Let $\{a_1,\cdots,a_n\}$ be $n$ linearly independent one-form in $\Omega(\mathbb{R}^n)$ and suppose that $\omega$ is a $k$- form such that $1 \leq k \leq n-1$.
Show that if $\omega \wedge a_i = 0$ for all $i \in \{1,2,\cdots,n\}$ then $\omega$ is the null $k$-form.
I know that $a_1 \wedge a_2 \wedge \cdots \wedge a_n \neq 0$ since they are linearly independent, and I also know that $a_i \wedge a_i = 0$. One other ideia is to decompose $\omega$ as a combination of wedge products of one-form factors and see if that, toghether with the fact that the exterior product of $\omega$ with each $a_i$ vanishes, give me that $\omega = 0$. Tried all that and couldn't get anywhere.
Can someone please help me here? Thanks!
As $I$ varies over increasing $k$-tuples, $a_I = a_{i_1}\wedge\dots\wedge a_{i_k}$ gives a basis for $\Lambda^k(V)$. Write $$\omega = \sum_I c_I a_I.$$ Consider $I^o$ fixed and choose any $i\notin I^o$. Wedge this expression with $a_i$. $$0=\omega\wedge a_i = \sum_I c_I a_{I}\wedge a_i = c_{I^o}a_{I^o}\wedge a_i + \dots.$$ These forms are linearly independent when $i\notin I$ (and the other terms are $0$). Thus, we get $c_{I^o}=0$.