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exp(At)

Started by Maddox, July 19, 2006, 06:58 PM

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Maddox

Evaluate eAt
Where A is an upper triangular matrix given by
1 2 3
0 1 2
0 0 1

What is the answer?  Determinent is obviously 1...

I just had that question on my diff eqns test I had no idea how to do it. I'm not sure if he talked about it in class since I don't go.  :-[

I took linear algebra about a year ago so I've forgotten a lot of stuff.
asdf.


Maddox

#2
Quote from: Yoni on July 20, 2006, 03:48 PM
http://mathworld.wolfram.com/MatrixExponential.html
ok... I think this is how you do it.


A is equal to
[ 0 2 3 ]    [ 1 0 0 ]
[ 0 0 2 ]  + [ 0 1 0 ]
[ 0 0 0 ]    [ 0 0 1 ]

Let B be the matrix on the left.
Then exp(At) is equal to exp(t(B+I)) where I is the identity matrix.
This then gives us exp(tB)exp(tI)


B is a nilpotent matrix, but I don't know how to get the matrix polynomial.  I think someone said it has to do with taylor polynomials.

Oh well...
asdf.

Yoni

Your evaluation is not correct. exp(A+B) != exp(A) + exp(B).

Maddox

Quote from: Yoni on July 21, 2006, 04:01 AM
Your evaluation is not correct. exp(A+B) != exp(A) + exp(B).
err, stupid mistake.   :-[

fixed.
asdf.

Maddox

Argh, I studied this for the final and he didn't put it on there.
asdf.