Matrix exponential

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  • Thread starter Josh1079
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Hi,

I'm kind of stuck with this theorem stating that: if A is an unipotent matrix, then exp(log A) = A and also if X is nilpotent then log(exp X) = X

Does anyone know any good approaches to prove this?

I know that for unipotent A, logA will be nilpotent and that for nilpotent X, exp(X) will be unipotent

Thanks!
 

Answers and Replies

  • #2
13,582
10,710
Hi,

I'm kind of stuck with this theorem stating that: if A is an unipotent matrix, then exp(log A) = A and also if X is nilpotent then log(exp X) = X

Does anyone know any good approaches to prove this?

I know that for unipotent A, logA will be nilpotent and that for nilpotent X, exp(X) will be unipotent

Thanks!
This depends a bit on what is given, resp. how the functions are defined. You could use a brute force method and simply insert everything into the series of ##\exp## and ##\log##. Or you could attack the problem with the Jordan Chevalley decomposition and / or the eigenvalues, resp. the characteristic polynomial. I haven't done it, but those keywords come to mind.
 

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