matrix^(-1/2)

On Nov 1, 2009, at 1:46 PM, spencerg wrote:

Hi, Chuck:

    Thanks very much, but why do I get "package 'expm' is not available"
    from install.packages("expm",repos="http://R-Forge.R-project.org")?

In my case I think it was it is because there is no 2.10 branch to either 
the:

http: //r-forge.r-project.org/bin/macosx/leopard/contrib/    ... or the
http: //r-forge.r-project.org/bin/macosx/universal/contrib/    ...trees.

I tried a variety of stems for the installer but got these messages:
Warning: unable to access index for repository
http://r-forge.r-project.org/bin/macosx/universal/contrib/latest/bin/macosx/leopard/contrib/2.10
Warning: unable to access index for repository
http://r-forge.r-project.org/bin/macosx/universal/contrib/bin/macosx/leopard/contrib/2.10
Warning: unable to access index for repository
http://r-forge.r-project.org/bin/macosx/leopard/contrib/2.10

So I wonder if the package installers' expectations for the r-forge 
repository are matching up with the tree structures.

sessionInfo()

I should also note that the matpow or "%^%" functions in expm would not 
address the OP's question since they require that the exponent be positive.

-- 
David.

    Best Wishes,
    Spencer Graves


Charles C. Berry wrote:

On Sun, 1 Nov 2009, spencerg wrote:

A question, a comment, and an alternative answer to matrix^(-1/2):

QUESTION:


What's the status of the "expm" package, mentioned in the email you 
cited from Martin Maechler, dated Apr 5 19:52:09 CEST 2008? I tried 
both install.packages('expm') and 
install.packages("expm",repos="http://R-Forge.R-project.org"), and got 
"package 'expm' is not available" in both cases.

Try

   http://r-forge.r-project.org/projects/expm/

HTH,

Chuck

COMMENT:


The solution proposed by Venables rests on Sylvester's matrix theorem, 
which essentially says that if a matrix A is diagonalizable with 
eigenvalue decomposition eigA <- eigen(A) and f: D ? C with D ? C 
be a function for which f(A) is well defined 
(http://en.wikipedia.org/wiki/Sylvester%27s_matrix_theorem), then f(A) 
= with(eigA, vectors %*% diag(f(values)) %*% solve(vectors)). Maechler 
and others have noted that this can be one of the least accurate and 
most computationally expensive ways to compute f(A).


ALTERNATIVE ANSWER:


For A^(-1/2), if A is symmetric and nonnegative definite, then 
solve(chol(A)) would be a very good way to compute it.


Hope this helps,
Spencer


David Winsemius wrote:

On Oct 31, 2009, at 9:33 PM, David Winsemius wrote:

  On Oct 31, 2009, at 4:39 PM, Kajan Saied wrote:

  Dear R-Help Team,

  as a R novice I have a (maybe for you very simple 
  question), how do I > >  get

  the following solved in R:

  Let R be a n x n matrix:
  \mid R\mid^{-\frac{1}{2}}
  solve(A) gives me the inverse of the matrix R, however not 
  the ^(-1/2) > >  of

  the matrix...
  GIYF: (and Bill Venables if friendly, too.)
  http://www.lmgtfy.com/?q=powers+of+matrix+r-project

I had assumed that the first hit I got:

https://stat.ethz.ch/pipermail/r-help/2008-April/160662.html

... would be the first hit anybody got, but that's not necessarily 
true
now and especially for the future. And further searching within the
results produced this more recent Maechler posting:

https://stat.ethz.ch/pipermail/r-devel/2008-April/048969.html

For the Mac users, there appears to be no binary, but the source 
compiles
without error on a 64-bit version of R 2.10.0:

install.packages("expm",repos="http://R-Forge.R-project.org",
type="source")

#The suggested code throws an error, so my very minor revision would 
be:

library(expm)
?"%^%"

Charles C. Berry                            (858) 534-2098
                                           Dept of Family/Preventive 
Medicine
E mailto:cberry at tajo.ucsd.edu                UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 
92093-0901

David Winsemius, MD
Heritage Laboratories
West Hartford, CT