# Wedge product and vectors

• Jan 23rd 2011, 12:38 PM
Sogan
Wedge product and vectors
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
• Jan 24th 2011, 05:16 AM
TheArtofSymmetry
Quote:

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? $v_1 \wedge...\wedge v_n=\sum_{\sigma \in S_n}sgn(\sigma)(v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)})$?
• Jul 8th 2015, 01:45 AM
Rebesques
Re: Wedge product and vectors
The (<=) direction can be proven by observing that $v_1\wedge v_2\wedge \ldots\wedge v_n$ is the linear form that corresponds to (a multiple of) the determinant of $\{v_1,\ldots,v_n\}$ when expressed for a given basis.
• Mar 6th 2016, 05:03 AM
Rebesques
Re: Wedge product and vectors
Quote:

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

Their determinant.