Let $V$ be a k-dimensional vector space and $m,n \in \mathbb{Z}$ are such that $m+n \le k$. We consider the wedge product: $\wedge: \Lambda^nV \times \Lambda^mV \to \Lambda^{m+n}V$ defined by $(v,w) \to \pi(v \otimes w)$ where $\pi$ is anti-symmetrisation projector. Is it true that if for fixed v we have $v\wedge w = 0$ for all $w \in \Lambda^mV$ then $v$ must be zero?
I'm trying to show this by taking $v_{i_1} \wedge ... \wedge v_{i_n}$ as a basis for $\Lambda^nV$ so that we can write $v = v^{i_1, ..., i_n}v_{i_1} \wedge ... \wedge v_{i_n}$. Now I want to be able to pick out particular elements from $\Lambda^mV$ to show that each $v^{i_1, ..., i_n}$ is zero. I'm struggling to do this.
Actually you dont have to pick particular elements of $\bigwedge^m V$, just pick any non-zero $v_{j_1} \wedge \dotsc \wedge v_{j_m}, ~ n+1 \leq j_1 < \dotsb j_m \leq k$ (This where you need $m+n \leq k$). Then the elements
$$v_{i_1} \wedge \dotsc \wedge v_{i_n} \wedge v_{j_1} \wedge \dotsc \wedge v_{j_m}, ~ 1 \leq i_1 < i_2 < \dotsc i_n \leq n$$
are linear independent in $\bigwedge^{n+m} V$, i.e. all coefficient vanish if the linear combination vanishes.