similar to: Matrix inversion

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

Hello. I am trying to invert a matrix, and I am finding that I can get different answers depending on whether I set LAPACK true or false using "qr". I had understood that LAPACK is, in general more robust and faster than LINPACK, so I am confused as to why I am getting what seems to be invalid answers. The matrix is ostensibly the Hessian for a function I am optimizing. I want to get

I haven't found a function that directly calculates the matrix inverse, does it exist? Otherwise what would be the fastest way to do it? Patrik Waldmann -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

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 methods of solving this system that I found online, two of which give me an error and one of which succeeds on the

> 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

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

On Thu, 19 Apr 2007, brech at delphioutpost.com wrote: > Full_Name: Christian Brechbuehler > Version: 2.4.1 Patched (2007-03-25 r40917) > OS: Linux 2.6.15-27-adm64-xeon; Ubuntu 6.06.1 LTS > Submission from: (NULL) (24.61.47.236) > > > Splus and R have different ideas about what qr.coef(qr()) should return, > which is fine... but I believe that R has a bug in that it is not

Hello, I have a question that is not directly related to R ... but I try to do it in R ;-) : I would like to generate a matrix Q satisfying (for a given Z, X and W) the two following conditions: t(Q)%*%Q=Z (1) XQ=W (2) where: Q is m rows and r columns X is p rows and m columns D is p rows and r columns C is r rows and r columns with m>p,r e.g: m=6, p=2 r=3

The solve function in r-patched has been changed so that it applies a tolerance when using Lapack routines to calculate the inverse of a matrix or to solve a system of linear equations. A tolerance has always been used with the Linpack routines but not with the Lapack routines in versions 1.7.x and 1.8.0. (You can use the optional argument tol = 0 to override this check for computational

Generalised Inverse: The Moore-Penrose Generalisied Inverse is probably better defined as a pseudo-Inverse that arises in solving least squares problems. Another well known pseudo-Inverse is the so-called Drazin pseudo-Inverse. If memory serves (and it's been 10-12 years!) it can be obtained via a diagonalisation. Anyway, I dare say Prof. Ripley (among others) probably has "all the

Dear R-listers, I have a dxr matrix Z, where d > r. And the product Z*Z' is a singular square matrix. The problem is how to get the left inverse U of this singular matrix Z*Z', such that U*(Z*Z') = I? Is there any to figure it out using matrix decomposition method? Thanks a lot for your help. Fred

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

R-users E-mail: r-help@r-project.org I have a quenstion on "gam()" in "gam" package. The help of gam() says: 'gam' uses the _backfitting algorithm_ to combine different smoothing or fitting methods. On the other hand, lm.wfit(), which is a routine of gam.fit() contains: z <- .Fortran("dqrls", qr = x * wts, n = n, p = p, y = y *

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

I just detected, that det() is not working on complex matrices any more, due to the fix to the bug reports noted above. I am not happy with this, as determinants are perfectly usable on complex matrices. AFAIUI the bugs resulted from less than optimal behaviour of qr() in certain cases. IMHO this is due to the unhappy decision to use a default for parameter tol to decide whether the the

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

Hi all, I wonder if this kind of questions are ok in this list... Quick question: What does it mean than the rank of the QR decomposition of a NxN matrix is N-1 ? m: NxN matrix qr(m)$rank equal to (N-1) Long version: I'm doing a manova on a matrix of 10 variables and 16 observations. > dim(tmp) [1] 16 10 > fit <- manova( tmp ~ treatment*mouse ) >results <-

Dear R people, I do not have much knowledge about linear algebra but currently I need to understand what the function qr.qty is actually doing. The documentation states that it calculates t(Q) %*% y via a previously performed QR matrix decomposition. In order to do that, I tried following basic example: m<-matrix(c(1,0,0,0,1,0,0,0,1,0,0,1),ncol=3) # 4x3 matrix

I have (below) an attempt at an R script to find the limit distribution of a continuous-time Markov process, using the formulae outlined at http://www.uwm.edu/~ziyu/ctc.pdf, page 5. First, is there a better exposition of a practical algorithm for doing this? I have not found an R package that does this specifically, nor anything on the web. Second, the script below will give the right

I think one needs an LU decomposition rather than QR. However, I couldn't find anything off the shelf to do an LU, other than learning that determinant() now uses LU instead of QR or SVD, so the code to do it must be in there for those that want it. You'll probably need to divide rows of U by the first entry if you insist on the unique reduced REF. However, I can't see any reason