When does the vanishing wedge product of two forms require one form to be zero? Let $\alpha$ and $\beta$ be two complex $(1,1)$ forms defined as:
$\alpha = \alpha_{ij} dx^i \wedge d\bar x^j$
$\beta= \beta_{ij} dx^i \wedge d\bar x^j$
Let's say, I know the following:
1) $\alpha \wedge \beta = 0$
2) $\beta \neq 0$
I want to somehow show that the only way to achieve (1) is by forcing $\alpha = 0$. 
Are there general known conditions on the $\beta_{ij}$ for this to happen?
The only condition I could think of is if all the $\beta_{ij}$ are the same. However, this is a bit too restrictive. I'm also interested in the above problem when $\beta$ is a $(2,2)$ form.

\begin{eqnarray}
\alpha\wedge\beta
&=&\sum_{i,j,k,l}\alpha_{ij}\beta_{kl}dx^i\wedge d\bar{x}^j\wedge dx^k\wedge d\bar{x}^l\cr
&=&(\sum_{i<k,j<l}+\sum_{i<k,l<j}+\sum_{k<i,j<l}+\sum_{k<i,l<j})\alpha_{ij}\beta_{kl}dx^i\wedge d\bar{x}^j\wedge dx^k\wedge d\bar{x}^l\cr
&=&\sum_{i<k,j<l}(-\alpha_{ij}\beta_{kl}+\alpha_{kj}\beta_{il}+\alpha_{il}\beta_{kj}-
\alpha_{kl}\beta_{ij})dx^i\wedge dx^k\wedge d\bar{x}^j\wedge d\bar{x}^l.
\end{eqnarray}
Thus
$$
\alpha\wedge\beta=0 \iff \alpha_{ij}\beta_{kl}+
\alpha_{kl}\beta_{ij}=\alpha_{kj}\beta_{il}+\alpha_{il}\beta_{kj}
\quad \forall \ i<k,j<l.
$$
