R: RE: R: Re: R: Re: chol( neg.def.matrix ) WAS: Re: Choleski and Choleski with pivoting of matrix fails
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: 25-nov-2009 18.55 A: <simona.racioppi at libero.it>, <P.Dalgaard at biostat.ku.dk> Cc: <r-help at r-project.org> Ogg: RE: [R] R: Re: R: Re: chol( neg.def.matrix ) WAS: Re: Choleski and
Choleski with pivoting of matrix fails
I do not understand what the problem is, as it works just fine for me: A <- matrix(c(0.5401984,-0.3998675,-1.3785897,-0.3998675,1.0561872, 0.8158639,-1.3785897, 0.8158639, 1.6073119), 3, 3, byrow=TRUE) eA <- eigen(A) chA <- eA$vec %*% diag(sqrt(eA$val+0i)) %*% t(eA$vec) all.equal(A, Re(chA %*% t(chA))) Y <- diag(c(1,2,3)) solve(chA %*% Y) Ravi.
----------------------------------------------------------------------------------- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvaradhan at jhmi.edu Webpage: http://www.jhsph.
edu/agingandhealth/People/Faculty_personal_pages/Varadhan.html
------------------------------------------------------------------------------------ -----Original Message----- From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of simona.racioppi at libero.it
Sent: Wednesday, November 25, 2009 9:59 AM To: P.Dalgaard at biostat.ku.dk Cc: r-help at r-project.org Subject: [R] R: Re: R: Re: chol( neg.def.matrix ) WAS: Re: Choleski and
Choleski with pivoting of matrix fails
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 to exploit the diagonal structure of Y by using 2*2 matrices for A and Y. In this way the problem mentioned above disappears (since all eigenvalues of A are positive) and when I perform the calculation above I
get
approximately the right answer. The approximation is quite good. However it
is
an approximation. Any suggestion? Thank you very much! Simon
----Messaggio originale---- Da: P.Dalgaard at biostat.ku.dk Data: 23-nov-2009 14.09 A: "simona.racioppi at libero.it"<simona.racioppi at libero.it> Cc: "Charles C. Berry"<cberry at tajo.ucsd.edu>, <r-help at r-project.org> Ogg: Re: R: Re: [R] chol( neg.def.matrix ) WAS: Re: Choleski and Choleski
with pivoting of matrix fails
simona.racioppi at libero.it wrote:
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?
No. That is theoretically impossible. If A = B'B, then x'Ax = ||Bx||^2 >= 0 for any x, which implies in particular that all eigenvalues of A should be nonnegative.
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 libero.it"<simona.racioppi at libero.it>, <r-help at r-
project.org>
Ogg: Re: [R] chol( neg.def.matrix ) WAS: Re: Choleski and Choleski
with
pivoting of matrix fails
Charles C. Berry wrote:
On Sat, 21 Nov 2009, simona.racioppi at libero.it wrote:
Hi Everyone,
I need to take the square root of the following matrix:
[,1] [,2] [,3]
[1,] 0.5401984 -0.3998675 -1.3785897
[2,] -0.3998675 1.0561872 0.8158639
[3,] -1.3785897 0.8158639 1.6073119
I tried Choleski which fails. I then tried Choleski with pivoting,
but
unfortunately the square root I get is not valid. I also tried eigen decomposition but i did no get far. Any clue on how to do it?!
If you want to take the square root of a negative definite matrix,
you
could use
sqrtm( neg.def.mat )
from the expm package on rforge:
http://r-forge.r-project.org/projects/expm/
But that matrix is not negative definite! It has 2 positive and one negative eigenvalue. It is non-positive definite. It is fairly easy in any case to get a matrix square root from the
eigen
decomposition:
v%*%diag(sqrt(d+0i))%*%t(v)
[,1] [,2] [,3] [1,] 0.5164499+0.4152591i -0.1247682-0.0562317i -0.7257079+0.3051868i [2,] -0.1247682-0.0562317i 0.9618445+0.0076145i 0.3469916-0.0413264i [3,] -0.7257079+0.3051868i 0.3469916-0.0413264i 1.0513849+0.2242912i
ch <- v%*%diag(sqrt(d+0i))%*%t(v) t(ch)%*% ch
[,1] [,2] [,3] [1,] 0.5401984+0i -0.3998675-0i -1.3785897-0i [2,] -0.3998675-0i 1.0561872+0i 0.8158639-0i [3,] -1.3785897-0i 0.8158639-0i 1.6073119-0i A triangular square root is, er, more difficult, but hardly impossible. -- O__ ---- Peter Dalgaard ?ster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45)
35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45)
35327907
-- O__ ---- Peter Dalgaard ?ster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907
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