Is the real matrix of complex structure skew symmetric？ Let $A\in\mathbb{R}^{2n\times 2n}$ with $A^2=-I$, where $I$ is the identity matrix in $\mathbb{R}^{2n\times 2n}$. It is clear that $L_A:\mathbb{R}^{2n}\rightarrow\mathbb{R}^{2n}, x\mapsto Ax$ is a complex structure on $\mathbb{R}^{2n}$. Is $A$ skew-symmetric?
I also want to know how the set
\begin{equation*}
\{A\in\mathbb{R}^{2n\times 2n}:A^2=-I\}
\end{equation*}
would be like. Is there any book talk about this thing?
 A: Let
$$ J = \begin{pmatrix} 0_n & -I_n \\ I_n & 0_n \end{pmatrix} \in M_{2n}(\mathbb{R})$$
be the "standard" complex structure and let $A \in M_{2n}(\mathbb{R})$ an arbitrary complex structure. Endow $\mathbb{R}^{2n}$ with the structure of a complex vector space by defining
$$ (a + ib)v = a\cdot v + bAv,\,\,\, (a + ib)\in\mathbb{C}, v\in\mathbb{R}^{2n} $$
and denote the resulting $n$-dimensional complex vector space by $(\mathbb{R}^{2n},A)$. Since all $n$-dimensional complex vector spaces are isomorphic to each other, there exists a complex isomorphism $L_{C} \colon (\mathbb{R}^{2n},J) \rightarrow (\mathbb{R}^{2n}, A)$ for some real matrix $C \in \operatorname{GL}_{2n}(\mathbb{R})$. Since $L_C$ is an isomorphic of complex vector spaces, we must have
$$ L_{C}(iv) = L_{C}(Jv) = C(Jv) = iL_{C}(v) = ACv $$
for all $v \in \mathbb{R}^{2n}$. That is, $CJ = AC$ or $A = CJC^{-1}$. In other words, $A$ is conjugate to the standard complex structure. One can easily verify that if $A$ is conjugate to the standard complex structure then $A^2 = -I$ and so $A$ is a also a complex structure. Hence,
$$ \{ A \in M_{2n}(\mathbb{R}) \, | \, A^2 = -I \} = \{ CJC^{-1} \, | \, C \in \operatorname{GL}_{2n}(\mathbb{R}) \} .$$
This also answers your first question. The standard complex structure $J$ is skew-symmetric. An arbitrary complex structure $CJC^{-1}$ will be skew-symmetric if (and only if) $C^T C$ is also complex (that is, it commutes with $J$). In particular, if $C$ is orthogonal, $CJC^{-1}$ will be skew-symmetric.
