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Topic: Computation of the matrix exponential
Replies: 6   Last Post: Feb 10, 2013 5:00 AM

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Chip Eastham

Posts: 2,341
Registered: 12/13/04
Re: Computation of the matrix exponential
Posted: May 24, 2010 10:53 PM
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On May 24, 8:37 pm, luca <luca.frammol...@gmail.com> wrote:
> Hi,
>
> i have the following problem: given a 3x3 real matrix, compute exp(A).
>
> I need a really fast way to do this. I have searched a bit with
> google, but it seems to me that
> computing the matrix exponential is not so simple, at least if your
> matrix does not have a special
> structure (for example A=diagonal matrix).
>
> I have found a simple method that use the diagonalization of A. If A
> has 3 distinct eigenvalues, than compute
> A=PDP^-1, where P is the matrix of the eigenvectors, D is a diagonal
> matrix (whose diagonal elements are
> the eigenvalues of A). Than, exp(A) = P exp(D) P^-1. Since P^-1 is
> fast enough and exp(D) is simple
> to compute, this should be a fast method.
>
> But, the problem is: i am not sure that the matrix A will always have
> 3 distinct eigenvalues...what happens
> if this does not happen? Can i use that formula even if 2 (or all
> three) eigenvalues are equal?
>
> Are there any other ways to compute exp(A) in a fast way?
>
> Thank you,
> Luca


If you have time to do the diagonalization
(by similarity), then of course that is a
way to get the matrix exponential.

If a matrix is not diagonalizable, then it
can be written in Jordan canonical form,
i.e. diagonal matrix plus some nilpotent
blocks corresponding to eigenvalues with
algebraic multiplicity greater than their
geometric multiplicity.

The application of power series to such
diagonal + nilpotent terms gives (esp.
in the 3x3 case) just a slight modification
to what you were already looking at for the
exponential of a purely diagonal matrix.

Let's consider an illustrative case, where
a eigenvalue r has geometric multiplicity
1 but algebraic multiplicity 3. Instead of
being diagonalizable, such a matrix is then
similar to J =

[ r 1 0 ]
[ 0 r 1 ]
[ 0 0 r ]

Since rI and the nilpotent part N of this
commute, exp(J) = exp(r)I * exp(N). The
nice thing about the power series exp(N)
is that it has only a (small) finite number
of terms:

exp(N) = I + N + (1/2)N^2 =

[ 1 1 0.5 ]
[ 0 1 1 ]
[ 0 0 1 ]

All the other cases are actually simpler
than this one.

regards, chip



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