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.

Solutions Collecting From Web of "Finding the basis of $\mathfrak{so}(2,2)$ (Lie-Algebra of $SO(2,2)$)"

The determinant condition implies that the trace of an element of the Lie algebra must be zero (see the Jacobi formula).

In your computations, you forgot to transpose $X$ (and you messed up the second matrix multiplication). You should get
$$\pmatrix{A^t&-C^t\\B^t&-D^t} + \pmatrix{A&B\\-C&-D} = 0.$$
You get that $A$ and $D$ must be antisymmetric (so the trace condition is automatically satisfied), and that $B^t = C$, so
$$X=\pmatrix{A&B\\B^t&D},$$
with $A,D\in\mathfrak{so}(2)$ and $B\in\mathfrak{gl_2} = Mat_2(\mathbb{R})$.

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