This is a [cross-post](https://math.stackexchange.com/questions/3090653/is-every-basis-for-bigwedgekv-satisfying-a-complementary-property-a-rescal).


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](https://math.stackexchange.com/questions/3089508/characterising-minors-of-diagonal-matrices).


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 $\text{span}(v\_{i\_1})=\text{span}(u\_{i\_1})$.



Is every basis for $\bigwedge^kV$ satisfying a "complementary" property a rescaling of a "standard" basis?
The $\omega\_{i\_1,i\_2}$ give six distinct two-dimensional spaces $L\_1,L'\_1,L\_2,L'\_2,L\_3,L'\_3$ such that each pair has a one-dimensional intersection except for $L\_1\cap L'\_1=L\_2\cap L'\_2=L\_3\cap L'\_3=0.$


Let $A=L\_1\cap L\_2,$ $B=L\_1\cap L'\_2,$ $C=L'\_1\cap L\_2,$ and $D=L'\_1\cap L'\_2.$ Then $L\_1$ contains $A$ and $B,$ but these are distinct because $L\_2\cap L'\_2=0.$ So $L\_1=A\oplus B$ (direct sum of one-dimensional spaces). Similarly $L'\_1=C\oplus D,$ $L\_2=A\oplus C,$ and $L'\_2=B\oplus D.$ The decomposition $V=A\oplus B\oplus C\oplus D$ will provide the required basis.


Suppose that $L\_3$ does not contain $A.$ Then $L\_3\cap L\_1$ and $L\_3\cap L\_2$ are distinct, because $L\_1\cap L\_2=A.$ So $L\_3=(L\_3\cap L\_1)\oplus (L\_3\cap L\_2)\subset A\oplus B\oplus C.$
Therefore $L\_3\cap L'\_1\subset (A\oplus B\oplus C)\cap (C\oplus D)=C.$


We have shown that $A\not\subset L\_3$ implies $C\subset L\_3.$ Swapping $L\_1,L'\_1,L\_2,L'\_2$ with $L'\_1,L\_1,L'\_2,L\_2$ respectively in this argument swaps $A,B,C,D$ with $D,C,B,A$ respectively. Swapping $L\_1,L'\_1,L\_2,L'\_2$ with $L\_2,L'\_2,L\_1,L'\_1$ respectively swaps $A,B,C,D$ with $A,C,B,D$ respectively. Swapping $L\_1,L'\_1,L\_2,L'\_2$ with $L'\_2,L\_2,L'\_1,L\_1$ respectively swaps $A,B,C,D$ with $D,B,C,A$ respectively. So either of $A\not\subset L\_3$ or $D\not\subset L\_3$ has the consequence that $B,C\subset L\_3.$ This means $L\_3=A\oplus D$ or $L\_3=B\oplus C.$ And we can swap $L\_3$ with $L'\_3$ in the argument to get the same conclusion for $L'\_3.$ This proves that $L\_3$ and $L'\_3$ are $A\oplus D$ and $B\oplus C$ in some order. This shows that the $\omega\_{i\_1,i\_2}$ are a rescaling of the basis induced by elements of $A,B,C,D$ respectively.


