Characterization of linear independence by wedge product Let $V$ be a vector space of finite dimension. Show that $x_1,...,x_k$ is linearly independent iff  $x_1\wedge ... \wedge x_k \neq 0$.
 A: I propose the following for the "$\Rightarrow$" direction (please correct me if I'm wrong):
Consider the $k-$th tensor power $V^{\otimes k}$ of $V$. Define the map
$\operatorname{Alt}^k: V^{\otimes k} \rightarrow V^{\otimes k}$ on simple tensors as 
$v_1\otimes \dots \otimes v_k \mapsto \frac{1}{k!}\sum_{\sigma \in S_k} \epsilon(\sigma)v_{\sigma(1)}\otimes \dots \otimes v_{\sigma(k)}$. Here, $S_k$ is the k-th symmetric group and $\epsilon(\sigma)$ is the sign of the permutation.
Then, the $k-$th exterior power of $V$ may be defined as $\bigwedge^kV = V^{\otimes k}/\operatorname{ker}(\operatorname{Alt}^k)$. The quotient map is $v_1\otimes \dots \otimes v_k \mapsto v_1\wedge \dots \wedge v_k$, and the map induced by $\operatorname{Alt}^k$ on the quotient is $v_1\wedge \dots \wedge v_k \mapsto \frac{1}{k!}\sum_{\sigma \in S_k} \epsilon(\sigma)v_{\sigma(1)}\otimes \dots \otimes v_{\sigma(k)}$, which is injective by construction.
Now, if $x_1,\dots, x_k$ are linearly independent, then $\{x_{\sigma(1)}\otimes \dots  \otimes x_{\sigma(k)}\}_{\sigma \in S_k}$ is a linearly independent collection in $V^{\otimes k}$. Therefore, $\frac{1}{k!}\sum_{\sigma \in S_k} \epsilon(\sigma)x_{\sigma(1)}\otimes \dots \otimes x_{\sigma(k)} \neq 0$. We conclude $x_1\wedge \dots \wedge x_k \neq 0$.
A: 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. 
