Finding a basis for $sp(4,\mathbb{C})$ and related basis. Let $$L = so_4(\mathbb{C})= \{x \in End(\mathbb{C}^4)|^txS + Sx = 0 \} \text{ where }S = \left(\begin{array}{cc} 0 & I_2 \\ -I_2 & 0 \end{array}\right)$$
Letting $x = \left(\begin{array}{cc} A & B \\ C & D \end{array}\right)$ I find $$\left(\begin{array}{cc} A & B \\ C & D \end{array}\right)^t\left(\begin{array}{cc} 0 & I_2 \\ -I_2 & 0 \end{array}\right) =\left(\begin{array}{cc} -C^t, A^t \\ -D^t, B^t \end{array}\right)$$ and after substituting $x$ in to $^txS + Sx = 0$  we end up with:
 $$\left(\begin{array}{cc} -C^t, A^t \\ -D^t, C^t \end{array}\right) = \left(\begin{array}{cc} C, D \\ -A, -B \end{array}\right)$$ If I haven't messed something up already.
First, if I say, for example $A = \left(\begin{array}{cc} a_{11}, a_{12} \\ a_{21}, a_{22} \end{array}\right)$ where $B,C,D$ are defined similarily, I end up with the 4x4 matrix:
 $$= \left(\begin{array}{cccc} a_{11} & a_{12} & b_{11} & b_{12}\\  a_{21} & a_{22} & b_{21} & b_{22} \\   c_{11} & c_{12} & d_{11} & d_{12}\\  c_{21} & c_{22} & d_{21} & d_{22} \end{array}\right)$$
and skipping past the algebra in trying to make $^txS + Sx = 0$ work I get:
$\left(\begin{array}{cc} A & b \\ c & A \end{array}\right)$ where $b = \left(\begin{array}{cc} 0 & b_{12} \\ b_{21} & 0 \end{array}\right)$ and $c = \left(\begin{array}{cc} 0 & -c_{12} \\ c_{21} & 0 \end{array}\right)$ Where $b_{12} = b_{21}$ and $-c_{12} = c_{21}$
Now if I am even close to being on the right track, I don't know what to do next in order to find a "basis for $L$"
$\textbf{Question 1}$ How do I find a basis for $L$ and prove that $Dim(L) = 10$ There is a hint to this question: (Hint; write $x = \left(\begin{array}{cc} A & B \\ C & D \end{array}\right)$ in block form with $A,B,C,D$ as blocks for order $2x2$.
Now I find that the Lie Algebra $L$ can be decomposed and written as a direct sum of weight vectors $L = L_0 \bigoplus_{\alpha \in \Phi}$, and $L_0 = \{x \in V; [x,h] = 0 \forall h \in H\}$ and we let $L_0 = H$. Now I don't quite know how that works, realize that $[x,h] = 0$ means that $xh = hx$ hence $H$ is abelian, and as any linear vector space can be written in Jordan form and each $h = d +n$ where $d$ is diagonal and $n$ is nilpotent, which because $h$ is diagonal already $h = d$ thus I am guessing that we can somehow decompose $h$ as a direct sum of eigenvectors - but I am not sure if this even the right approach. That being said:
$\textbf{Question 2}$: How do I find a basis for $H = \{h \in L | h \text{ is diagonal }$ and prove that $H = 2$?
This last part, I don't know how to approach. If we are taking the adjoint action of $H$ then we are taking $ad(h)(-) = [h,-]$ am I correct in that I am just looking at the adjoint action on all $x \in V$, where as above $V$ is the vector space representation of $L$?
$\textbf{Question 3}$: Find a basis of $L$ given by weight vectors for the adjoint action of $H$. How do these weights look as linear combinations of functionals $\varepsilon_i$ such that $\varepsilon(diag(a_1,a_2,a_3,a_4)) = a_i$ for $i = 1,2$?
An aside on $\textbf{Question 1}$: In Erdmann and Wildon's book on Lie Algebras, exercise 4.2 has a 2x2 matrix which: $\left(\begin{array}{cc} m & p \\ q & -m^t \end{array}\right)$ where $p$ and $q$ are symmetric; it looks similar to the matrix I came up with above: $\left(\begin{array}{cc} A & b \\ c & A \end{array}\right)$. Is my matrix supposed to be the same as theirs?
I hope I didn't say too much in my trying to explain my thoughts. Once again, any help is appreciated and thanks in advance.

The elements of $\mathfrak{sp}_4(\mathbb{C})$ can be written in block form
$$
X=\begin{pmatrix} A & B \\ C & -A^t \end{pmatrix}
$$
with $2\times 2$ blocks, where $B^t=B$ and $C^t=C$. From here we count that the dimension of the Lie algebra equals $10$, because $A$ has $4$ parameters, and $B$ and $C$ have three parameters (so that $4+3+3=10$). Now just take these parameters $0$ respectively $1$ to obtain a basis with $10$ elements. Usually we write $H_1=diag(1,0,-1,0)$, $H_2=diag(0,1,0,-1)$, $X_{12}=E_{12}-E_{43}$ etc. There are many references, e.g., here.
