# Show that $v_1\wedge\dots\wedge v_k = x_1\wedge\dots\wedge x_k \implies \text{span}\{v_1,\dots, v_k\} = \text{span}\{x_1,\dots, x_k\}$

Let $$V$$ be an $$n$$-dimensional space and $$v_1,\dots, v_k \in V$$ are linearly independent. It is clear that if $$x_1,\dots, x_k \in V$$ have the same span as $$v_1\dots v_k \in V$$ then there is a scalar $$t$$ such that $$v_1\wedge\dots\wedge v_k = t\ x_1\wedge\dots\wedge x_k$$ But does the converse also holds ?, i.e. (wlog) if we have $$v_1\wedge\dots\wedge v_k = x_1\wedge\dots\wedge x_k$$ does it imply that $$\text{span}\{v_1,\dots ,v_k\} = \text{span}\{x_1,\dots, x_k\}$$ ?

My attempt:

Complete $$\{v_1,\dots, v_k\}$$ to a basis $$\{v_1,\dots, v_k,\dots,v_n\}$$ then we have a basis $$\{v_{i_1}\wedge\dots\wedge v_{i_k}\}$$ of $$\Lambda^k V$$. Now expand the vectors $$x_j = x_j^l v_l$$, so we get

\begin{align} v_1\wedge\dots\wedge v_k &= x_1\wedge\dots\wedge x_k \\ &= x_1^{i_1}\dots x_k^{i_k}\ v_{i_1}\wedge\dots\wedge v_{i_k}\\ &= \sum_{1\le i_1<\dots Which implies that all the dets are zero except the first one (with $$(i_1,\dots,i_k) = (1,\dots,k)$$). Now we have to show that all the components $$x_r^s = 0$$ for $$s>k$$. For a particular $$s$$ we consider all deteminants that have $$w_s = (x_1^s,\dots,x_k^s)$$ as the last row vector, where that first $$k-1$$ row vectors belong to the $$k$$ row vectors in the first determinant. Since all these determinants vanish, then $$w_s$$ is a linear combination of (the linearly independent) $$\{w_1,\dots,w_k\}$$ in the following manner $$w_s = a_{(j)}^1 w_1 + \dots + \widehat{a_{(j)}^j w_j} + \dots + a_{(j)}^k w_k, \qquad 1\le j \le k$$ where the hat indicates omition of the $$j$$-th term. It is now possible to see $$w_s = 0$$, by observing that $$a_{(1)}^2 w_2 + \dots + a_{(1)}^k w_k = a_{(2)}^1 w_1 + a_{(2)}^3 w_3 + \dots + a_{(2)}^k w_k = \dots.$$ that, all e.g. $$a_{(1)}^j = 0$$, and so $$w_s = 0$$.

I'd like to know if there is a shorter way to prove it.

• How about: $y \in \operatorname{span} \{ x_1, \ldots, x_k \}$ if and only if $y \wedge x_1 \wedge \cdots \wedge x_k = 0$? Commented Oct 19, 2021 at 20:54
• Nice!...really. Shorter than I thought Commented Oct 19, 2021 at 20:56

$$\DeclareMathOperator{span}{span}$$If we have independent vectors $$c_1, \dots, c_n$$, then $$y \in \span(\{c_1, \ldots, c_n\})$$ if and only if $$c_1 \land \cdots \land c_n \land y = 0$$ (thanks to u/DanielSchlepler in the comments for pointing this out).
If $$t \cdot (x_1 \land \cdots \land x_n) = v_1 \land \cdots \land v_n$$, then because the $$v$$s are independent, $$t \neq 0$$ and therefore $$x_1 \land \cdots \land x_n \neq 0$$. Therefore, the $$x$$s are independent.
Then we see that $$y \in \span(\{v_1, \ldots, v_n\})$$ if and only if $$v_1 \land \cdots \land v_n \land y = 0$$ if and only if $$t \cdot x_1 \land \cdots \land x_n \land y = 0$$ if and only if $$x_1 \land \cdots \land x_n \land y = 0$$ if and only if $$y \in \span(\{x_1, \ldots, x_n\})$$.
So $$\span(\{v_1, \ldots, v_n\}) = \span(\{x_1, \ldots, x_n\})$$.