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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.

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

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

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