Show that for every alternating $m-1$-tensor on $\mathbb R^m$ there exists a unique $v \in \mathbb R^m$ such that for every linear function from $\mathbb R^m$ into $\mathbb R$ and for every $v_1, \dots, v_m \in \mathbb R^m$, $(\omega \wedge f)(v_1, \dots, v_m)=f(v)\det(v_1, \dots, v_m)$.
I have reduced the problem to the following:
Show that for every alternating $m-1$-tensor on $\mathbb R^m$ there exists a unique $v \in \mathbb R^m$ such that for every linear function from $\mathbb R^m$ into $\mathbb R$ , $(\omega \wedge f)(e_1, \dots, e_m)=f(v)$.
I have accomplished this, but my proof was very ugly (I calculated the left side using the definition of the wedge product, using the Alt function). I'd like to know if there is a better way to accomplish it.
Edit:
It's very simple to prove this result if there is an easy way to prove that $(\phi_1\wedge\dots\wedge \phi_{k-1}\wedge \phi_{k+1}\wedge \dots \wedge\phi_m\wedge f)(e_1, \dots, e_m)=f(e_k)$, where the $\phi_i(e_j)=\delta_{ij}$. Is there an elegant way to show this?
Uniqueness of $v$ (if it exists) is easy, so the function $$v \mapsto [(v_1,...,v_m,f) \mapsto f(v) \mathrm{det}(v_1,...,v_m)]$$ is injective. You need to show that the space of functions of this form on $(\mathbb{R}^m)^m \times (\mathbb{R}^m)^*$ (alternating on the $m$ vectors and linear in the functional) is also $m$-dimensional.