Q:

Can we derive that $A$ commutes with $B$ from this?

Based on some Physics backgrounds, I want to confirm the following thing. 
Let $[A,B]:=AB-BA$, where $A,B$ are matrices. Now the question is as follows:
If for any real number $\lambda$, $[A,e^{\lambda B}]=0$, then is $[A,B]=0$ true? Where $A,B$ are matrices.
If the above statement is true, how to give a rigorous proof ?
So far the approach by myself is: $[A,e^{\lambda B}]=\lambda[A,B]+\frac{\lambda^2}{2}[A,B^2]+...=0\Rightarrow\frac{1}{\lambda}[A,e^{\lambda B}]=[A,B]+\frac{\lambda}{2}[A,B^2]+...=0 $(for any nonzero $\lambda$), then $\lim_{ \lambda\rightarrow 0}\frac{1}{\lambda}[A,e^{\lambda B}]=[A,B]=0$.
Is my proof rigorous from the Math viewpoint? Other excellent method is welcome! Thanks!

A:

$e^{\lambda B}$ is an analytic function of $\lambda$. Therefore, $$\frac{d}{d\lambda }e^{\lambda B}=Be^{\lambda B}=e^{\lambda B}B.$$
We write
$$0=\frac{d}{d\lambda }[A,e^{\lambda B}] = ABe^{\lambda B}-e^{\lambda B}BA ,$$
hence for $\lambda=0$ we obtain $[A,B]=0$.
Previous version
$$0=\frac{d}{d\lambda }[A,e^{\lambda B}] = Ae^{\lambda B}B-e^{\lambda B}BA =e^{\lambda B}(AB- BA),$$
hence by the fact that $ e^{\lambda B} $ is non-singular we obtain that $[A,B]=0$.
Possible generalisation of the problem
If $[A,e^{\lambda B}]$ is a constant matrix with respect to $\lambda$ on a non-empty interval $(\lambda_-,\lambda_+)$, then $[A,B]=0$.
Clarification
A matrix is called non-singular if its determinant is non-zero (i.e. the matrix is invertible or the matrix doesn't have zero eigenvalues). In our case, the eigenvalues of the matrix $e^{\lambda B}$ have the form $e^{\lambda \mu_j}\ne 0$ where $\mu_j$ are eigenvalues of $B$. The determinant of the matrix is a product of its eigenvalues taken with multiplicity, hence we can conclude that $\det e^{\lambda B}\ne 0$.
As for analyticity, I suggest you read these two articles on wiki: Jordan normal form and 
Jordan matrix, especially functions of matrices and spectral mapping theorems.

