simona.racioppi at libero.it
2009-Nov-25 14:59 UTC
[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 Choleskiwith 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 convertit>> 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 theeigen>>> 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 > >
Ravi Varadhan
2009-Nov-25 17:55 UTC
[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 Choleskiwith 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 convertit>> 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 theeigen>>> 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 > >______________________________________________ R-help at 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.
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