[R] Matrix multiplication precision

[Nair, Murlidharan T]

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: dmbates at gmail.com [mailto:dmbates at gmail.com] On Behalf Of Douglas Bates
Sent: Wednesday, July 15, 2009 7:29 AM
To: Nair, Murlidharan T
Cc: r-help at r-project.org
Subject: Re: [R] Matrix multiplication precision

On Wed, Jul 15, 2009 at 3:42 AM, Nair, Murlidharan T<mnair at 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
>
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