Wedge product and vectors

**Sogan** · January 23rd 2011, 12:38 PM

Hello,

i want to proof this claim:

Let V be a finite vector space.
$v_1,..,v_n$ are not linearly indipendent <=> $v_1 \wedge...\wedge v_n=0$ .

One direction(=>) wasn't so difficult. But i have problems with the other direction.
So we have $0=v_1 \wedge...\wedge v_n=\sum_{\sigma \in S_n}sgn(\sigma)(v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)})$ .

Can you see, why they have to be linearly dependent?

Regards

**TheArtofSymmetry** · January 24th 2011, 05:16 AM

Originally Posted by Sogan

Hello,

i want to proof this claim:

Let V be a finite vector space.
$v_1,..,v_n$ are not linearly indipendent <=> $v_1 \wedge...\wedge v_n=0$ .

One direction(=>) wasn't so difficult. But i have problems with the other direction.
So we have $0=v_1 \wedge...\wedge v_n=\sum_{\sigma \in S_n}sgn(\sigma)(v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)})$ .

Can you see, why they have to be linearly dependent?

Regards

Just wondering, how do you derive this? $v_1 \wedge...\wedge v_n=\sum_{\sigma \in S_n}sgn(\sigma)(v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)})$ ?

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