I have a Lagrangian subspace $L$ of $\mathbb{R}^{2n}$ (i.e. $L^{\perp_{\omega_0}} = L$ where $\omega_0$ is the natural symplectic form.
I want to show that, given the natural almost complex structure $J_0$ on $\mathbb{R}^{2n}$, $J_0(L) = L^\perp$ (where here, $\perp$ is the orthogonal space using the euclidian metric).
if $n = 1$, this is clear. However, for $n = 2$, it is less clear.
Indeed, if $n = 2$, the Lagrangian condition reduces to
$$v_1w_2+v_3w_4 = w_1v_2+w_3v_4$$
for $v,w\in L$.
and $J_0(v)\cdot w$ for $v,w\in L$ reduces to
$$v_3w_1+v_4w_2-v_1w_3-v_2w_4$$
How can these two equations be linked?
Use the formula $\omega(x,J_0(y))=b(x,y)$ where $b$ is the Euclidean metric.
If $x,y\in L$, $\omega(x,J_0(J_0(y))=b(x,J_0(y))=-\omega(x,y)=0$ since $J_0^2=-Id$. This implies that $J_0(L)$ is orthogonal to $L$.