Symplectic lie algebra Can anyone explain me why, in the symplectic lie algebra, which is defined as $ sp(n)=\{X \in  gl_{2n}:X^tJ+JX=0\}$ where $J=\begin{pmatrix}
        0 & I  \\
        -I & 0  \\
        \end{pmatrix} $  we can write its elements, in block form  $X=\begin{pmatrix}
        A & B  \\
        C & -A^t  \\
        \end{pmatrix} $ where $ A,B,C \in M_{n\times n}$ and $B=B^t,C=C^t$ .How does it proved?

Hint: The block decomposition used for $J$ (as well as the answer) suggest writing a generic matrix $X \in \mathfrak{gl}_{2n}$ as
$$X = \begin{pmatrix} A & B \\ C & D \end{pmatrix}.$$
To produce the block matrix description of $\mathfrak{sp}_{2n}$, simply substitute the block expressions for $X$ and $J$ in the definition of that algebra:
\begin{align}
X^t J + J X &= 0 \\
\begin{pmatrix} A & B \\ C & D \end{pmatrix}^t \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} + \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} &= 0 \textrm{.}
\end{align}
Now, simplify the l.h.s. of the equation to produce algebraic conditions that $A, B, C, D$ must satisfy.
