TITLE: Why is an alternating $2$-form decomposable if and only if its self-wedge vanishes?
QUESTION [6 upvotes]: Given a vector space $V$, and a $2$-tensor $w$ in the second exterior power $\Lambda^2 V$. Assume that $w \wedge w=0$. Why is $w$ decomposable?
Thanks for your help!

REPLY [3 votes]: A proof not requiring a basis and canonical form: Evaluate the interior product 
with any vector $V$ for which $i_Vw\ne0$: $0=i_V(w\wedge w)=(2i_Vw)\wedge w$. So 
$w=(i_Vw)\wedge u$, for some 1-form $u$.