I have a quick algebra question :).
Say that $V$ is an $n-$dimensional vector space with basis $v_1,...,v_n$, and let $v_1^*, ..., v_n^*$ be the dual basis for $V^*$.
I am trying to prove that the $k-$th exterior power of $V^*$ : $\Lambda^k V^*$ is spanned by vectors of the form $v_{i_1}^*\wedge...\wedge v_{i_k}^*$ for $1 \leq i_1 < ... < i_k \leq n$. Where $\Lambda^k V^*$ is defined as the set of alternating $k-$forms $V^k \to \mathbb{R}$.
I looked up the proof on wikipedia and other sites, but they take it as a definition that $\Lambda^k V^*$ is the set of $\omega_1\wedge...\wedge\omega_n$ where the $\omega_i'$s are elements of $V^*$ (from which the claim follows easily).
However, I don't know how to get to that point. Is there a way to show that every $k$-form on $V$ can be expressed as the exterior product of $k$ $1-$forms? Or is there something easier that I am missing? I am a novice in the subject so any help would be appreciated.
Note: the definition of exterior product I use is that for $\alpha \in \Lambda^k V^*$, $\beta \in \Lambda^p V^*$, then:
$\alpha\wedge\beta(x_1,...,x_{k + p}) := \displaystyle\sum \text{sign}(\sigma) \alpha(x_{\sigma(1)},..., x_{\sigma(k)})\beta(x_{\sigma(k+1)}, ..., x_{\sigma(k+p)})$.
Thank you for your help :)
The set of alternating $k$-forms is a vector space. Let's figure out its dimension. If two alternating forms agree on all tuples of $(v_{i_1}, \ldots... v_{i_k})$ with $1 \leq i_1 < ... < i_k \leq n$, then, by the alternating condition, they agree on all tuples of distinct basis vectors (not necessarily ordered), and, since they both vanish on tuples with repeated vectors, they agree on all tuples of basis vectors, and hence, by multilinearity, on all tuples of vectors -- and so they are equal. This shows that the linear map sending the form to its values on all the tuples $(v_{i_1}, \ldots... v_{i_k})$ with $1 \leq i_1 < ... < i_k \leq n$ -- of which there are $\binom{n}{k}$ -- is injective, so the dimension of the space of alternating forms is at most $\binom{n}{k}$.
On the other hand, the $\binom{n}{k}$ forms $v_{i_1}^*\wedge...\wedge v_{i_k}^*$ with $1 \leq i_1 < ... < i_k \leq n$ are alternating and linearly independent (each of them sends the ordered tuple $(v_{i_1},..., v_{i_k})$ with same indexes to $1$, all others to $0$, so a linear combination can only send everything to $0$ if all the coefficients are zero). This means that the dimension is exactly $\binom{n}{k}$ -- and the linearly independent set of the $v_{i_1}^*\wedge...\wedge v_{i_k}^*$s is actually a basis. QED.