Wedge product and vectors

Hello,

i want to proof this claim:

Let V be a finite vector space.

$\displaystyle v_1,..,v_n$ are not linearly indipendent <=> $\displaystyle v_1 \wedge...\wedge v_n=0$.

One direction(=>) wasn't so difficult. But i have problems with the other direction.

So we have $\displaystyle 0=v_1 \wedge...\wedge v_n=\sum_{\sigma \in S_n}sgn(\sigma)(v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)})$.

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

Regards

Re: Wedge product and vectors

The (<=) direction can be proven by observing that $\displaystyle v_1\wedge v_2\wedge \ldots\wedge v_n$ is the linear form that corresponds to (a multiple of) the determinant of $\displaystyle \{v_1,\ldots,v_n\}$ when expressed for a given basis.

Re: Wedge product and vectors

Quote:

Originally Posted by

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

Their determinant.