Matrix Exponential

  • #1
Theorem
Let A be a square matrix nXn then exp(At) can be written as

[tex]
exp(At)=\alpha_{n-1}A^{n-1}t^{n-1} + \alpha_{n-2}A^{n-2}t^{n-2} + ... + \alpha_1At + \alpha_0 I
[/tex]

where [tex]\alpha_0 , \alpha_1 , ... , \alpha_{n-1}[/tex] are functions of t.
Let define

[tex] r(\lambda)=\alpha_{n-1}\lambda^{n-1} + \alpha_{n-2}\lambda^{n-2} + ... + \alpha_1\lambda + \alpha_0 [/tex].

If [tex]\lambda_i[/tex] is an eigenvalue of At with multiplicity k, then

[tex] e^{\lambda_i }= r(\lambda_i) [/tex]
[tex] e^{\lambda_i} = \frac{dr}{d\lambda}|_{\lambda=\lambda_i} [/tex]
etc

Does anyone know any reference where it gives a proof for this theorem? I only know how to prove this theorem intuitively using the Cayley-Hamilton theorem. I need a formal proof. The book (Schaum Outline Series) that I got it only state the theorem.

This theorem will allowed me later to solve system of linear differential equations.

Please help.
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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The Cayley-Hamilton theorem is rigorous. The difficulty is that the a_i are all seies which you must show converge. Do you know the Jordon decomposition theorem?

Jordon decomposition theorem
Let A be a linear operator over the field C (complex numbers can be generalizes)
The exist an invertable linear operator S (made up of the generalized eigenvectors of A) such that SA=BS where B is a direct sum of Jordan blocks.

let ' denot inverse
now we have reduced the problem to one on Jordon blocks
J=aI+H
exp(J)=exp(a)exp(H)
H^n=0 so the infinite sum has all higher terms zero
QED
 
  • #3
We are computing the matrix exponential in an undergraduate engineering mathematics class. As such we never came across the Jordon decomposition theorem before. No wonder I have difficulty in finding the literatures for the proof.
Now there is a hint. I will again search the literatures or ask one of our professor of algebra.
Thanks lurflurf.


I will come back to this thread later, especially because I don't understand a word in the proof. :cry:
let ' denot inverse
now we have reduced the problem to one on Jordon blocks
J=aI+H
exp(J)=exp(a)exp(H)
H^n=0 so the infinite sum has all higher terms zero
QED
If only someone could proved the theorem using the Cayley-Hamilton theorem !!!:smile:
 
  • #4
That professor of algebra is on leave. It's chinese new year holiday.

:smile: Searching the internet, I found the proof that I wanted.
http://web.mit.edu/2.151/www/Handouts/CayleyHamilton.pdf

Along the way I get in love with that jordan decomposition. :!!) Still reading.
In fact jordan decomposition may refer to quite different concept
http://en.wikipedia.org/wiki/Jordan_decomposition

The one given by lurflurf is Jordan normal form.

I think computing matrix exponential using Jordan matrix ( ref: http://en.wikipedia.org/wiki/Matrix_exponential ) is less efficient compare to that using the theorem that I had stated.
 
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