Thanks for the explanation. I am not too deep into linear algebra. T.mat, Rz.mat and Q.mat are square matrices 4 x 4.
To be more clear, here is what they look like. h=3.39/2; T.mat<-matrix(c(1,0,0,0,0,1,0,0,0,0,1,0,0,0,-h,1), nrow=4) # translation of the system alpha<-36*pi/180 cos.alpha<-cos(alpha) minus.sin.alpha<- -1*sin(alpha) sin.alpha<-sin(alpha) Rz.mat<-matrix(c(cos.alpha,minus.sin.alpha,0,0,sin.alpha,cos.alpha,0,0,0,0,1,0,0,0,0,1), nrow=4) Rx.mat<-matrix(c(1,0,0,0,0,cos.alpha, minus.sin.alpha,0,0,sin.alpha,cos.alpha,0,0,0,0,1), nrow=4) Q.mat is a product of Rz.mat and Rx.mat with different angles. The resultant matrix Mn.mat is used to computed the next coordinates of a helix. center.Of.bases<-matrix(c(0.0,0.0,0.0,1),nrow=4) Mn.mat<-qr.solve(compute.Mn.mat(T.mat,Rz.mat,Q.mat)) # This function computes the Mn.mat matrix using specific angles. Which is used to get the new coordinates. New.center.Of.bases<-as.numeric(Mn.mat %*% center.Of.bases) Does these lines of code tell you anything about the complexity of the problem? Let me know if I should do anything different. I was really glad hear from you since you are an expert in the area. Cheers../Murli -----Original Message----- From: dmba...@gmail.com [mailto:dmba...@gmail.com] On Behalf Of Douglas Bates Sent: Wednesday, July 15, 2009 7:29 AM To: Nair, Murlidharan T Cc: r-help@r-project.org Subject: Re: [R] Matrix multiplication precision On Wed, Jul 15, 2009 at 3:42 AM, Nair, Murlidharan T<mn...@iusb.edu> wrote: > Hi!! > I am trying to multiply 5 matrices and then using the inverse of that matrix > for further computation. I read about precision problems from the archives > and the suggestion was to use as.numeric while computing the products. Hmm. I'm not sure what the origins of that advice might be but I don't think it would apply in general. > I am still having problems with the results. Here is how I am using it > #Mn.mat<-(T.mat %*% Rz.mat %*% Q.mat %*% Rz.mat %*% T.mat) # I was doing > this in one step which I have broken down into multiple steps as below. > Mn1.mat<-matrix(as.numeric(T.mat %*% Rz.mat),nrow=4) > Mn2.mat<-matrix(as.numeric(Mn1.mat %*% Q.mat),nrow=4) > Mn3.mat<-matrix(as.numeric(Mn2.mat %*% Rz.mat),nrow=4) > Mn.mat<-matrix(as.numeric( Mn3.mat %*% T.mat),nrow=4) I doubt very much that the as.numeric would have any effect on precision in these cases. You are simply taking a numeric result in its matrix form, converting it to a vector then converting the vector back to a matrix., > mm <- matrix(round(rnorm(8), 3), nrow = 4) > mm [,1] [,2] [1,] -0.110 2.195 [2,] 0.616 0.353 [3,] 0.589 0.970 [4,] 1.028 0.857 > as.numeric(mm) [1] -0.110 0.616 0.589 1.028 2.195 0.353 0.970 0.857 The key in situations like this is to analyze the structure of the computation and decide whether you can group the intermediate results. You have written your product as T %*% Rz %*% Q %*% Rz %*% T What are the characteristics of T, Rz, and Q? Are they square and symmetric, square and triangular, diagonal? Is Q orthogonal (the factors in an orthogonal - triangular decomposition are often written as Q and R)? Did you happen to skip a few transposes in that formula - often formulas like that generate symmetric matrices. And the big lesson, of course, is the first rule of numerical linear algebra., "Just because a formula is written in terms of the inverse of a matrix doesn't mean that is a good way to calculate the result; in fact, it is almost inevitably the worst way of calculating the result". You only calculate the inverse of a matrix after you have investigated all other possible avenues for calculating the result and found them to be fruitless. You haven't told us what you plan to do with the inverse and that is an important consideration., If, for example, it represents a variance-covariance matrix, and you want to express the result as standard deviations and correlations you would not compute the variance-covariance matrix but stop instead at the Cholesky factor of the inverse. You may want to check the facilities of the Matrix package for expressing your calculation. It is a recommended package that is included with binary versions of R from 2.9.0 and has many ways of expressing and exploiting structure in matrices. > For getting the inverse I am doing the following > > Mn.mat.inv<-qr.solve(Mn.mat) > > > Will I run into precision problems when I do the above? > > Thanks ../Murli > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
