1
$\begingroup$

I wanted to use the definition of a wedge product which says $λ_1λ_2\cdots λ_k(v_1,v_2,\ldots,v_k)=\det(λ_i(v_j))$ with $1<i,j<k$ but I'm not sure if that even can work

$\endgroup$
2
$\begingroup$

This is essentially by the definition you wrote. It is assumed that $x_1, \ldots, x_n$ is a basis, and by definition $dx_1, \ldots, dx_n$ are the dual basis. Thus if you expand out $v_j=a_{j1}v_1+\cdots + a_{jn}v_n$ in this basis, then $dx_i(v_j)=a_{ji}$.

Also in this basis $(v_1, \ldots, v_n)=(a_{ij})$. So since $\det(a_{ij})=\det(a_{ij}^T)=\det(a_{ji})=dx_1\cdots dx_n(v_1, \ldots , v_n)$ we are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.