Wedge product/Finite-dimensional/Basis/Fact/Proof

We show first that we have a generating system. Because the elements of the form ${\displaystyle {}w_{1}\wedge \ldots \wedge w_{n}}$ form a generating system of ${\displaystyle {}\bigwedge ^{n}V}$ due to fact  (1), it is enough to show that these elements can be represented. For every ${\displaystyle {}w_{j}}$, there exists a representation ${\displaystyle {}w_{j}=\sum _{i=1}^{m}a_{ij}v_{i}}$; therefore, according to fact  (4). we can write the ${\displaystyle {}w_{1}\wedge \ldots \wedge w_{n}}$ as linear combinations of wedge products of the basis elements; however, every ordering may occur. Hence, let ${\displaystyle {}v_{k_{1}}\wedge \ldots \wedge v_{k_{n}}}$ be given, with ${\displaystyle {}k_{j}\in \{1,\ldots ,m\}}$. By swapping neighboring vectors, using fact  (3), we may achieve (maybe with another sign) that the indices are (not necessarily strict) increasing. If an index appears twice, the wedge product is ${\displaystyle {}0}$, due to fact  (2). Hence, no index occurs twice, and this wedge product is in the form asked for.

To show that the family is linearly independent, we show, using fact, that for every subset ${\displaystyle {}I=\{i_{1},\ldots ,i_{n}\}\subseteq \{1,\ldots ,m\}}$ with ${\displaystyle {}n}$ elements (where ${\displaystyle {}i_{1}<\ldots <i_{n}}$), there exists a ${\displaystyle {}K}$-linear mapping

${\displaystyle \bigwedge ^{n}V\longrightarrow K}$

such that ${\displaystyle {}v_{i_{1}}\wedge \ldots \wedge v_{i_{n}}}$ is not mapped to ${\displaystyle {}0}$, but all other wedge products in the family are mapped to ${\displaystyle {}0}$. To show this, it is enough, by fact, to give an alternating multilinear mapping

${\displaystyle \triangle \colon V^{n}\longrightarrow K}$

satisfying ${\displaystyle {}\triangle {\left(v_{i_{1}},\ldots ,v_{i_{n}}\right)}\neq 0}$ but ${\displaystyle {}\triangle {\left(v_{j_{1}},\ldots ,v_{j_{n}}\right)}=0}$ for every other strictly increasing index tuple. Let ${\displaystyle {}U}$ be the linear subspace generated by ${\displaystyle {}v_{i}}$, ${\displaystyle {}i\neq i_{k}}$, of ${\displaystyle {}V}$, and let ${\displaystyle {}W=V/U}$ denote the residue class space. Then the images of the ${\displaystyle {}v_{i_{k}}}$, ${\displaystyle {}k=1,\ldots ,n}$, form a basis of ${\displaystyle {}W}$, and the images of all other subsets with ${\displaystyle {}n}$ elements of the given basis do ot form a basis of ${\displaystyle {}W}$, because at least one element is mapped to ${\displaystyle {}0}$. We consider now the composed mapping

${\displaystyle \triangle :V^{n}\longrightarrow W^{n}\cong (K^{n})^{n}{\stackrel {\det }{\longrightarrow }}K.}$

This mapping is multilinear and alternating, due to fact and fact. Due to fact, we have ${\displaystyle {}\triangle (z_{1},\ldots ,z_{n})=0}$ if and only if the images of ${\displaystyle {}z_{i}}$ in ${\displaystyle {}W}$ do not form a basis.