condition for a bivector to be simple

Question

If $B$ is a bivector and $x$ is a vector, I have to prove that

Bivector $B$ is simple with $x$ (i.e. there's some vector $y$ such that $B=x\wedge y$) iff $B\wedge x = 0$

Now, for necessity part:

$B = x\wedge y \Longrightarrow B\wedge x=x\wedge y\wedge x=0$ (for definition of outer product)

And I'm stuck on the converse part.

Answer 1

If $B \wedge x = 0$, then $Bx^{-1} = B \cdot x^{-1}$ is some vector. $Bx^{-1} x = B$ on the one hand, but using a different associative grouping for the geometric product, you get $(B x^{-1}) \wedge x$ instead. Hence, $y = -Bx^{-1}$.