How do I prove the following: If $V$ is $n$-dimensional, and $\alpha \in \Lambda ^{k} V$, if $\alpha \land \beta =0$ for all $\beta \in \Lambda ^{n-k}$ then $\alpha =0$.
For $k=1$, then we can form a basis of $V$ by $\alpha, v_{2}, ... , v_{n}$ and take $\beta = v_{2} \land ... \land v_{n}$ and then we get $\alpha \land \beta =\det\left(\alpha, v_{2}, ... , v_{n}\right)\neq 0$. But this approach doesn't help for $k>1$.
Any thoughts?
You can use basically the same idea. Fix $e_1,..,e_n$ a basis of $V$, then $(e_I)_{|I|=k}$ is a basis of $\wedge^k V$, where $e_{\{ i_1<..<i_k\}}=e_{i_1} \wedge ... \wedge e_{i_k}$.
So there exists coefficients $(\lambda_I)_{|I|=k}$ such that $$\alpha=\sum_{I\subset \{1,...,n\},|I|=k} \lambda_I e_I.$$ Now, choose some $I$, applying the hypothesis for $\beta=e_{I^c}$, where $I^c$ is the complement of $I$ in $\{1,...,n\}$, we have $$0=\alpha \wedge e_{I^c}= \pm \lambda_I,$$ the sign being given by the signature of an appropriate permutation. In any case, $\lambda_I=0$ for all $I$, so $\alpha=0$.