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.

  • $\begingroup$ 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

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.

  • 1
    $\begingroup$ Thank you! is there a way to deduce this from multiplicative Jordan decomposition of $e^X$? $\endgroup$ – PerelMan Apr 10 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.