# I Matrix exponential

1. Jul 20, 2017

### Josh1079

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!

2. Jul 20, 2017

### Staff: Mentor

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.