For the question below, I have the following definitions and concepts in mind:
The $k^{th}$ exterior power of a real vector space $V$, denoted $\Lambda^k(V)$ can be realized as the quotient
of the tensor product $\bigotimes^k V$ with the subspace of $\bigotimes^k V$ generated by all elements of the form
$v_1 \otimes \dots \otimes v_k$ where $v_i = v_j$ for some $i \neq j$. The equivalence class of $(v_1, \dots, v_k)$
in $\Lambda^k(V)$ is denoted by $v_1 \wedge \cdots \wedge v_k$; it can be thought of as the image of $(v_1, \dots, v_k)$
under the canonical alternating multilinear map that sends each element in $V^k$ to its equivalence class in $\Lambda^k(V)$.
By the universal property of the exterior product, I understand that every alternating multilinear form defined on $V^k$
can identified with a unique linear form with domain $\Lambda^k(V)$. Finally, I know that the determinant of an endomorphism $T:V\rightarrow V$ can be defined as the unique real number $\det T$ such that
$$
Tv_1 \wedge \cdots \wedge Tv_n = (\det T)v_1 \wedge \cdots \wedge v_n
$$
My Question: It is a fact that if $\phi^i, \dots, \phi^k$ are linear forms on $V$ then
$$
\phi^1 \wedge \cdots \wedge \phi^k(v_1, \dots, v_k) = \det[\phi^i(v_j)]
$$
Can this be proved using the facts outlined above without resorting to the combinatorial definition of the determinant/wedge product?
Best Answer
Please refer to my answer on my own question :) I had pretty much the same question from a homework and I eventually figured out a proof for that statement.
algebraic manipulation of differential form