Let $V$ be a vector space over the field $F.$ Let $\{v_1,\ldots,v_4\}$ be linearly dependent set. Then how do I show that $w=v_1 \wedge v_2 + v_3 \wedge v_4$ is decomposable ?
Note that to show $w$ is decomposable we have to show that $w=x \wedge y$ for some $x,y \in V.$ What I tried: Atleast one of the $v_i$ can be written as linear combination of the others, so wlog let $v_1=c_2v_3 + c_3 v_3 + c_4 v_4$ then $w= v_3\wedge (c_3v_2+v_4)+v_4\wedge c_4 v_2.$ I think I need some other trick. Please help. Thanks.
By your work, we have $$ w = c_3 v_2 \wedge v_3 + c_2 v_4 \wedge v_2 + 1 \ v_3 \wedge v_4 $$ Now, it suffices to find two vectors $u_1,u_2 \in \Bbb F^3$ whose "cross-product" is $$ u_1 \times u_2 = (c_3,c_2,1) $$ for instance, we can take $u_1 = (1,0,-c_3)$ and $u_2 = (0,1,-c_2).$ With that in mind, try computing $$ (v_4 - c_3 v_2) \wedge (v_3 - c_2 v_2) $$