# Is every decomposable basis for $\bigwedge^kV$ “standard”?

This is a curiosity:

Let $$V$$ be a $$d$$-dimensional real vector space, and let $$1. 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 $$\lambda_{i_1,\ldots,i_k}$$?

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 $$c=(x+y)\wedge z$$.

• This is a perfect answer! extremely elegant. Thank you. – Asaf Shachar Jan 27 at 9:43