Symplectic lie algebra

Question

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?

Answer 1

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.