Infinitesimal generator flux of Lorentz trasformations in spacetime I'm considering the following matrixs which I know that they form a flux of Lorentz trasformation in spacetime. 
I want to know how to calculate the infinitesimal generator of this flux. Unfortunately I have no particular knowledge of Lie algebra for this reason I need an explanation that does not assume the whole knowledge of it. 
$$\begin{pmatrix}
\frac{4- \cos(\rho)}{3} & \frac{2- 2\cos(\rho)}{3} & 0 & -\frac{\sin(\rho)}{\sqrt{3}} \\
\frac{2\cos(\rho) - 2}{3} & \frac{4- \cos(\rho)}{3} & 0 & \frac{2\sin(\rho)}{\sqrt{3}}\\ 
0 & 0 & 1 & 0 \\
-\frac{\sin(\rho)}{\sqrt{3}} & -\frac{2\sin(\rho)}{\sqrt{3}} & 0 & \cos(\rho) \\
\end{pmatrix}$$ 
Thank you so much for your help
 A: I'm a bit suspicious of the 22 entry of the matrix you write down ,
 $$M=\begin{pmatrix}
\frac{4- \cos(\rho)}{3} & \frac{2- 2\cos(\rho)}{3} & 0 & -\frac{\sin(\rho)}{\sqrt{3}} \\
\frac{2\cos(\rho) - 2}{3} & \frac{4- \cos(\rho)}{3} & 0 & \frac{2\sin(\rho)}{\sqrt{3}}\\ 
0 & 0 & 1 & 0 \\
-\frac{\sin(\rho)}{\sqrt{3}} & -\frac{2\sin(\rho)}{\sqrt{3}} & 0 & \cos(\rho) \\
\end{pmatrix}$$ whose logarithm you are invited to take. I suspect that entry to be something like $(4\cos \rho -1)/3$ instead---see below. Your time appears to be in the 4th component, unlike the first one in the conventional notation.
In any case, observe $M=\mathbb{1}$ as $\rho\to 0$, so to find its logarithm, we expand in the first two powers of ρ,
$$
M=\mathbb{1} -\frac{\rho}{\sqrt{3}}\begin{pmatrix}
 0 &  0 & 0 &  1\\
 0 &  0 & 0 &  -2\\ 
0 & 0 & 0 & 0 \\
 1 &  2& 0 &  0 \\
\end{pmatrix}+ \frac{\rho^2}{6}  \begin{pmatrix}
 1 &  2 & 0 &  0 \\
  - 2  &  1& 0 &  0\\ 
0 & 0 &  0 & 0 \\
 0 &0  & 0 &  -3 \\
\end{pmatrix}           
+Ο(\rho^3).
$$
Let us call the first big matrix A and the second one B.  Note
$$A^2=\begin{pmatrix}
 1 &  2 & 0 &  0 \\
  - 2  &  -4& 0 &  0\\ 
0 & 0 &  0 & 0 \\
 0 &0  & 0 &  -3 \\
\end{pmatrix}       .    $$
Now, if the 22 entry of B were -4 instead of 1, we'd have $A^2=B$, and thus $A^3=-3A$, $A^4=-3A^2$, etc... so (glory!) you can confirm
$$
M=\mathbb{1} -\frac{\rho}{\sqrt{3}}A+ \frac{1}{2} \left(-\frac{\rho A}{\sqrt{3}}\right)^2 +...=e^{-\frac{\rho }{\sqrt{3}}A},
$$
since the expansion of the exponential reduces to 
$$
=\mathbb{1} -\sin\rho ~ \frac{A}{\sqrt{3}} +\frac{1-\cos\rho}{3} A^2 ,
$$
by the above recursive rules.
You would then, indeed, call this logarithm $A/\sqrt{3}$ of the exponential, up to the parameter -ρ, the generator of the group element M .  In your specific case, you see it is a linear combination of a spacetime rotation (antisymmetric elements) and a boost-like strain (symmetric elements). 
However, as it stands, your B is problematic, which is why I  am convinced it is wrong, and should be my proposed expression, instead.
