# An Expression for the Wedge Product

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?

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Could you say a word about how you're viewing this $\phi_1 \wedge \cdots \wedge \phi_k$ and its action on $k$-tuples? –  Dylan Moreland Jul 8 '12 at 1:12
@Dylan Well, I understand from the combinatorial perspective that the $k-$fold exterior product of forms acts like a k-linear alternating function and that, for instance, given a $p$-form $\alpha$ and a $q$-form $\beta$, $$\alpha \wedge \beta (v_1, \dots, v_{p+q}) = \frac{1}{p!q!} \sum_{\sigma \in S_k} \epsilon(\sigma) \alpha \otimes \beta (v_{\sigma(1)}, \dots v_{\sigma(p+q)})$$ But there are many other equivalent definitions... –  ItsNotObvious Jul 8 '12 at 1:33