Exponential of nonconstant matrix

  • Thread starter ranoo
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  • #1
9
0
how to compute the nonconstant matrix exponential؟

i.e. if there is a matrix A(t), how i can find exp(A(t)) ???
 

Answers and Replies

  • #2
22,089
3,291
This is easy if your matrix is nilpotent or diagonizable.
If it is neither of those, then you will want to triangulate your matrix to write it as a sum of a diagonizable matrix and a nilpotent matrix.

Did you have any particular matrixx in your mind?
 
  • #3
1,444
4
It does not matter whether you matrix is "constant" or "non-constant". You define A=A(t) and calculate exp(A).

Added: Unless you have in mind so called http://en.wikipedia.org/wiki/Ordered_exponential" [Broken]
 
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  • #4
9
0
the 2x2 matrix is A(t)=[0 1]
0 t

I can't write the matrix but tha first row 0,1 and the second row 0,t
 
  • #5
1,444
4
You have , for [itex]n\geq 1[/itex]
[tex]A(t)^n=\begin{pmatrix}0&t^{n-1}\\0&t^n\end{pmatrix}[/tex]

and

[tex]e^A(t)=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\sum_{i=1}\frac{1}{n!}\begin{pmatrix}0&t^{n-1}\\0&t^n\end{pmatrix}[/tex]

I hope you will be able to finish. But better check the above. I could have made a mistake!
 
  • #6
679
2
Remember exp(A) is defined as a taylor expansion exponential function, the series actually converge for any matrix A, so in principle we can always express exp(A(t)) in this way, then each entry of exp(A) is an infinite series of numbers, so you can try and work out the sum to get a closed form. There're better ways to find the closed form of exp(A), if A is diagonalizable just diagonlize it, if not you can always use a Jordan decomposition, it works in a similar manner.
 

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