similar to: How to get the pseudo left inverse of a singular square matrix?

>I'm rusty, but not *that* rusty here, I hope. > >If W (=Z*Z' in your case) is singular, it can not have >inverse, which by >definition also mean that nothing multiply by it will >produce the identity >matrix (for otherwise it would have an inverse and >thus nonsingular). > >The definition of a generalized inverse is something >like: If A is a >non-null

I'm rusty, but not *that* rusty here, I hope. If W (=Z*Z' in your case) is singular, it can not have inverse, which by definition also mean that nothing multiply by it will produce the identity matrix (for otherwise it would have an inverse and thus nonsingular). The definition of a generalized inverse is something like: If A is a non-null matrix, and G satisfy AGA = A, then G is called

I am using the ginv function from MASS and have run across this problem that I do not understand. If I define the matrix A as below, its g-inverse does not satisfy the Moore-Penrose condition A %*% ginv(A) %*% A = A. The matrix A is X'WX in a quadratic regression using some very large dollar values. The much simpler matrix B does satisfy the MP condition. Am I doing something wrong? Is

Dear R-devel: I could use some advice about matrix calculations and steps that might make for faster computation of generalized inverses. It appears in some projects there is a bottleneck at the use of svd in calculation of generalized inverses. Here's some Rprof output I need to understand. > summaryRprof("Amelia.out") $by.self self.time self.pct

Thorsten is right. There is a direct formula for computing the Moore-Penrose inverse using the singular value composition of a matrix. This is incorporated in the following: mpinv <- function(A, eps = 1e-13) { s <- svd(A) e <- s$d e[e > eps] <- 1/e[e > eps] return(s$v %*% diag(e) %*% t(s$u)) } Hope it helps. Dietrich

Dear R-users, I have one question about using SVD to get an inverse matrix of covariance matrix Sometimes I met many singular values d are close to 0: look this example $d [1] 4.178853e+00 2.722005e+00 2.139863e+00 1.867628e+00 1.588967e+00 [6] 1.401554e+00 1.256964e+00 1.185750e+00 1.060692e+00 9.932592e-01 [11] 9.412768e-01 8.530497e-01 8.211395e-01 8.077817e-01 7.706618e-01 [16]

Is there any routine to obtain a g-inverse of a matrix in R or S-PLUS? Tapio Nummi University of Tampere Finland -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To:

Dear all, The background is that I'm trying to fix this bug in the geometry package: https://r-forge.r-project.org/tracker/index.php?func=detail&aid=1993&group_id=1149&atid=4552 Boiled down, the problem is that there exists at least one matrix X for which det(X) != 0 and for which solve(X) fails giving the error "system is computationally singular: reciprocal condition

i am trying to use matinv from the Design package to compute the generalized inverse of the normal equations of a 3x3 design via the sweep operator. That is, for the linear model y = ? + x1 + x2 + x1*x2 where x1, x2 are 3-level factors and dummy coding is being used the matrix to be inverted is X'X = 9 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 3 3 0 0 1 1 1 1 0 0 1 0 0 1 0 0 3 0 3 0 1 1 1 0 1 0 0 1

Hi, R-listers I tried to plot several graphs in a sigle x-y coordinate settings, like the following: |(y) s | ****** s | ***** s | sssssssssssssssssss |_______________________________(x) where "*" and "s" denote two diffrent plots. However, when I used plot(data1); % data1 is the data points of "*"

> On 20 Apr 2016, at 13:22, A A via R-help <r-help at r-project.org> wrote: > > > > > I have a situation in R where I would like to find any x (if one exists) that solves the linear system of equations Ax = b, where A is square, sparse, and singular, and b is a vector. Here is some code that mimics my issue with a relatively simple A and b, along with three other

Hi, is there a lm which will implement the moore-pensrose generalized inverse. Thanks. Harsh --------------------------------- [[alternative HTML version deleted]]

Hey, R-listers I was recommended to try using smooth.spline function for estimating 2-Dimensinal curve given a data set. So will you please tell me where to get this R function? Or which package provides this function? Thanks for your point. Fred

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

Howdy, I am trying to invert a matrix for the purposes of least squares. I have tried a number of things, and the variety of results has me confused. 1. When I try solve() I get the following: >Error in solve.default(t(X) %*% X) : system is computationally singular: reciprocal condition number = 3.76391e-20 2. When I try qr.solve(), I get: >Error in qr.solve(t(X) %*% X) : singular matrix

Hey, I want to draw several plots sequently, but have to make them dispaly in one figure. So how to achieve this? Thanks. Fred

I cannot find what is the function label for matrix inversion in R. I have found 'ginv' for the moore-penrose in the MASS package, but there is probably a simple inversion operator in the base package. Where can I find it? ____________________________________________ Yvonnick Noel, PhD. University of Lille 3 Department of Psychology F-59653 Villeneuve d'Ascq Cedex (+33) 320 41 63 48

Why do x<-b%*%ginv(A) and x<-solve(A,b) give different results?. It seems that I am missing some basic feature of matrix indexing. e.g.: A<-matrix(c(0,-4,4,0),nrow=2,ncol=2) b<-c(-16,0) x<-b%*%ginv(A);x x<-solve(A,b);x Thanks in advance, Angel

Thanks for the help. Sorry, I am not sure why it looks like that in the mailing list - it looks much more neat on my end (see attached file). On Wednesday, April 20, 2016 2:01 PM, Berend Hasselman <bhh at xs4all.nl> wrote: > On 20 Apr 2016, at 13:22, A A via R-help <r-help at r-project.org> wrote: > > > > > I have a situation in R where I would like to

Dear R group, I have to solve a hessian matrix 40*40, called M, in order to obtain the standart deviations of estimators. When I use the function solve(M), I have the following error message: "Error in solve.default(M) : Lapack routine dgesv: system is exactly singular" Do you know an alternative approach which could succeed? I have found some information about the function