Deducing the additive Jordan decomposition

Question

In $M_n(\Bbb C)$, I could prove that the additive Jordan decomposition of $X=D+N$ with $D$ diagonalizable and $N$ nilpotent gives a multiplicative Jordan decomposition $e^X=e^De^N$.

Is that true the other way around? My goal is to calculate all solutions of the equation $e^X = I_n$.

I am not comfortable with using the logarithm on $e^X$ because I know the exponential is not even surjective on $M_n(\Bbb C)$. Any suggestions to tackle this questions?

Thank you for your help.

Answer 1

Since $I$ is diagonalizable, necessarily $X$ is diagonalizable. Then

$X=Pdiag( \lambda_j)P^{-1}$ where $\lambda_j=2k_j i\pi$ and $k_j$ is an integer.

Edit. Answer to PerelMan. $e^De^N=I$ implies that $e^N=e^{-D}=I+N+\cdots + N^{n-1}/(n-1)!$ is diagonalizable.

Then $R=N(I+\cdots+N^{n-2}/(n-1)!)$ is diagonalizable; yet $R^n=N^n (...)^n=0$ and, consequently, $R=0$. That implies $N=0$ because the second factor, in the definition of $R$, is invertible.