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.