Matrix Exponential Theorem: Proving Formally with Cayley-Hamilton

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

<br /> 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<br />

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

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

If \lambda_i is an eigenvalue of At with multiplicity k, then

e^{\lambda_i }= r(\lambda_i)
e^{\lambda_i} = \frac{dr}{d\lambda}|_{\lambda=\lambda_i}
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.
 
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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
 
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:
lurflurf said:
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:
 
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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