Are all $k$-vectors in $\mathbb{R}^3$ simple? The second comment of the accepted answer here, claims that all $k$-vectors in $\mathbb{R}^3$ are simple, that is, they can be written as the wedge product of k vectors.
I tried to show this for $2$-vectors in $\mathbb{R}^3$, using properties of the wedge product to rewrite an arbitrary $2$-vector $a e_1 \wedge e_2 + b e_1 \wedge e_3 + c e_2 \wedge e_3 $, where $a,b,c$ are arbitrary constants, as something of the form $v_1 \wedge v_2$, where $v_1,v_2$ are arbitrary vectors $\in \mathbb{R}^3$.
If this is true, how do we decompose an arbitrary $2$-vector in $\mathbb{R}^3$ into the wedge product of three vectors?
 A: No, no, every $2$-vector can be written as the wedge product of two vectors. For example, assuming $c\ne 0$, you can write
$$ae_1\wedge e_2+ be_1\wedge e_3+ce_2\wedge e_3 = (be_1+ce_2)\wedge (-\frac ace_1+e_3).$$
A: Yes, and one can see this abstractly and geometrically using the usual cross product map,
$$\times : \Bbb R^3 \times \Bbb R^3 \to \Bbb R^3.$$
Since $\times$ is skew (that is, since $X \times Y = - Y \times X$ for all $X, Y \in \Bbb R^3$), by the Universal Property of Exterior Algebras it descends to a (linear) map
$$\ast : \Lambda^2 \Bbb R^3 \to \Bbb R^3$$
characterized by the commutativity of the diagram
$$
\begin{array}{ccc}
\Bbb R^3 \times \Bbb R^3 & & \\
\wedge\downarrow \,\,\,\,\, & \stackrel{\times}{\searrow} & \\
\Lambda^2 \Bbb R^3 & \stackrel{\ast}{\rightarrow} & \Bbb R^3
\end{array} .
$$
(I've deliberately used $\ast$ to denote this map, because it is a special case of the Hodge Star operator on an oriented, finite-dimensional inner product space.) Now, $\times$ is surjective, so for any $\zeta \in \Lambda^2 \Bbb R^3$ there are vectors $X, Y \in \Bbb R^3$ such that $X \times Y = \ast \zeta$. But the surjectivity of $\times$ implies that $\ast$ is surjective, and hence (because $\dim \Lambda^2 \Bbb R^3 = \dim \Bbb R^3$) that it is an isomorphism. Then, by construction, $\zeta = \ast^{-1} (X \times Y) = X \wedge Y$, that is, $\zeta$ is decomposable.
Alternatively, the claim follows from the general and occasionally useful fact that on any vector space $\Bbb V$, a bivector $\zeta \in \Lambda^2 \Bbb V$ is simple iff $\zeta \wedge \zeta \in \Lambda^4 \Bbb V$ is zero; of course, if $\dim \Bbb V = 3$, $\Lambda^4 \Bbb V$ is trivial, and so all bivectors are decomposable.
