Matrix equation involving commutator Given two matrices $N\times N$ $A,B$, is there some method to solve the matrix equation:
$$e^{[A,B]}=A$$
where the symbol $[A,B]$ means the commutator of the two matrices: $[A,B]=AB-BA$?
Thanks in advance.
 A: In the $2 \times 2$ case, the only solutions are when $A = \mathrm{Id}$, and concievably some other solution when $B$ is of the form $\left( \begin{smallmatrix} \lambda & 1 \\ 0 & \lambda \end{smallmatrix} \right)$. (I doubt it, but I didn't check this case.)
Since the problem is unvariant under conjugation, we may assume that $B$ is in Jordan canonical form. Let's do the case where $B$ is diagonalizable: $B = \left( \begin{smallmatrix} \lambda & 0 \\ 0 & \mu \end{smallmatrix} \right)$ and $A = \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right)$ then
$$[A,B] =(\lambda - \mu)  \begin{pmatrix} 0 & - x \\  y & 0 \end{pmatrix} \ \mathrm{and} \  [A,B]^2 = - (\lambda-\mu)^2 xy \ \mathrm{Id}.$$
So 
$$e^{[A,B]} = \cosh \left( (\lambda-\mu)^2 xy \right) \mathrm{Id} - (\lambda-\mu) \sinh  \left( (\lambda-\mu)^2 xy \right) \begin{pmatrix} 0 & -x \\ y & 0 \end{pmatrix}. \quad (\dagger)$$
Comparing off diagonal entries:
$$\begin{array}{rl}
x =& \phantom{-} (\lambda-\mu) \sinh  \left( (\lambda-\mu)^2 xy \right) x \\
y =& - (\lambda-\mu) \sinh  \left( (\lambda-\mu)^2 xy \right) y
\end{array}$$
These equations are only consistent if at least one of $x$ and $y$ is $0$. In this case, $(\lambda - \mu)xy=0$ so $(\dagger)$ collapses to $e^{[A,B]} = \mathrm{Id}$ and $A=\mathrm{Id}$.
