Suppose $V$ is a (degree $1$) finite dimensional vector space over a field $k$, and consider the exterior algebra $A=\Lambda^\bullet(V)$ on $V$. This means that $A$ is the vector space spanned by elements $v_1\wedge\dots\wedge v_n$ ($v_i\in V$) with associative multiplication satisfying $v\wedge v=0$ ($v\in V$).
Now, let $X$ be a codimension $2$ subspace of $A_2$. Is it true that $X\wedge V=A_3$?
I have proved that if $X$ is spanned by pure vectors $v\wedge w$, then the assertion is true, but I can't prove it in the general case.