Wedge product/Base change/Computation/Remark

For bases ${\displaystyle {}v_{1},\ldots ,v_{m}}$ and ${\displaystyle {}w_{1},\ldots ,w_{m}}$ of a ${\displaystyle {}K}$-vector space ${\displaystyle {}V}$, with the relations

${\displaystyle {}v_{j}=\sum _{i=1}^{m}a_{ij}w_{i}\,,}$

we obtain, between the bases

${\displaystyle v_{i_{1}}\wedge \ldots \wedge v_{i_{n}}{\text{ with }}1\leq i_{1}<\ldots <i_{n}\leq m\,\,{\text{ and }}\,\,w_{i_{1}}\wedge \ldots \wedge w_{i_{n}}{\text{ with }}1\leq i_{1}<\ldots <i_{n}\leq m}$

of ${\displaystyle {}\bigwedge ^{n}V}$, the relation

${\displaystyle v_{j_{1}}\wedge \ldots \wedge v_{j_{n}}=\sum _{1\leq i_{1}<\cdots <i_{n}\leq m}{\left(\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{s=1}^{n}a_{i_{s}j_{\pi (s)}}\right)}w_{i_{1}}\wedge \ldots \wedge w_{i_{n}}\,.}$

This rests, according to fact  (4), on

${\displaystyle {}{\begin{aligned}v_{j_{1}}\wedge \ldots \wedge v_{j_{n}}&={\left(\sum _{i=1}^{m}a_{ij_{1}}w_{i}\right)}\wedge \ldots \wedge {\left(\sum _{i=1}^{m}a_{ij_{n}}w_{i}\right)}\\&=\sum _{1\leq i_{1}<\cdots <i_{n}\leq m}{\left(\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{s=1}^{n}a_{i_{s}j_{\pi (s)}}\right)}w_{i_{1}}\wedge \ldots \wedge w_{i_{n}}.\end{aligned}}}$