similar to: (Off topic.) Observed Fisher information.

1. Martin Maechler's comments should be taken as replacements for anything I wrote where appropriate. Any apparent conflict is a result of his superior knowledge. 2. 'eigen' returns the eigenvalue decomposition assuming the matrix is symmetric, ignoring anything in m[upper.tri(m)]. 3. The basic idea behind both posdefify and nearPD is to compute the

Dear R-users, Is there a preferred method for testing whether a real symmetric matrix is positive definite? [modulo machine rounding errors.] The obvious way of computing eigenvalues via "E <- eigen(A, symmetric=T, only.values=T)$values" and returning the result of "!any(E <= 0)" seems less efficient than going through the LU decomposition invoked in

Hi, Given a positive integer N, and a real number \lambda such that 0 < \lambda < 1, I would like to generate an N by N stochastic matrix (a matrix with all the rows summing to 1), such that it has the second largest eigenvalue equal to \lambda (Note: the dominant eigenvalue of a stochastic matrix is 1). I don't care what the other eigenvalues are. The second eigenvalue is

Dear Peter, thank you very much for your answer. My problem is that I need to calculate the following quantity: solve(chol(A)%*%Y) Y is a 3*3 diagonal matrix and A is a 3*3 matrix. Unfortunately one eigenvalue of A is negative. I can anyway take the square root of A but when I multiply it by Y, the imaginary part of the square root of A is dropped, and I do not get the right answer. I tried

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

Dear all, I encounter some covariance matrix with quite small eigenvalues (around 1e-18), which are smaller than the machine precision. The dimension of my matrix is 17. Here I just fake some small matrix for illustration. a<-diag(c(rep(3,4),1e-18)) # a matrix with small eigenvalues b<-matrix(1:25,ncol=5) # define b to get an orthogonal matrix b<-b+t(b) bb<-eigen(b,symmetric=T)

$ 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

Hello, I have a short question about the prcomp function. First I cite the associated help page (help(prcomp)): "Value: ... SDEV 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). ROTATION the matrix of variable loadings

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

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

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]

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

It works! But Once I have the square root of this matrix, how do I convert it to a real (not imaginary) matrix which has the same property? Is that possible? Best, Simon >----Messaggio originale---- >Da: p.dalgaard at biostat.ku.dk >Data: 21-nov-2009 18.56 >A: "Charles C. Berry"<cberry at tajo.ucsd.edu> >Cc: "simona.racioppi at

Dear all, I am currently working on the calculation of eigenvalues (and -vectors) of large matrices. Since these are mostly sparse matrices and I remember some specific functionalities in MATLAB for sparse matrices, I started a research how to optimize the calculation of eigenvalues of a sparse matrix. The function eigen itself works with the LAPACK library which has no special handling for

Hello Sir I am getting problem in plotting in CCA . Could you please help me? I wrote the below command but I don't know why it is taking only first 5 env data rather than all 9. > strain.data <- read.xlsx("Dee rhiz.xlsx", sheetName="strain", header = T, row.names = 1) > env.data <- read.xlsx("Dee rhiz.xlsx", sheetName="env", header = T,

Hello I am trying to determine wether a given matrix is symmetric and positive matrix. The matrix has real valued elements. I have been reading about the cholesky method and another method is to find the eigenvalues. I cant understand how to implement either of the two. Can someone point me to the right direction. I have used ?chol to see the help but if the matrix is not positive definite it

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]]

Is it possible to avoid the loop in the following function (or make the function otherwise more efficient) and can someone point me to a possible solution? (It would be great if hours could be reduced to seconds :-). # --------------------------------------------- RanEigen=function(items=x,cases=y,sample=z) { X=matrix(rnorm(cases*items),nrow=cases,byrow=F) S=crossprod(X-rep(1,cases) %*%

On April 15th, Elizabeth wrote: <snip> > In execises 39-42, determine if the columns of the matrix span > R4: <snip> >(or x <- matrix(data=c(7, -5, 6, -7, 2, -3, 10, 9, -5, > 4, -2, 2, 8, -9, 7, 15), nrow=4, ncol=4) > >That is the whole of the question <snip> Have you tried det(x) and/or eigen(x) ? A zero determinant (within

Dear Friends, Here is an algorithm for finding concave hulls: http://get.dsi.uminho.pt/local/ Has anyone implemented such an algorithm in R? RSiteSearch('concave hull') didn't reveal one (I think). _____________________________ Professor Michael Kubovy University of Virginia Department of Psychology Postal Address: P.O.Box 400400, Charlottesville, VA 22904-4400 Express Parcels