Is every decomposable basis for $\bigwedge^kV$ "standard"?

Question

This is a curiosity:

Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. Set

Let $\omega^{i_1,\ldots,i_k}$ be a basis for $\bigwedge^kV$, whose elements are all decomposable. Is $\omega^{i_1,\ldots,i_k}$ necessarily "induced" by a standard basis up to scaling?

i.e. does there exist a basis $v_i$ for $V$, such that $\omega^{i_1,\ldots,i_k}=\lambda_{i_1,\ldots,i_k}v^{i_1} \wedge \ldots \wedge v^{i_k}$ for some real scalars $\lambda_{i_1,\ldots,i_k}$?

Answer 1

No. For instance, let $d=4$ and $k=2$. Note that every standard basis for $\bigwedge^2 V$ (up to scaling) has the property that for any basis element $b$, there is exactly one basis element $c$ such that $b\wedge c\neq 0$. On the other hand, if $\{w,x,y,z\}$ is a basis for $V$, then consider the following decomposable basis: $$w\wedge x, w\wedge y, w\wedge z,x\wedge y, (x+y)\wedge z,y\wedge z.$$ Here, the basis element $b=w\wedge x$ has two different $c$ such that $b\wedge c\neq 0$, namely $c=y\wedge z$ and $c=(x+y)\wedge z$.