I'm trying to solve the following problem:
Given real $2$-forms $\omega_1, \omega_2, \omega_3\in\Lambda^{1,1}\mathbb C^2$ satisfying
$\omega_1\wedge\omega_1=\omega_2\wedge\omega_2=\omega_3\wedge\omega_3=0$
The forms $\omega_1\wedge\omega_2$ and $\omega_2\wedge\omega_3$ are proportional to the standard volume form of $\mathbb R^4=\mathbb C^2$ with positive coefficient.
Prove that $\omega_1\wedge\omega_3$ is also proportional with nonnegative coefficient.
What I did was write the forms in the basis $\omega_i=a_idz_1d\overline{z_1}+b_idz_1d\overline{z_2}+c_idz_2d\overline{z_1}+d_i dz_2d\overline{z_2}$ and define a quadratic form on the space of real $(1,1)$ forms by
$$q(a,b)=\frac{a\wedge b}{dz1d\overline{z_1}dz_2d\overline{z_2}}=\frac{a\wedge b}{-4\text{Vol}}.$$
In particular $q(\omega_i, \omega_j)=a_1d_2-b_1c_2-b_2c_1+d_1a_2$ so now we just have a multilinear algebra problem i.e. we want to choose the $a_i, b_i, c_i, d_i$ such that $q(\omega_i, \omega_i)=0$ and $q(\omega_1, \omega_2)<0$ and $q(\omega_2, \omega_3)<0$ and conclude that we must have $q(\omega_1, \omega_3)\leq 0$. But choosing randomly I found the coefficients
[0.9999732190030922, -0.007318557002146543, 0.5057424794163562, -0.8626844988254956]
[-0.8774094614749035, -0.47974226092175776, -0.1365943806841771, -0.9906270615955866]
[0.7318502897798106, 0.6814654454550189, 0.8814419385032646, 0.47229239782957255]
(they are [a_i,b_i,c_i,d_i] in order and written in a basis such that q is diagonal with entries 1,1,-1,-1) which have all the inequalities as before except $q(\omega_1, \omega_3)>0$. Am I misunderstanding something?