Can the Baker-Campbell-Hausdorff formula for $\ln(AB)$ be simplified for similar, diagonizable matrices? Given two similar, diagonizable square matrices $A$ and $B$ that do not commute, can the Baker-Campbell-Hausdorff formula be simplified exploiting the similarity to obtain a nice expression for $\ln(AB)$? (It's probably not simply $\ln A + \ln B$ due to the lack of commutivity)
 A: I would say you can simply use Baker-Campbell-Hausdorf in this way:
$$
\log AB = \log e^{\tilde A} e^{\tilde B} = \log e^{\tilde A + \tilde B + \frac{1}{2}[\tilde A,\tilde B] + \dots} =  \tilde A + \tilde B + \frac{1}{2}[\tilde A,\tilde B] + \dots
$$
where $\tilde A = \log A$ and $\tilde B = \log B$. I dont think you can simplify much more. 
A: I read many mistakes. Assume that $\log(.)$ is the principal logarithm, that is $\log(re^{i\theta})=\log(r)+i\theta$ where $\theta\in (-\pi,\pi)$.


*

*When we write $\log(AB)=\log(A)+\log(B)$, we implicitly assume that $A,B,AB$ have no negative eigenvalues.

*The previous equality is not true when $AB=BA$; in fact, it is false even when $n=1$ over $\mathbb{C}$ : $\log(e^{2i\pi/3}e^{2i\pi/3})=-2i\pi/3\not= 2i\pi/3+2i\pi/3$ and when $n=2$ over $\mathbb{R}$: cf. $\log(Rot(2\pi/3)Rot(2\pi/3))$.

*About the Enrico's post, $\log(e^X)\not= X$ because $e^{0_2}=exp(\begin{pmatrix}0&2\pi\\-2\pi&0\end{pmatrix})$.


EDIT. @ Tobias Kienzler.
Let $A,B\in M_n(\mathbb{C})$ s.t. $AB=BA$ and $A,B,AB$ have no eigenvalues in $(-\infty,0]$; let $spectrum(A)=(\lambda_i)_i$. Note that $A,B$ are not necessarily diagonalizable.
Proposition: There is a permutation $(\mu_i)_i$ of the eigenvalues of $B$ s.t. $\log(AB)-\log(A)-\log(B)$ is similar to $diag((\log(\lambda_i\mu_i)-\log(\lambda_i)-\log(\mu_i))_i)$.
