Generalized Complex Linear Structure on $\mathbb{C}^n$ The complex $n$-dimensional space $\mathbb{C}^n$ is also a real $2n$-dimensional space. The action of scalar multiplication by $i$, as a complex $n\times n$ matrix, is simply the scalar matrix with $i$'s on the diagonal. Now, let $\{e_1, \dots, e_n\}$ be a basis of $\mathbb{C}^n$. Then $\{e_1,\dots,e_n,ie_1,\dots,ie_n\}$ form a basis for the corresponding real space. If I order the basis like this, then the corresponding real $2n\times 2n$ matrix, call it $J$, would look like
$$
J_{2n} = \begin{bmatrix}
0 & -I_n \\
I_n & 0
\end{bmatrix}
$$
Now, I am just wondering if it is possible to obtain a generalized version of $J$, i.e. suppose there is an orthogonal transformation $f$ of a real space $V$ such that $f^2 = -I$. Does this imply that $\dim{V} = 2n$? If it does, then can I decompose $V = U \oplus U^\perp$ into two perpendicular $n$-dimensional subspaces $U$ and $U^\perp$ and find an isometry $g: U \to U^\perp$ and write $f$ as
$$
f = \begin{bmatrix}
0 & -g \\
g & 0
\end{bmatrix}
$$
instead of $I_n$ above? That way, $f$ would describe the action of multiplying by $i$ on a complex
vector space, where $U$ and $U^\perp$ would be the real and imaginary parts, respectively. I think this has to work, maybe with a different choice of basis. But I cannot figure out why exactly this is true. Can someone give me some directions?
EDIT: I just realized that $\dim V$ has to be even since $(\det f)^2 = (-1)^{\dim V} \ge 0$. But I am still having trouble finding the isometry $g$ to write $f$ in block-antidiagonal form. Can someone give me some directions?
 A: These are all standard results.
It can be shown that $\dim V = 2n$. Because $J^2=-I$, the eigenvalues of $J$ can only be $\pm i$. If the dimension of $V$ is odd, then $\det J=\pm i\not\in\mathbb R$. But this fact can also be established in finding $V$ and $V^{\perp}$ as follows.
Consider the complexification $V^c=V\otimes_{\mathbb R}\mathbb C\cong V\oplus iV$ upon which $J$ naturally acts as a complex linear transformation. Then $V^c=\ker(I-iJ)\oplus\ker(I+iJ)$ as the minimal polynomial of $J$ has no repeated eigenvalues. Let $u_1+iv_1, \cdots, u_n+iv_n$ be a basis of $\ker(I-iJ)$. Then it's easy to see that $u_1-iv_1, \cdots, u_n-iv_n$ form a basis of $\ker(I+iJ)$, and $Ju_j=v_j, \forall j=1, 2, \cdots, n$. Therefore $u_1, \cdots, u_n, v_1, \cdots, v_n$ form a basis of $V^c$ and henceforth $V$. Let $U=\text{span}(u_1, \cdots, u_n)$ and $U^{\perp} = \text{span}(v_1, \cdots, v_n)$. We are done.
Be warned that this decomposition is not unique: Take $n=1$, $J=J_2$, then any $1$-dimensional subspace of $\mathbb R^2$ can serve as $U$ with $JU$ as $U^{\perp}$.
Note that in the above discussion, we have never used the fact that $V$ has an inner product and $J$ is orthogonal and we didn't justify the notation $U^{\perp}$ truly as the orthogonal complement of $U$. Under the extra conditions, we have an easier argument. Pick an arbitrary $u_1\in V$, then $(u_1, Ju_1)=(Ju_1, J^2u_1)=-(u_1, Ju_1)$ hence $(u_1, Ju_1)=0$. And by $J$ is orthogonal, $\text{span}(u_1, Ju_1)^{\perp}$ is $J$-invariant and we may continue the process, eventually we find $u_1, u_2, \cdots, u_n, v_1, \cdots, v_n$ that form an orthogonal basis of $V$, with the property $Ju_j = v_j$, in which case it's justified that $U^{\perp}=\text{span}(v_1, \cdots, v_n)$ is indeed the orthogonal complement of $U:=\text{span}(u_1, \cdots, u_n)$.
A: A general complex structure on any $2n$ dimensional vector space $V$ (real) is given by a map $J\in End(V)$ such that $J\circ J = -I$. So in matrix form $J^2=-I$. That definition includes the matrices you've written down in your prompt. A vector space given a complex structure is called $(V,J)$.
This construction works because there exists a linear isomorphism $\phi : V \to \mathbb{C}^n$ such that $\phi(Jv)=i\phi(v)$.
