Finding the basis of $\mathfrak{so}(2,2)$ (Lie-Algebra of $SO(2,2)$) I want to calculate the Basis of the Lie-Algebra $\mathfrak{so}(2,2)$. My idea was, to use a similar Argument as in this Question. The $SO(2,2)$ is defined by:
$$
SO(2,2) := \left\{ X \in Mat_4(\mathbb R): X^t\eta X = \eta,\; \det(X) = 1 \right\}
$$
(With $\eta = diag(1,1,-1,-1)$)
With the argument from the link, i get the following equation:
$\forall X \in \mathfrak{so(2,2)}$:
$$
X^t\eta + \eta X = 0.
$$
My idea was to use the block decomposition:
$$
X = \left(\begin{matrix}
A & B\\
C & D \end{matrix} \right), \; 
\eta = \left(\begin{matrix}
\mathbb I & 0\\
0 & -\mathbb I\\ \end{matrix}\right).
$$
I get the following equation:
$$
\left(\begin{matrix}
A^t & -B^t\\
C^t & -D^t\\ \end{matrix}\right) +
\left(\begin{matrix}
A & B\\
C & D
\end{matrix}\right) = 0.
$$
Is this correct? I also don't really know, what to do with the $det(X) = 1$ condition.

It follows that
$$
\begin{split}
\mathfrak{so(2,2)} = span\left\{ 
\left(\begin{matrix} 
 0 &  1 &  0 &  0\\
-1 &  0 &  0 &  0\\
 0 &  0 &  0 &  0\\
 0 &  0 &  0 &  0\\
\end{matrix}\right),
\left(\begin{matrix} 
 0 &  0 &  0 &  0\\
 0 &  0 &  0 &  0\\
 0 &  0 &  0 &  1\\
 0 &  0 & -1 &  0\\
\end{matrix}\right), \\
\left(\begin{matrix} 
 0 &  0 &  1 &  0\\
 0 &  0 &  0 &  0\\
-1 &  0 &  0 &  0\\
 0 &  0 &  0 &  0\\
\end{matrix}\right),
\left(\begin{matrix} 
 0 &  0 &  0 &  1\\
 0 &  0 &  0 &  0\\
 0 &  0 &  0 &  0\\
 1 &  0 &  0 &  0\\
\end{matrix}\right),
\left(\begin{matrix} 
 0 &  0 &  0 &  0\\
 0 &  0 &  1 &  0\\
 0 &  -1 &  0 &  0\\
 0 &  0 &  0 &  0\\
\end{matrix}\right),
\left(\begin{matrix} 
 0 &  0 &  0 &  0\\
 0 &  0 &  0 &  1\\
 0 &  0 &  0 &  0\\
 0 & -1 &  0 &  0\\
\end{matrix}\right),
\right\}
\end{split}
$$
Therefore $\dim_{\mathbb R}(\mathfrak{so(2,2)}) = 6$?
