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Let $A,B \in M_{n\times n}(\Bbb{C})$ with $e^A =e^B$, Assume that $A,B$ are diagonizable,therefore they have eigenspace decomposition as $\Bbb{C}^n = \bigoplus E_\lambda(A) = \bigoplus E_\mu(B)$.

Is it possible to prove $B$ has same eigenspace decomposition as $A$ with eigenvalue differ by a constant, that is if $v \in E_\lambda(A)$ i.e. $Av = \lambda v$ then $v$ is also eigenvector of $B$ with $Bv = (\lambda + 2\pi ik)v$ for some integer $k \in \Bbb{Z}$?

(A related question) I have checked this post here, which shows eigenspace decomposition of $e^A$ does not necessarily gives the eigenspace decomposition of $A$.

Is it possible that my statement is true? I don't have enough tools in my head to handle this problem.

yi li
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    Could you take $A=2\pi \begin{pmatrix} 0 & 1\ -1 & 0\end{pmatrix}$ and $B=0$? – Giulio R May 04 '23 at 09:47
  • Oh thank you Giulio , I ask a silly question. I need some result like this. – yi li May 04 '23 at 09:52
  • Hey @Giulio , I realize it's ok in the example you provided? Although the eigenspace decomposition not the same however for each eigenvector of $A$ say $Av = (2\pi i) v$ this must also be the eigenvector of $Bv = (2\pi i - 2\pi i) v$ with $k =1$? – yi li May 04 '23 at 10:09

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