similar to: bug in det using method="qr" (PR#1244) (PR#4450)

Full_Name: Armin Roehrl Version: 1.4.0(2001-12-19) OS: Linux; Suse 7.3 Submission from: (NULL) (217.84.18.204) Hello, On a given matrix det returns different results whether I use the method "qr" or "eivenvalues". The matrix: > m2 [,1] [,2] [,3] [1,] 9.822616e+09 3.841723e+09 79790.09 [2,] 3.841723e+09 1.502536e+09 31251.82 [3,]

Hello, When I use det() and qr() on complex matrices the result is in some cases indeterministic. The documentation speaks of numeric matrices (and not of complex matrices) but det() uses qr() which should handle complex matrices correctly. I've also tried using only qr() with similar results. det() returns a value that is not the determinant of the complex matrix (in accordance with

Hello, Currently, determinant(A) calculates the determinant of 'A' by factorizing A=LU and computing prod(diag(U)) [or the logarithm of the absolute value]. The factorization is done by LAPACK routine DGETRF, which gives a status code INFO, documented [1] as follows: *> INFO is INTEGER *> = 0: successful exit *> < 0: if INFO = -i, the i-th

Hi uRsers, when inverting a 2 by 2 matrix using solve, I encountered a error message: solve.default(sigma, tol = 1e-07) : system is computationally singular: reciprocal condition number = 1.7671e-017 and then I test the determinant of this matrix: 6.341393e-06. In my program, I have a condition block that whether a matrix is invertible like this: if(det(sigma)<1e-7) return NULL;

r-devel list, I have recently written an R package that solves a linear least squares problem, and computes the multivariate normal density function. The bulk of the code is written in C, with interfacing code to the BLAS and Lapack libraries. The motivation here is speed. I ran into a problem computing the determinant of a symmetric matrix in packed storage. Apparently, there are no explicit

Hi, Which R function I should use to obtain determinant of a matrix with real(and complex) numbers? Kalin --------------------------------- Never miss a thing. Make Yahoo your homepage. [[alternative HTML version deleted]]

Hello all. First of all, I have only been using are a short time and I'm not an expert in statistics either. I have the following problem. I'm working with measurements of physical samples, each measurement has about 4000 variables. I have 33 of those samples. From those 400 variables I deduced through non-statiscal means that I needed about 200 of them. I read those into a data.frame

2017-06-22 19:49 GMT+02:00 Uwe Ligges <ligges at statistik.tu-dortmund.de>: > On 22.06.2017 17:11, Bernd Funovits wrote: >> >> Hello, >> >> I experienced some unexpected behaviour while determining the rank of matrices (sometimes 1x1 matrices): >> base::qr(matrix(1e-20))$rank returns 1 (incorrect) >> base::qr(diag(c(1, 1e-20)))$rank returns 2

Two friend reported me a problem, which I can't solve: (I run R-1.0.0, Debian Linux) They hava a function "corr.matrix" (see end of mail), and when they create a 173x173 matrix with this function V <- corr.matrix(0.3, n=173) V1 <- qr.solve(V) reports: Error in qr(a, tol = tol) : NA/NaN/Inf in foreign function call (arg 1) For n < 173, qr.solve returns the correct

I apologise for not including a reproducible example with this query but I hope that I can make things clear without one. I am fitting some finite mixture models to data. Each mixture component has p parameters (p=29 in my application) and there are q components to the mixture. The number of data points is n ~ 1500. I need to select a good q and I have been considering model selection methods

Hi, I have noticed different rank values calculated by qr() depending on LAPACK parameter. When it is FALSE (default) a true rank is estimated and returned. Unfortunately, when LAPACK is set to TRUE, the min(nrow(A), ncol(A)) is returned which is only occasionally a true rank. Would not it be more consistent to replace the rank in the latter case by something based on the following pseudo code ?

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

[Hope this is the right list where to send...] An attempt to update package 'mnormt' involves the addition of a small new function called 'pd.solve'. When I come to the package checking stage, an error occurs in parsing pd.solve.Rd. The full transcript of the outcome is copied below (it includes details on my installation) but the critical point is where the \examples{} section

2017-06-22 20:31 GMT+02:00 Uwe Ligges <ligges at statistik.tu-dortmund.de>: > > > On 22.06.2017 20:09, I?aki ?car wrote: >> >> 2017-06-22 19:49 GMT+02:00 Uwe Ligges <ligges at statistik.tu-dortmund.de>: >>> >>> On 22.06.2017 17:11, Bernd Funovits wrote: >>>> >>>> >>>> Hello, >>>> >>>> I

### An numeric problem in R ######## ###I have two matrix one is########## A <- matrix(c(21.97844, 250.1960, 2752.033, 29675.88, 316318.4, 3349550, 35336827, 24.89267, 261.4211, 2691.009, 27796.02, 288738.7, 3011839, 31498784, 21.80384, 232.3765, 2460.495, 25992.77, 274001.6, 2883756, 30318645, 39.85801, 392.2341, 3971.349, 40814.22, 423126.2,

Dear R-users, a question concerning sparse matrices in package "spam" (spam_0.29-2). On one hand I have a spam object (n X n) from which I cannot compute the inverse. On the other hand, if I convert this object in a plain matrix, I can find the inverse without any problem. Specifically I get the following error message: Error in chol.spam(a, ...) : Singularity problem when

Can somebody help me with the following issue (SEM in R), please:   When I run the model (includes second order models) in R, it gives me the following:   1)       In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars,  :   Could not compute QR decomposition of Hessian. Optimization probably did not converge.   2)       I have aliased parameters and NaNS   or sometimes when

Le 22/01/2018 ? 17:40, Keith O'Hara a ?crit?: > This behavior is noted in the qr documentation, no? > > rank - the rank of x as computed by the decomposition(*): always full rank in the LAPACK case. For a me a "full rank matrix" is a matrix the rank of which is indeed min(nrow(A), ncol(A)) but here the meaning of "always is full rank" is somewhat confusing. Does it

Dear R-help list, Pls I have this problem. Suppose I have a matrix of size nxn say, generated as follows   z<-matrix(rnorm(n*n,0,1),nrow=n)   I want to write a function such that for i in 1:n, I will remove the rows and columns corresponding to i (so, will be left with n-1*n-1 submatrix in each cases). Now I need the sum of the determinant of each of this submatrices. As an example, if n=3, it

In summary.manova the qr decomposition of a NxN matrix is calculated and for some cases is giving me a rank < N. However, following suggestions of professor Ripley to calculate the rank of a Matrix On 7 Jun 2002, Brian Ripley wrote: > For a more reliable answer, look at the SVD > (function svd) and look at the > singular values. For example (from lda.default) X.s <-