Basis for $\mathbb{R}^{2n}$ with an almost complex structure

Question

The question itself is nothing but linear algebra:

Let $\{x_1,\cdots,x_n\}$ be n linearly independent vectors (not necessarily orthogonal) in $\mathbb{R}^{2n}$ and $J^2=-1$ is the almost complex structure, we want to show that $\{x_1,\cdots, x_n, Jx_1,\cdots, Jx_{n}\}$ span $\mathbb{R}^{2n}$.

To check that they are linearly independent, we assume:\begin{equation}\begin{aligned} \sum_i a_ix_i+\sum_j b_jJx_j=0 \\ \end{aligned} \end{equation}

Act $J$ on both side, we have:\begin{equation} \sum_i a_i Jx_i-\sum_jb_jx_j=0 \end{equation}

Then I saw somebody concluded right away the following which I can not figure out why: \begin{equation} \sum_i (a_i^2+b_i^2)x_i=0 \end{equation}

Thanks for your help!

Answer 1

But it's not true. That is, not as stated. If you take just any old $x_1,\ldots,x_n$ that is. Consider $\mathbb R^4$, and pick some $x_1$ and for $x_2$ pick $J x_1$, then you only span a 2 dimensional subspace still. So you definitely want $\operatorname{span} \{ x_1,\ldots,x_n \} \cap \operatorname{span} \{ J x_1, \ldots, J x_n \}$ to be trivial, otherwise what you are saying is simply not true. But then this follows by dimension, since $J$ had better be invertible and so it preserves dimensions of subspaces.

The idea is that you need to pick a so-called maximally totally-real linear subspace. If your initial subspace $X$, the span of the $x_1,\ldots,x_n$, has any complex structure, then the span of $X$ and $JX$ is not everything.

Answer 2

In fact, choose any $x_1 \neq 0$. Then $x_1$ and $Jx_1$ are linearly independent since otherwise $J$ has a real eigenvalue. Next, choose $$ x_2 \not\in \mathop{\mathrm{span}}\{x_1,Jx_1\}. $$ Then $x_1$, $x_2$, $Jx_1$, $Jx_2$ are linearly independent since otherwise $V = \mathop{\mathrm{span}}\{x_1,x_2,Jx_1\}$ is invariant for $J$ and then $J|_V$ must have a real eigenvalue $\lambda$ since $\dim V = 3$. Then $\lambda$ is also an eigenvalue for $J$, which is a contradiction.

Proceeding in this way, you find a basis of the form $x_1$, $\dots$, $x_n$, $Jx_1$, $\dots$, $Jx_n$.