multiplication. The fundamental equation is: Ax = lx. Where 'l' is the eigen value and x is the eigen vector corresponding to the eigenvalue. 'R' returns the eigenvalues as a vector (e <- eigen(A); e$values). So in order to 'check' the result I would multiply the eigenvalues

In the manual page for prcomp(), it says that sdev is "the standard deviations of the principal components (i.e., the square roots of the eigenvalues of the covariance/correlation matrix, though the calculation is actually done with the singular values of the data matrix)." ?However, this is not what I'm finding. ?The values appear to be the standard deviations of a reprojection of

Hi, I was wondering if function eigen() does something different from the function call eigen() in SAS. I'm in the process of translating a SAS code into a R code and the values of the eigenvectors and eigenvalues of a square matrix came out to be different from the values in SAS. I would also appreciate it if someone can explain the difference in simple terms. I'm pretty new to both

Dear All, how do I compute the left eigenvector of a matrix? I gather that "eigen" computes the right eigenvectors... Regards, Federico Calboli -- ================================= Federico C. F. Calboli PLEASE NOTE NEW ADDRESS Dipartimento di Biologia Via Selmi 3 40126 Bologna Italy tel (+39) 051 209 4187 fax (+39) 051 251 208 f.calboli at ucl.ac.uk

Hello. Consider the following matrix: mp <- matrix(c(0,1/4,1/4,3/4,0,1/4,1/4,3/4,1/2),3,3,byrow=T) > mp [,1] [,2] [,3] [1,] 0.00 0.25 0.25 [2,] 0.75 0.00 0.25 [3,] 0.25 0.75 0.50 The eigenvectors of the previous matrix are 1, 0.25 and 0.25 and it is not a diagonalizable matrix. When you try to find the eigenvalues and eigenvectors with R, R responses: > eigen(mp) $values [1]

Inicio del mensaje reenviado: > De: Arnau Mir <arnau.mir@uib.es> > Fecha: 14 de noviembre de 2011 13:24:31 GMT+01:00 > Para: Martin Maechler <maechler@stat.math.ethz.ch> > Asunto: Re: [R] How to compute eigenvectors and eigenvalues? > > Sorry, but I can't explain very well. > > > The matrix 4*mp is: > > 4*mp > [,1] [,2] [,3] > [1,]

Dear all, I've used the 'prcomp' command to calculate the eigenvalues and eigenvectors of a matrix(gg). Using the command 'principal' from the 'psych' package  I've performed the same exercise. I got the same eigenvalues but different eigenvectors. Is there any reason for that difference? Below are the steps I've followed: 1. PRCOMP #defining the matrix

$ R --version R 1.6.1 (2002-11-01). So I would like to perform principal components analysis on a 16X16 correlation matrix, [princomp(cov.mat=x) where x is correlation matrix], the problem is princomp complains that it is not non-negative definite. I called eigen() on the correlation matrix and found that one of the eigenvectors is close to zero & negative (-0.001832311). Is there any way

Hi, How the eigenvectors output by the eigen() function are ordered. The first column corresponds to the largest eigenvalue? or is the last column as in Octave? I'm performing a spatial-temporal analysis of some climatic variables so my matrices are MxN (locations*time)and I'm looking for the leading EOF's. As I have understand the eigenvectors columns represent those EOF's

Hi everyone, I want to convert an R function into VBA for calculating the eigenvectors and eigenvalues of a matrix using the "Power Method". The function is: PowerMethod <- function(x, tolerance) {     my.mat <- var(x[,-1], na.method="available")     matSize <- dim(my.mat)[1]     eigenVec <- matrix(NA, nrow=matSize, ncol=matSize)     eigenVal <- rep(NA,

Folks: I'm trying to port some code from python over to R, and I'm running into a wall finding R code that can solve a generalized eigenvalue problem following this function model: http://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.eig.html Any ideas? I don't want to call python from within R for various reasons, I'd prefer a "native" R solution if one

Hey, All In principal component analysis (PCA), we want to know how many percentage the first principal component explain the total variances among the data. Assume the data matrix X is zero-meaned, and I used the following procedures: C = covriance(X) %% calculate the covariance matrix; [EVector,EValues]=eig(C) %% L = diag(EValues) %%L is a column vector with eigenvalues as the elements percent

Dear R-listers Is there anyone who knows why I get different eigenvectors when I run MatLab and R? I run both programs in Windows Me. Can I make R to produce the same vectors as MatLab? #R Matrix PA9900<-c(11/24 ,10/53 ,0/1 ,0/1 ,29/43 ,1/24 ,27/53 ,0/1 ,0/1 ,13/43 ,14/24 ,178/53 ,146/244 ,17/23 ,15/43 ,2/24 ,4/53 ,0/1 ,2/23 ,2/43 ,4/24 ,58/53 ,26/244 ,0/1 ,5/43) #R-syntax