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Let $\omega\in \Lambda^nV$ be an $n$-form for some real linear space $V$. It is known that for $v_1, \dots,v_n\in V$:

$$\omega(v_1,\dots,v_n)=0\iff \{v_1, \dots,v_n\} \ \text{is linearly dependent.}$$

The $\Leftarrow$ implication is easily done. I am, however, having trouble proving $\Rightarrow$.

Can anyone give either a hint or the proof.

Stijn D'hondt
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  • The zero form is an $n$-form, is it not? Is it meant to say $\forall\omega\in\Lambda^nV:\omega(v_1,\ldots,v_n)=0$? – Arthur Sep 01 '19 at 12:34
  • @Arthur, yes, that is implicit by choosing $\omega$ to be a random $n$-form. – Stijn D'hondt Sep 01 '19 at 17:04
  • There is a huge difference between $$\forall \omega(\omega(v_i) = 0\iff v_i\text{ linearly dependent})$$ and $$(\forall\omega(\omega(v_i) = 0)\iff v_i\text{ linearly dependent}$$The way you phrased it, it reads like the former. And that's just not true. – Arthur Sep 01 '19 at 17:18
  • Is it not true that if an $n$-form maps $n$ vectors to zero, the $n$ vectors must be linearly dependent? – Stijn D'hondt Sep 01 '19 at 17:30
  • Maybe it isn't true, if so, I think it $\textit{is}$ true if $\omega$ is a top-form? That is to say, if $\text{dim}(V)=n$? Because then there is (up to a constant) just the one top-form, and then your latter statement is immediately true :). – Stijn D'hondt Sep 01 '19 at 17:35
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    No. It is true that if the vectors are linearly dependent, then any alternating form must send them to $0$. On the other hand, the form that always sends everything to $0$ is an alternating form, and it even sends linearly independent vectors to $0$. – Arthur Sep 01 '19 at 17:36
  • @Arthur, alright, say $\omega$ isn't the zero-form, and say it sends these $n$ vectors to $0$. Is it then true that these $n$ vectors must be linearly dependent? – Stijn D'hondt Sep 01 '19 at 17:39
  • @Arthur, I think the answer to my last question might be "no", but that if we make $n=\text{dim}(V)$, the answer will be "yes". (Which is the question before my last question) – Stijn D'hondt Sep 01 '19 at 17:40

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