Does log of a matrix factor through similarity? Is it a bijection up to branch choice? When taking the log of a matrix we have various choices, but fixing a particular choice, we should have
$$P^{-1}\log{(A)} P = \log(P^{-1}AP),$$
right? (Here $P \in GL$.)
It is supported by the notion that we can exponentiate both sides and it comes out true. Is there some snag here that I'm missing?
Also, once we choose a branch of logarithm, do we always have a bijection between $M_n(\mathbb{C})$ and $GL(n, \mathbb{C})$ given by $A \mapsto e^A$?
More specifically, given a Jordan block associated with a nonzero eigenvalue $\lambda$
$$\left( \begin{matrix} \lambda & 1 & & & \\  & \lambda &1 & & \\ &&\ddots&\ddots \\ &&&\ddots&1 \\ &&&&\lambda \end{matrix}  \right)$$
We can choose a basis so that the above linear operator has the form
$$\left( \begin{matrix} \lambda & \lambda & \lambda/2! & \dotsb & \lambda/n!\\  & \lambda &\lambda & \ddots & \vdots \\ &&\ddots&\ddots& \lambda/2! \\ &&&\ddots&\lambda \\ &&&&\lambda \end{matrix}  \right)$$
and then a log of such a matrix is 
$$\left( \begin{matrix} \log{\lambda} & 1 & & & \\  & \log{\lambda} &1 & & \\ &&\ddots&\ddots \\ &&&\ddots&1 \\ &&&&\log{\lambda} \end{matrix}  \right),$$
where we have to choose a value for $\log{\lambda}$ (in principle we could choose a different value of $\log{\lambda}$ for each diagonal entry).
Once we fix a choice of branch of log, it seems that we can get a unique log for each matrix in $GL_n(\mathbb{C})$. 
 A: As far as I know, the logarithm is defined after the exponential is. Because the serie 
$$e^A=\sum_{n\in\mathbb{N}} \frac{A^n}{n!}$$
has a radius of convergence $R=+\infty$.
What you can show is (absolutely non trivial) : $\exp(M_n(\mathbb{C}))=GL_n(\mathbb{C})$, but for any $B\in GL_n(\mathbb{C})$ there exists infinitely many $A$ such that $e^A=B$.
Then how to define the logarithm from here ? It would be some function such that 
$$\exp(\log(A))=A$$
On a good neighborhood ($||Id-A||_F<1$ I think with Froebenius norm) of the identity it could be :
$$\log(A)=\sum_{n\geq 1} \frac{(I-A)^k}{k}$$
If you work on diagonalisable matrices, then you could apply the log to the diagonal numbers...
In any case I think that everything depend on how you define the log but ideally yes, it should hold that :
$$P^{-1}\log(A)P=\log(P^{-1} A P)$$
It holds true for the two examples I gave you but for some fancy definition it might not be the case I think.
A: Yes, given any section $L$ (for logarithm) of $\def\C{\Bbb C}\exp:\C\to\C^\times$ (for instance defined using a branch cut), you can define a section $L_n$ of $\exp:M_n(\C)\to GL_n(\C)$ such that the spectrum of $L_n(A)$ is the image by $L$ of the spectrum of $A$. One idea is to use the multiplicative Jordan-Chevalley decomposition $A=A_{ss}A_u$, use $L$ to define a semisimple logarithm of the semisimple part $A_{ss}$, and to use the fact that the restriction of $\exp$ to nilpotent complex matrices defines a bijection to the unipotent ones, with an inverse given by the power series for $\log(1+X)$ (of which in fact only $n$ terms are used!). Since $A=A_{ss}$ and $A_u$ commute, so do $L_n(A_{ss})\in \C[A_{ss}]$ and $L_n(A_u)\in \C[A_u]$, and one can then just take $L_n(A_{ss}A_u)=L_n(A_{ss})+L_n(A_u)$.
I can give more details, but have no time for them right now.
