similar to: Spectral Decomposition

Hi: I create a hermitian matrix and then perform its singular value decomposition. But when I put it back, I don't get the original hermitian matrix. I am having the same problem with spectral value decomposition as well. I am using R 1.7.0 on Windows. Here is my code: X <- matrix(rnorm(16)+1i*rnorm(16),4) X <- X + t(X) X[upper.tri(X)] <- Conj(X[upper.tri(X)]) Y <-

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

I thought that the function eigen(A) will return a matrix with eigenvectors that are independent of each other (thus forming a base and the matrix being invertible). This seems not to be the case in the following example A=matrix(c(1,2,0,1),nrow=2,byrow=T) eigen(A) ->ev solve(ev$vectors) note that I try to get the upper triangular form with eigenvalues on the diagonal and (possibly) 1 just

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

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

Hi, I am using eigen to get an eigen decomposition of a square, symmetric matrix. For some reason, I am getting a column in my eigen vectors (the 52nd column out of 601) that is a column of all NAs. I am using the option, symmetric=T for eigen. I just discovered that I do not get this behavior when I use the option EISPACK=T. With EISPACK=T, the 52nd eigenvector is (up to rounding error) a

Hello all, I have the square symetric matrix A: 2 1 1 1 2 1 1 1 2 My first question is what is the easiest way to enter this matriz in R? Second, matrix A has an eigenvalue with multiplicity 2, in this case, how could I find the two related ortogonal eigenvectors given below by R, without the help of R, I mean, I want to know how R calculate this eigenvectors related to the same eigenvalue.

Hi everyone, I am using eigen() to extract the 2 major eigenpairs from a large real square symmetric matrix. The procedure is already rather efficient, but becomes somehow slow for real time needs with moderately large matrices (few thousand lines). The R implementation statically extracts all eigenvalues (and optionally associated eigenvectors). I heard about optimizations of the eigen

R-3.0.1 rev 62743, binary downloaded from CRAN just now; macosx 10.8.3 Hello, eigen(symmetric=TRUE) behaves strangely when given complex matrices. The following two lines define 'A', a 100x100 (real) symmetric matrix which theoretical considerations [Bochner's theorem] show to be positive definite: jj <- matrix(0,100,100) A <- exp(-0.1*(row(jj)-col(jj))^2) A's being

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 dear R-users I try to reproduce the steps included in a LDA. Concerning the eigenvectors there is a difference to SPSS. In my textbook (Bortz) it says, that the matrix with the eigenvectors V usually are not normalized to the length of 1, but in the way that the following holds (SPSS does the same thing): t(Vstar)%*%Derror%*%Vstar = I where Vstar are the normalized eigenvectors. Derror

Hi, I try to calculate the angle between two first eigenvectors of different covariance matrices of biological phenotypic traits for different populations. My issue here is, that all possibilities to do so seem to normalize the eigenvectors to length 1. Although the helpfile of eigen() states, that using eigen(, symmetric = FALSE, EISPACK =TRUE) skips normalization this is (I guess) not applicable

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 users, How to get a symmetric square root of a positive definite matrix? I have tried using spectral decomposition, but some eigen values come out to be complex. Is there any function in R that can give the symmetric square root of a pd matrix? -- Arindam Fadikar M.Stat Indian Statistical Institute. New Delhi, India [[alternative HTML version deleted]]

hello, i want to compute the top k eigenvalues+eigenvectors of a (large) real symmetric matrix. since it doesn't look like any top-level R function does this, i'll call LAPACK from a C shlib and then use .Call. the only LAPACK function i see to do this in R_ext/Lapack.h is dsyevx. however, i know that in LAPACK dsyevr can also return a partial eigendecomposition. why is dsyevr not

hello, i want to compute the top k eigenvalues+eigenvectors of a (large) real symmetric matrix. since it doesn't look like any top-level R function does this, i'll call LAPACK from a C shlib and then use .Call. the only LAPACK function i see to do this in R_ext/Lapack.h is dsyevx. however, i know that in LAPACK dsyevr can also return a partial eigendecomposition. why is dsyevr not

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]

Say a matrix of size of thousands? I am looking for an eigen-value decomposition algo in R to give good eigenvalues... Is that a hopeful thing? Thank you! [[alternative HTML version deleted]]

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 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]]