Why is $e_1\wedge e_2\dots \wedge e_n \ne 0$ in $\wedge^n V$?

Question

I want to show $e_1\wedge e_2\dots \wedge e_n \ne 0$ in $\wedge^n V$, where $e_1,\dots e_n$ are linearly independent in $V$.

Inspired by this answer, it suffices to show that there exists a linear space $U$ and alternating multi-linear map $f:V\times \cdots \times V \to U$ such that $f(e_1,e_2,\cdots e_n)\ne 0$. But I have trouble constructing $U$ and this map. Maybe this approach is ineffective.

Answer 1

We can identify $V$ with $k^n$ where $k$ is the ground field. And then we can use $e_1=(1,0,\ldots,0),\ldots,e_n=(0,\ldots,0,1)$. Consider the map $T:V^n\to k$ defined as: $$T(v_1,\ldots,v_n)=\mbox{det}\begin{pmatrix}\alpha_{11} & \cdots & \alpha_{1n} \\ \alpha_{21} & \cdots & \alpha_{2n}\\ \vdots & \ddots & \vdots \\ \alpha_{n1} & \cdots & \alpha_{nn}\end{pmatrix},$$ where $v_j=(\alpha_{1j},\alpha_{2j},\ldots,\alpha_{nj})$, with $\alpha_{ij}\in k$. alternating multilinear and such that $T(e_1,\ldots,e_n)=1\neq 0$.

Answer 2

$\textbf{Proposition}$. We have $v_1\wedge \ldots \wedge v_k=0$ if and only if the collection $v_1, v_2,\ldots, v_k$ is linearly dependent.

$\textbf{Proof}$. Suppose $v_1, v_2,\ldots, v_k$ is linearly dependent. Without loss of generality, assume $v_1$ is in the span of $v_2,\ldots, v_k$. Then $$ v_1 = a_2 v_2 + \ldots + a_k v_k, $$ where not all $a_2,\ldots, a_k$ equal zero. Then \begin{align*} v_1\wedge \ldots \wedge v_k &= (a_2 v_2 + \ldots + a_k v_k)\wedge \ldots \wedge v_k \\ &= \sum_{i=2}^{k} a_i v_i \wedge v_2\wedge \ldots \wedge v_i \wedge \ldots \wedge v_k \\ &=0 \hspace{4mm}\mbox{ since }v_i \mbox{ occurs twice in each term. }\\ \end{align*} Now suppose $v_1,v_2,\ldots, v_k$ are linearly independent. Then we can extend it to a basis $v_1,v_2,\ldots, v_n$ of $V$. This means the collection of the form $$ v_{i_1}\wedge \ldots \wedge v_{i_k}, \mbox{ where }i_1 < i_2 < \ldots< i_k, $$ form a basis for $\bigwedge \!{}^k V$. Since $v_1\wedge \ldots\wedge v_k$ is an element of this basis, $v_1\wedge \ldots\wedge v_k\not=0$. $\hspace{1.5cm}$ $\square$

$\textbf{Corollary}$. Let $V$ be an $n$-dimensional vector space. For $k>n$, $$ \bigwedge \!{}^k V= \{ 0\} $$ and for $k=n$, $$ \dim \bigwedge \!{}^n V=1, $$ with basis $v_1\wedge \ldots\wedge v_n$ for any basis $v_1,v_2,\ldots, v_n$ of $V$.