# Deducing the additive Jordan decomposition

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.

• Did you downvote my post? Incredible. – loup blanc Apr 16 at 21:22
• @loupblanc no I didn't? I don't see any post on this question now? – PerelMan Apr 17 at 8:21

## 1 Answer

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.

• Thank you! is there a way to deduce this from multiplicative Jordan decomposition of $e^X$? – PerelMan Apr 10 at 16:50