This is a cross-post.

Let $V$ be a $4$-dimensional real vector space. Let $\omega_{i_1,i_2}$
be a basis for $\bigwedge^2V$, where each $\omega_{i_1,i_2}$ is decomposable. Suppose that for every $\omega_{i_1,i_2}$, there is **exactly one** other basis element $\omega_{j_1,j_2}$ such that $\omega_{i_1,i_2} \wedge \omega_{j_1,j_2} \neq 0$.

Is $\omega_{i_1,i_2}$ necessarily a rescaling of a basis that is induced by a basis of $V$? i.e. do there exist a basis $v_i$ for $V$ and $\lambda_{i_1,i_2} \in \mathbb R$, such that $\lambda_{i_1,i_2}\omega_{i_1,i_2}=v_{i_1} \wedge v_{i_2}$?

We must allow a possible rescaling of the $\omega_{i_1,i_2}$: The "complementary" property is scale-invariant, but being an "induced basis" is not invariant:

Indeed, if $\omega^{i_1,\ldots,i_k}$ is an "induced basis" for $\bigwedge^kV$, and $\lambda_{i_1,\ldots,i_k} \omega^{i_1,\ldots,i_k} $ is also induced, then the $\lambda_{i_1,\ldots,i_k}$ **must be the $k$-minors of some diagonal $d \times d$ matrix.** In other words, we have $\lambda_{i_1,\ldots,i_k}=\sigma_{i_1}\cdot \ldots\cdot\sigma_{i_k}$ for some $\sigma_1,\ldots,\sigma_d \in \mathbb{R}$. This implies that the $\lambda_{i_1,\ldots,i_k}$ cannot be chosen freely; there are non-trivial relations.

Thus, the rescalings of induced bases which remain induced are restricted.

The question can be asked for any even $d$ and $k=d/2$. I thought it would be easier to start with the simplest case.

*If you are interested, here is a proof for the rigidity of induced bases:*

We shall prove that $\lambda_{i_1,\ldots,i_k}$ must be the $k$-minors of some diagonal $d \times d$ matrix.

Suppose that $ \omega^{i_1,\ldots,i_k} =v^{i_1} \wedge \ldots \wedge v^{i_k}$ and $\lambda_{i_1,\ldots,i_k} \omega^{i_1,\ldots,i_k} =u^{i_1} \wedge \ldots \wedge u^{i_k}$ for some bases $u_i,v_i$ of $V$. Then, we have $\text{span}(v_{i_1},\dots,v_{i_k})=\text{span}(u_{i_1},\dots,u_{i_k})$, for every $1 \le i_1 < \ldots < i_k \le d$. This implies that $u_i \in \text{span}(v_i)$: Indeed, by switching between $i_k$ and $i_{k+1}$ in $$\text{span}(v_{i_1},\dots,v_{i_{k-1}},v_{i_k})=\text{span}(u_{i_1},\dots,u_{i_{k-1}},u_{i_k}), \tag{1}$$ we obtain

$$\text{span}(v_{i_1},\dots,v_{i_{k-1}},v_{i_{k+1}})=\text{span}(u_{i_1},\dots,u_{i_{k-1}},u_{i_{k+1}}). \tag{2}$$

By intersecting (1) and (2), we see that

$$\text{span}(v_{i_1},\dots,v_{i_{k-1}})=\text{span}(u_{i_1},\dots,u_{i_{k-1}}). \tag{3}$$

In the passage from $(1)$ to $(3)$ we have "removed" the last vectors $v_{i_k},u_{i_k}$. Continuing in this way, we can remove all vectors until we reach $\text{span}(v_{i_1})=\text{span}(u_{i_1})$.