Almost Complex Structure 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!

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 $x_1$, $\dots$, $x_n$, $Jx_1$, $\dots$, $Jx_n$.
