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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<i_k\le n} \det \begin{bmatrix} x_1^{i_1} & \dots & x_k^{i_k} \\ \vdots & \ddots & \vdots \\ x_1^{i_1} & \dots & x_k^{i_k} \end{bmatrix} v_{i_1}\wedge\dots\wedge v_{i_k} \end{align} 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.

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

1 Answer 1

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$\DeclareMathOperator{span}{span}$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 $\span(\{v_1, \ldots, v_n\}) = \span(\{x_1, \ldots, x_n\})$.

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    $\begingroup$ It seems that the result is not restricted to finite-dimensional vector spaces $\endgroup$
    – Physor
    Commented Oct 19, 2021 at 21:04
  • $\begingroup$ @Physor This does indeed seem to be the case. $\endgroup$ Commented Oct 19, 2021 at 21:05

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