Exterior Algebra and linearly independent vectors Suppose that $v_1,\cdots,v_r$ are linearly independent vectors in some vector space $V$. I want to try and show that for any $w \in \bigwedge^p(V)$ that
$$
w = \sum_{i=1}^{r} v_i \wedge \psi_i
$$
for some $\psi_i \in \bigwedge^{p-1}(V)$ if and only if
$$
v_1\wedge v_2\wedge \cdots \wedge v_r\wedge w = 0.
$$
The forward direction is trivial by writing $w$ as the sum and extending the wedge product linearly. It's the second implication that's giving me some trouble.
If we assume that $v_1\wedge v_2\wedge \cdots \wedge v_r\wedge w = 0$, then I want to conclude that I can write $w$ in the appropriate form by examining well-chosen alternating, multi-linear forms from $V^{p+r}$ into some vector space so that I can use the universal property of $\bigwedge^{p+r}(V)$, and evaluate the induced map at $v_1\wedge v_2\wedge \cdots \wedge v_r\wedge w$ and get $0$.
The problem I'm having is that $w$ is not necessarily an elementary wedge product, so I don't have a canonical way of thinking of it as an element of $V^p$.
Any ideas for this backward direction would be greatly appreciated.
 A: Let $\{e_1,\ldots, e_k\}$ be a basis of $V$ such that $v_i=e_i$ for $1\le i\le r$. $w\in \bigwedge^p(V) \implies$
$$w = \sum_{\alpha\in P}f_{\alpha}e_{\alpha_1}\wedge\ldots \wedge e_{\alpha_s}$$
Where $P = \{(i_1,\ldots, i_s) \mid 1 \le  i_1 < i_2 < \cdots < i_s \le k, s\leq p\}$ and I will use $|\alpha|$ to denote the number of elements in the tuple. Clearly
$$v_1\wedge \cdots \wedge v_r =  e_{1}\wedge\cdots \wedge e_{r}$$
So
\begin{align*}
&v_1\wedge  \cdots \wedge v_r\wedge w = 0\\ 
\implies&  e_{1}\wedge\cdots \wedge e_{r}\wedge  \sum_{\alpha\in P}f_{\alpha}e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_s} = 0\\
\implies& \forall \alpha\in P, f_\alpha \neq 0 \implies \exists l_\alpha \leq |\alpha|, \alpha_{l_\alpha} \leq r \text{ (Let $l_\alpha$ denote the least such value)}\\
\implies& w = \sum_{\alpha\in P, f_\alpha \neq0}f_{\alpha}e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_m}\wedge e_{l_\alpha}\wedge e_{\alpha_n}\wedge \cdots \wedge e_{\alpha_s} \space\space(\alpha_m<l_\alpha<\alpha_n)\\
\implies& w = \sum_{\alpha\in P, f_\alpha \neq0}f_{\alpha}(-1)^m e_{l_\alpha}\wedge e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_m}\wedge e_{\alpha_n}\wedge \cdots \wedge e_{\alpha_s} \space\space(\alpha_m<l_\alpha<\alpha_n)\\
\implies& w = \sum_{i=1}^rv_{i}\wedge\sum_{\alpha\in P, f_\alpha \neq0,l_\alpha = i}f_{\alpha}(-1)^m\wedge e_{\alpha_1}\wedge\cdots \wedge e_{\alpha_m}\wedge  e_{\alpha_n}\wedge \cdots \wedge e_{\alpha_s} \space\space(\alpha_m<l_\alpha<\alpha_n)
\end{align*}
Might have made a mistake somewhere but the idea should be clear. If you have a notation that you suggest I use for clarity, please feel free to comment!
