Exponential matrix proof

  • Thread starter GregoryGr
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  • #1
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I have to prove the inverse of the matrix e^A. we haven't studied exponential matrices in uni but he gave us the definition of it with the series e^A= I+A+A^2 ... A^k where A^k=0, even for numbers greater than k.
I have tried to think of a way to prove it, but neither my classmates or I found something. I looked up google etc, but all the proofs were with things that we didn't learn. Any help is welcome.
 

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  • #2
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Consider this:if [itex] A={a_{ij}} [/itex] then for [itex] B=e^A[/itex] we have [itex] b_{ij}=e^{a_{ij}} [/itex].
 
  • #3
HallsofIvy
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Prove what? Not that e^A= I+ A+ A^2/2+ A^3/3!+ ... because you were given that as a definition. And not that [tex]A^n= 0[/tex] for n greater than some k because that is not generally true.

Perhaps your teacher was saying that if there exist some k, such that if A^n = 0 if n> k, then e^A= 1+ A+ A^2/2!+ A^3/3!+ ...+ A^k/k!

Either your teacher gave e^A= I+ A+ A^2/2+ A^3/3!+ ... (an infinite sum) from which it follows that if A^n= 0 for n> k that e^A= I+ A+ A^2/2+ A^3/3!+ ... + A^k/k! or your teacher, wishing to avoid the technical complications involved in defining infinite sums of matrices, just said that if A^n= 0 for n> k, then e^A= I+ A+ A^2/2+ A^3/3!+ ... + A^k/k!

In either case, it does not require "proof" because it is a definition.
 
  • #5
Office_Shredder
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Consider this:if [itex] A={a_{ij}} [/itex] then for [itex] B=e^A[/itex] we have [itex] b_{ij}=e^{a_{ij}} [/itex].
This is not the matrix exponential. For example if A is diagonal then eA has the form bjj = ea_jj, but off the diagonal you still get all zeros (whereas your formula would give 1s).

Gregory, if you've been assigned an exercise you should post the problem in the homework forum, with the full problem statement as you have been given it and the work you have done to try to solve it.
 

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