similar to: about the Choleski factorization

Dear All, My question is simple but I need someone to help me out. Suppose I have a positive definite matrix A. The funtion chol() gives matrix L, such that A = L'L. The inverse of A, say A.inv, is also positive definite and can be factorized as A.inv = M'M. Then A = inverse of (A.inv) = inverse of (M'M) = (inverse of M) %*% (inverse of M)' = ((inverse of

Ladies and Gentlemen, using > A <- structure(c(1, 0, 0, 3, 2, 1, 4, 5, -3, -2, 1, 0), .Dim = as.integer(c(3,4))) I get > dim(A) [1] 3 4 > qr.R(qr(A),complete=TRUE) [,1] [,2] [,3] [,4] [1,] -1 -3.000000 -4.000000 2.0000000 [2,] 0 -2.236068 -3.130495 -0.8944272 [3,] 0 0.000000 -4.919350 -0.4472136 > qr.R(qr(A),complete=FALSE) [,1]

Dear All i have a little puzzle about eigenvector in the R. As we know that the eigenvector can be displayed on several form. For example A=matrix(c(1,2,4,3),2,2) if we want to get the eigenvalue and eigenvector, the code followed eigen(A) $values [1] 5 -1 $vectors [,1] [,2] [1,] -0.7071068 -0.8944272 [2,] -0.7071068 0.4472136 however, we also can calculate the vector matrix

Hi, All: What tools exist for diagnosing singular gradient problems with 'nls'? Consider the following toy example: DF1 <- data.frame(y=1:9, one=rep(1,9)) nlsToyProblem <- nls(y~(a+2*b)*one, DF1, start=list(a=1, b=1), control=nls.control(warnOnly=TRUE)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial

hi all i have a question that about the eigen analysis found in R and in eviews. i used the same data set in the two packages and found different answers. which is incorrect? the data is: aa ( a correlation matrix) 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 0.9801 0.9801 0.9801 0.9801 0.9801 1 now > svd(aa) $d [1] 4.9204

I have been playing around with sparse matrices in the Matrix package, in particularly with the Cholesky factorization of matrices of class dsCMatrix. And BTW, what a fantastic package. My problem is that I have to carry out repeated Cholesky factorization of a spares symmetric matrices, say Q_1, Q_2, ...,Q_n, where the Q's have the same non-zero pattern. I know in this case one does

I'm trying to use a Xorcom Astribank wth Ubuntu 8.04, but got no success. I can see the channel bank with lsusb, but when I tried to use zaptel_hardware, or when I try the /etc/init.d/script, they don't see my Channel Bank. I compiled the latest Zaptel 1.4.10, with Astribank's dependecies, fxload and libusb-dev. Anyone have a similiar experience ? Best Regards, -- Guilherme Loch G?es

What's the difference between the TE121 and TE122. I read the description on Digium's site and it isn't clear to me. Best regards, -- Guilherme Loch G?es Visite nossa loja virtual: http://www.shopvoip.com.br Not?cias e F?rum sobre VoIP com software livre: http://www.asteriskexperts.com.br -------------- next part -------------- An HTML attachment was scrubbed... URL:

Hi, with small matrices eigen works as expected: > eigen(cbind(c(1,4),c(4,7)), only.values = TRUE) $values [1] 9 -1 $vectors NULL > eigen(cbind(c(1,4),c(4,7))) $values [1] 9 -1 $vectors [,1] [,2] [1,] 0.4472136 -0.8944272 [2,] 0.8944272 0.4472136 > eigen(cbind(c(1,-1),c(1,-1))) $values [1] -3.25177e-17+1.570092e-16i -3.25177e-17-1.570092e-16i $vectors

The following snippet suggests that there is either a bug in qr(,LAPACK=T), or some bug in my understanding. Note that the detected rank is correct (= 2) using the default LINPACK qr, but incorrect (=3) using LAPACK. This is running on Linux Redhat 9.0, using the lapack library that comes with the Redhat distribution. I'm running R 1.7.1 compiled from the source. If the bug is in my

I have a server with 8 Xorcom Astribank, they're connected to the server via a USB hub, my problem is: when a reboot the server the ID of the Channel Bank changes, cousing a big mess on my server. Anybody know how can I solve this ? Best regards, -- Guilherme Loch G?es Visite nossa loja virtual: http://www.shopvoip.com.br Not?cias e F?rum sobre VoIP com software livre:

Dear List, My colleague has been having a problem with the following data and plotting commands. The example below is part of a larger set of plots, but I've isolated the problem to this example using this small dataset (below), which kills rgui consistently. My version info > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status

Hi All, I am looking for help in rotating species titles produced using the strat.plot( ) function in the rioja package. This function produces a stratigraphic plot of paleoenvironmental data. Currently the titles of each species are plotted vertically while they are typically plotted at a 45 degree angle in other programs. Does anyone have any idea of how to rotates these titles? Below is an

Currently, LogLevel must be set to VERBOSE to see the fingerprint of an accepted key, and the default LogLevel is INFO. Since this is useful security information, I would like to propose that the 'Accepted publickey' message be modified to include the fingerprint of the accepted key. Is this a reasonable solution? Here is an example log snippet with LogLevel VERBOSE: Oct 1 15:23:24

Hello everybody, I am trying to do a logistic regression model with lrm() from the design package. I am comparing to groups with different medical outcome which can either be "good" or "bad". In the help file it says that lrm codes al responses to 0,1,2,3, etc. internally and does so in alphabetical order. I would guess this means bad=0 and good=1. My question: I am trying to

Mythical Asterisk Creatures, oft-discussed, rarely seen: 1) An "advanced" graphical user interface 2) An IAX2 hardware device 3) A Radius CDR report module 4) A live-method, robust SQL-based dialplan 5) LDAP/SQL/Radius authentication for SIP phones 6) Robust R2 signalling support 7) Multilingual language recordings of all existing * .gsm files 8) Free exchange of PSTN gateways in

It works! But Once I have the square root of this matrix, how do I convert it to a real (not imaginary) matrix which has the same property? Is that possible? Best, Simon >----Messaggio originale---- >Da: p.dalgaard at biostat.ku.dk >Data: 21-nov-2009 18.56 >A: "Charles C. Berry"<cberry at tajo.ucsd.edu> >Cc: "simona.racioppi at

Every half year or so, I probably will repost this list, adding and subtracting as the community makes advances (or ignores what isn't required.) >Date: Thu, 9 Oct 2003 04:51:23 -0400 >To: asterisk-users-lists.digium.com >From: John Todd <jtodd@loligo.com> >Subject: Sasquatch, the Loch Ness Monster, UFOs and... > >Mythical Asterisk Creatures, oft-discussed, rarely

Thanks for your message! Actually it works quite well for me too. If I then take the trace of the final result below, I end up with a number made up of both a real and an imaginary part. This does not probably mean much if the trace of the matrix below givens me info about the degrees of freedom of a model... Simona >----Messaggio originale---- >Da: RVaradhan at jhmi.edu >Data:

Dear Peter, thank you very much for your answer. My problem is that I need to calculate the following quantity: solve(chol(A)%*%Y) Y is a 3*3 diagonal matrix and A is a 3*3 matrix. Unfortunately one eigenvalue of A is negative. I can anyway take the square root of A but when I multiply it by Y, the imaginary part of the square root of A is dropped, and I do not get the right answer. I tried