Why the generators of boosts transform like a vector under rotation? $$\left[J_i,J_j \right]=i\epsilon_{ijk}J_k$$
$$\left[J_i,M_j \right]=i\epsilon_{ijk}M_k$$
$$\left[M_i,M_j \right]=-i\epsilon_{ijk}J_k$$
where $J_i$ is the generator of rotation of Lorentz group, $M_i$ is the generator of boost of Lorentz group
In many textbook of QFT, they say that the second one implies that the generators of the boosts transform like a vector under rotations. But I can't see it explicitly. Can anyone give 
me the explanation.
 A: Vectors transform linearly,
\begin{equation} 
x _i  \rightarrow A _{ ij } x _j  
\end{equation} 
through some transformation matrix $A$.
Now consider the transformation of $ M _i $:
\begin{align} 
e ^{ i \theta _j J _j } M _i  e ^{ - i J _j \theta _j } & = \left( 1 + i \theta _j J _j - ... \right) M _i  \left( 1 - i \theta _j J _j - ... \right) \\ 
& = M _i  + i \theta _j \left[ J _j , M _i  \right] + ...
\end{align} 
Now if $ \left[ J _j , M _i \right] $ is only proportional to $ M _j $ as above then infinitesimally,
\begin{align} 
e ^{ i \theta _j J _j } M _i  e ^{ - i J _j \theta _j } & = M _i  + i \theta _j \epsilon _{ jik}  M _k \\ 
&= (\delta_{ik}+i\theta_j\epsilon_{jik})M_k\\
&= A _{i,j}M _j 
\end{align} 
for some transformation matrix $A$ as required.
A: The important fact is that the change of an object $O$ under an infinitesimal transformation generated by a generator $G$ can be written in terms of their commutator (infinitesimal parameter $\alpha$):
$$O\longrightarrow O + \delta O\enspace\,\,\text{where}\enspace\,\, \delta O = i\alpha [G,\,O]$$
(to prove this, see JeffDror's answer).
To interpret your first commutator $[J_i, J_j] = i\epsilon_{ijk} J_k$, put object $O=J_j$ and generator $G=J_i$, and the result on the right hand side is $\delta J =  i\alpha_j\epsilon_{ijk}\,J_i$ tells you how the object $J$ transforms under the action of the generator $J$.  This defines a vector.  More generally, we learned
 $$\delta O =  i\alpha_j\epsilon_{ijk}O_k\qquad\text{under $J_i$}$$
 is how a vector $O$ transforms under the action of $J$.  From now on, any $O$ satisfying the above is a vector.
Now move on to the next commutator $[J_i, M_j] = i\epsilon_{ijk} M_k$.  To interpret this, we identify $M$ as our object and $J$ as the generator.  The result on the right hand side $\delta M_i =  i\alpha_j \epsilon_{ijk}M_k$ tells you that $M$ transforms precisely in the same way a vector changes.
