similar to: matrix of eigenvalues

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]

DeaR list, I happily use eigen() to compute the eigenvalues and eigenvectors of a fairly large matrix (200x200, say), but it seems over-killed as its rank is limited to typically 2 or 3. I sort of remember being taught that numerical techniques can find iteratively decreasing eigenvalues and corresponding orthogonal eigenvectors, which would provide a nice alternative (once I have the

Hello, Can you get eigenvalues in addition to eigevectors using prcomp? If so how? I am unable to use princomp due to small sample sizes. Thank you in advance for your help! Rebecca Young -- Rebecca Young Graduate Student Ecology & Evolutionary Biology, Badyaev Lab University of Arizona 1041 E Lowell Tucson, AZ 85721-0088 Office: 425BSW rlyoung at email.arizona.edu (520) 621-4005

from ?eigen symmetric: if 'TRUE', the matrix is assumed to be symmetric (or Hermitian if complex) and only its lower triangle is used. If 'symmetric' is not specified, the matrix is inspected for symmetry. I think that could mislead a naive reader as it suggests that, with symmetric=TRUE, the result of eigen() (vectors and values) depends only on

Hi, It is my understanding that the eigenvectors of a circulant matrix are given as follows: 1,omega,omega^2,....,omega^{p-1} where the matrix has dimension given by p x p and omega is one of p complex roots of unity. (See Bellman for an excellent discussion on this). The matrix created by the attached row and obtained using the following commands indicates no imaginary parts for the

Hello Does anybody know how one can compute d largest eigenvalues/eigenvectors in R, like in MATLAB eigs function ? eigen function computes all eigenvectors/eigenvalues, and they are slightly different than those generated by matlab eigs. Thanks in advance -- View this message in context: http://www.nabble.com/matlab-eigs-function-in-R-tp17619641p17619641.html Sent from the R help mailing list

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

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

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]]

Looks like the files did not go through again. In any case, here is the kinv: please cut and paste and save to a file: -1.16801E-03 -2.24310E-03 -1.16864E-03 -2.24634E-03 -1.17143E-03 -2.25358E-03 -1.17589E-03 -2.26484E-03 -1.18271E-03 -2.27983E-03 -1.19124E-03 -2.29896E-03 -1.20164E-03 -2.32206E-03 -1.21442E-03 -2.34911E-03 -1.22939E-03 -2.38073E-03

I have been looking at the "eigen" function and have reintroduced the ability to compute (right) eigenvalues and vectors for non-symmetric matrices. I've also made "eigen" complex capable. The code is based on the eispack entry points RS, RG, CH, CG (which is what S appears to use too). The problem with both the S and R implementations is that they consume huge amounts

Full_Name: Pierre Legendre Version: 2.1.1 OS: Mac OSX 10.4.3 Submission from: (NULL) (132.204.120.81) I am reporting the mis-behaviour of the function 'eigen' in 'base', for the following input matrix: A <- matrix(c(2,3,4,-1,3,1,1,-2,0),3,3) eigen(A) I obtain the following results, which are incorrect for eigenvalues and eigenvectors 2 and 3 (incorrect imaginary portions):

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

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

I'm using R version 2.0.1 on a Windows 2000 operating system. Here is some actual code I executed: > test [,1] [,2] [1,] 1000 500 [2,] 500 250 > eigen(test, symmetric=T)$values [1] 1.250000e+03 -3.153033e-15 > eigen(test, symmetric=T)$values[2] >= 0 [1] FALSE > eigen(test, symmetric=T, only.values=T)$values [1] 1250 0 > eigen(test, symmetric=T,

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, I have the following problem. It has two parts. 1. I need to calculate the stationary probabilities of a Markov chain, eg if the transition matrix is P, I need x such that xP = x in other words, the left eigenvectors of P which have an eigenvalue of one. Currently I am using eigen(t(P)) and then pick out the vectors I need. However, this seems to be an overkill (I only need a single

Hi, I have a question for my simulation problem: I would like to generate a positive (or semi def positive) covariance matrix, non singular, in wich the spectral decomposition returns me the same values for all dimensions but differs only in eigenvectors. Ex. sigma [,1] [,2] [1,] 5.05 4.95 [2,] 4.95 5.05 > eigen(sigma) $values [1] 10.0 0.1 $vectors [,1]

Dear all, It is not a R related problem rather than statistical/mathematical. However I am posting this query hoping that anyone can help me on this matter. My problem is to get the Geometrical Interpretation of Eigen value and Eigen vector of any square matrix. Can anyone give me a light on it? Thanks and regards, Arun [[alternative HTML version deleted]]

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