Dimension of $\mathfrak{sp}(2n, \mathbb{R})$

Question

I'm computing the dimension of $\mathrm{Sp}(2n, \mathbb{R})$ and I've shown that its Lie algebra is $\mathfrak{sp}(2n, \mathbb{R})=\{X\in\mathrm{Mat}_{2n}(\mathbb{R})\mid XJ+JX^\mathrm{T}=0\}$, where $$J=\begin{pmatrix} 0&-I_n\\ I_n&0 \end{pmatrix}.\tag{1}$$ Now I try to find the form of the elements in $\mathfrak{sp}(2n, \mathbb{R})$, so let $$X=\begin{pmatrix} A&B\\ C&D \end{pmatrix},\tag{2}$$ where $A$, $B$, $C$, $D$ are all $n\times n$ matrices. It follows from some calculations that $$C=B\quad\text{and}\quad D=-A.\tag{3}$$ Then the form of the elements in $\mathfrak{sp}(2n, \mathbb{R})$ is as $$X=\begin{pmatrix} A&B\\ B&-A \end{pmatrix},\tag{4}$$ i.e., $X$ is determined by two matrices $A$ and $B$. Since both $A$ and $B$ are in $\mathrm{Mat}_n(\mathbb{R})$, the dimension of $\mathfrak{sp}(2n, \mathbb{R})$ equals $2n^2$. However, I know that fact $\dim \mathfrak{sp}(2n, \mathbb{R})=2n^2+n$, so where did my derivation go wrong?

Answer 1

OP should probably recheck their calculation (3). According to Wikipedia it should be $$D~=~-A^T,\qquad B~=~B^T,\qquad C~=~C^T,$$ leading to $$\dim \mathfrak{sp}(2n, \mathbb{R})~=~n^2+2\frac{n(n+1)}{2}~=~n(2n+1).$$