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NoIdea
science forum beginner
Joined: 15 Feb 2006
Posts: 28
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 Posted: Fri Mar 31, 2006 9:36 am Post subject:
finding matrix exponentials using jordan decomposition
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is it possible to use the jordan decomposition rather than the
eigen/spectral to find matrix exponentials/logarithms?
i know that using the eigen decomposition this is fairly straight forward to
acheive, except in the case of a degenerate matrix where the eigen
decomposition does not exist. In this case could a jordan decomposition be
used instead? |
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Duncan Muirhead
science forum addict
Joined: 08 Oct 2005
Posts: 70
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| Posted: Fri Mar 31, 2006 10:36 am Post subject:
Re: finding matrix exponentials using jordan decomposition
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On Fri, 31 Mar 2006 09:36:12 +0000, NoIdea wrote:
| Quote: | is it possible to use the jordan decomposition rather than the
eigen/spectral to find matrix exponentials/logarithms?
i know that using the eigen decomposition this is fairly straight forward to
acheive, except in the case of a degenerate matrix where the eigen
decomposition does not exist. In this case could a jordan decomposition be
used instead?
Matrix Computations (Golub and Van Loan) discusses this briefly. They |
prefer the Schur Decomposition for this
Duncan |
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Martin Eisenberg
science forum beginner
Joined: 11 May 2005
Posts: 28
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| Posted: Fri Mar 31, 2006 7:35 pm Post subject:
Re: finding matrix exponentials using jordan decomposition
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NoIdea wrote:
| Quote: | is it possible to use the jordan decomposition rather than the
eigen/spectral to find matrix exponentials/logarithms?
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Moler and Van Loan say it's impractical in their article "Nineteen
Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five
Years Later":
http://www.math.unm.edu/~kapitula/courses/math512/MatrixExponential.pdf
Martin
--
Quidquid latine scriptum sit, altum viditur. |
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Herman Rubin
science forum Guru
Joined: 25 Mar 2005
Posts: 730
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| Posted: Sat Apr 01, 2006 3:53 pm Post subject:
Re: finding matrix exponentials using jordan decomposition
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In article <pan.2006.03.31.10.36.31.380200@this.address>,
Duncan Muirhead <noone@this.address> wrote:
| Quote: | On Fri, 31 Mar 2006 09:36:12 +0000, NoIdea wrote:
is it possible to use the jordan decomposition rather than the
eigen/spectral to find matrix exponentials/logarithms?
i know that using the eigen decomposition this is fairly straight forward to
acheive, except in the case of a degenerate matrix where the eigen
decomposition does not exist. In this case could a jordan decomposition be
used instead?
|
If by the Jordan decomposition you mean the decomposition
into blocks of the form hI + J, where J is one above the
main diagonal, I must have read this more than 60 years
ago. Expand f(hI + J) in a Taylor series, which only has
a finite number of non-zero terms.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 |
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Lou Pecora
science forum beginner
Joined: 07 Jun 2005
Posts: 33
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| Posted: Sat Apr 01, 2006 5:43 pm Post subject:
Re: finding matrix exponentials using jordan decomposition
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In article <e0m7m4$40s2@odds.stat.purdue.edu>,
hrubin@odds.stat.purdue.edu (Herman Rubin) wrote:
| Quote: | If by the Jordan decomposition you mean the decomposition
into blocks of the form hI + J, where J is one above the
main diagonal, I must have read this more than 60 years
ago. Expand f(hI + J) in a Taylor series, which only has
a finite number of non-zero terms.
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Yes, J is what I think is called nilpotent (hope I got that right).
Meaning that only a finite number of self multiplications will result in
a non-zero matrix. It's a good way to get analytical expressions for
exponentiation of matrices. I was wondering, is there a way to generate
the analytical expression just given an arbitrary J? Or is it necessary
to just multiply until the matrix becomes zero? Seems like the number
of terms could be predicted just by the dimension of J (or size of its
nonzero part). Jumping to an analytical formula would make it very fast
for large problems.
-- Lou Pecora (my views are my own) REMOVE THIS to email me. |
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Peter Spellucci
science forum Guru
Joined: 29 Apr 2005
Posts: 702
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| Posted: Mon Apr 03, 2006 9:49 am Post subject:
Re: finding matrix exponentials using jordan decomposition
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In article <pecoraREMOVE-643C65.12430501042006@ra.nrl.navy.mil>,
Lou Pecora <pecoraREMOVE@THISanvil.nrl.navy.mil> writes:
| Quote: | In article <e0m7m4$40s2@odds.stat.purdue.edu>,
hrubin@odds.stat.purdue.edu (Herman Rubin) wrote:
If by the Jordan decomposition you mean the decomposition
into blocks of the form hI + J, where J is one above the
main diagonal, I must have read this more than 60 years
ago. Expand f(hI + J) in a Taylor series, which only has
a finite number of non-zero terms.
Yes, J is what I think is called nilpotent (hope I got that right).
Meaning that only a finite number of self multiplications will result in
a non-zero matrix. It's a good way to get analytical expressions for
exponentiation of matrices. I was wondering, is there a way to generate
the analytical expression just given an arbitrary J? Or is it necessary
to just multiply until the matrix becomes zero? Seems like the number
of terms could be predicted just by the dimension of J (or size of its
nonzero part). Jumping to an analytical formula would make it very fast
for large problems.
-- Lou Pecora (my views are my own) REMOVE THIS to email me.
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jordan block = lambda*I + J
and since these two commute the binomial formula applies. this plus nilpotency
of J gives a closed formula for the powers.
hth
peter |
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Derek R O'Connor
science forum beginner
Joined: 10 Apr 2006
Posts: 2
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| Posted: Mon Apr 10, 2006 12:19 am Post subject:
Re: finding matrix exponentials using jordan decomposition
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Nick Higham has just completed a three-year reasearch project on Matrix Functions and is writing a book on the subject.
See his papers on this topic at
http://www.maths.man.ac.uk/~higham/
Regards,
Derek O'Connor |
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