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Let $V$ be a vector space of finite dimension. Show that $x_1,...,x_k$ is linearly independent iff $x_1\wedge ... \wedge x_k \neq 0$.

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What are you having trouble with? (It's not necessary that $V$ have finite dimension.) Can you prove one direction? –  Qiaochu Yuan May 20 '11 at 13:15
    
I'm having trouble with proving that $x_1\wedge ... \wedge x_n = 0$ implies that they are linearly dependent. I guess I'm just missing some way to rewrite it... –  user3620 May 21 '11 at 8:47

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Hint for one direction: if there is a linear dependence, one of the $x_i$ is a linear combination of the others. Then substitute into $x_1\wedge\cdots \wedge x_k$.

Hint for the other direction: You can do row operations $x_i\mapsto x_i+rx_j$ for $i\neq j$ without affecting the wedge $x_1\wedge\cdots\wedge x_k$. Similarly you can divide any $x_j$ by a scalar without affecting whether $x_1\wedge\cdots\wedge x_k$ is nonzero. I'm not sure what properties you already know about the wedge. If you know that wedges $e_{i_1}\wedge\cdots \wedge e_{i_k}$, $i_1<i_2<\cdots<i_k$ form a basis for $\wedge^kV$, when $e_i$ is a basis for $V$, then you're home free.

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Thanks! I should however been more to the point, it was the other direction I had trouble with. Do you have to use some strong property of the exterior product or is it enough to just rewrite it in some clever way? –  user3620 May 21 '11 at 8:50
    
@user3260: I added a hint for the other direction, but I'm not sure what you get to assume about the wedge product. Let me know if you need further hints. –  Grumpy Parsnip May 21 '11 at 12:02
    
@user3620: you do need to use a "strong property." In one direction you only need the universal property but in the other direction you actually need to know that the exterior powers are nontrivial in a particular way. –  Qiaochu Yuan May 21 '11 at 12:17
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Thanks, I think I got it now! –  user3620 May 21 '11 at 13:16

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