Which $n$-forms are pullbacks of top forms on $\Bbb R^n$ 
Let $V$ be a finite-dimensional vector space.  I write $F_n(V)$ for the $n$th exterior power of the dual vector space.  Which elements of $F_n(V)$ can be pulled back from a top form along a linear map $V \to R^n$?

When $n = 0$ or $1$, the answer is all of them.  It is also true for $n > $ dimension of $V$.  Is it true for all $n,V$?
 A: Equivalently, you are asking when an element $\omega\in F_n(V)$ can be written in the form $\alpha_1\wedge\dots\wedge \alpha_n$ for $\alpha_1,\dots,\alpha_n\in V^*$; call such an $\omega$ simple.  When $1<n<\dim V-1$, it turns out that not every $n$-form is simple.  For instance, it is clear that if $\omega$ is simple then $\omega\wedge\omega=0$.  It follows that if $e_1,e_2,e_3,e_4\in V^*$ are linearly independent, then $\omega=e_1\wedge e_2+e_3\wedge e_4$ cannot be simple (since $\omega\wedge\omega=2e_1\wedge e_2\wedge e_3\wedge e_4\neq 0$).  In fact, if $V$ is 4-dimensional, it can be shown that $\omega\in F_2(V)$ is simple iff $\omega\wedge\omega=0$.  In general, the simple forms can be characterized as those which satisfy a certain collection of quadratic equations called the "Plücker relations" which are a bit complicated to state.
Note that $\omega$ is simple iff any nonzero scalar multiple of $\omega$ is simple, so it makes sense to talk about the collection of simple forms as a subset of the projective space $\mathbb{P}(F_n(V))$ (if you're not familiar with that, it's just the set of nonzero elements of $F_n(V)$ where elements which differ by a scalar multiplication are identified).  This subset is known as the image of the Plücker embedding, and can canonically be identified with the set of $n$-dimensional subspaces of $V^*$ (given $\omega=\alpha_1\wedge\dots\wedge\alpha_n$, take the subspace spanned by $\{\alpha_1,\dots,\alpha_n\}$; conversely, given a subspace, let $\{\alpha_1,\dots,\alpha_n\}$ be a basis, and take $\omega=\alpha_1\wedge\dots\wedge\alpha_n$).
This gives a heuristic argument that there should be non-simple $n$-forms when $1<n<\dim V-1$.  Namely, the dimension of $F_n(V)$ is $\binom{\dim V}{n}$; let's figure out what the "dimension" of the set of simple forms should be (here I'm using "dimension" in an intuitive geometric sense, as the set of simple forms is not a linear subspace!).  A simple form is essentially the same as an $n$-dimensional subspace of $V$ together with a scalar factor, so their dimension should be $1+d$, where $d$ is the dimension of the set of $n$-dimensional subspaces of $V$.  To give an $n$-dimensional subspace of $V$, you have to give $n$ linearly independent vectors of $V$, which gives $n\cdot \dim V$ degrees of freedom.  But given a subspace, there is an $n^2$-dimensional set of possible bases for it (since given one basis, any invertible $n\times n$ matrix gives you another basis).  So we should expect that $d=n\cdot\dim V-n^2=n(\dim V-n)$.
When $1<n<\dim V-1$, you can compute that $1+d=1+n(\dim V-n)$ is strictly smaller than $\binom{\dim V}{n}$.  That is, the set of simple $n$-forms has smaller dimension than the set of all $n$-forms.  In particular, not all $n$-forms are simple.  (As I said, this is just a heuristic argument, since I have not rigorously defined what I mean by "dimension".  However, with some machinery from algebraic geometry, this argument can be made rigorous.)
A: Hint Any top form $\alpha \in \Bbb R^n$ is decomposable, that is, we can write it as a wedge product of $1$-forms:
$$\alpha = \beta^1 \wedge \cdots \wedge \beta^n.$$
Then, for any linear map $T: V \to \Bbb R^n$, we have
$$T^* \alpha = T^* (\lambda \beta^1 \wedge \cdots \wedge \beta^n) = \lambda T^* \beta^1 \wedge \cdots \wedge T^* \beta^n.$$ In particular, any form that is the pullback under a linear map of a top form is itself decomposable.
On the other hand, one can show constructively (i.e., by constructing a suitable top form on $\Bbb R^n$ and linear map $T: V \to \Bbb R^n$) that this condition is also sufficient.
Remark If $\dim V \geq 4$, then for any cobasis $(e^a)$ of $V$, the $2$-form $\gamma := e^1 \wedge e^2 + e^3 \wedge e^4$ satisfies $\gamma \wedge \gamma \neq 0$, and in particular $\gamma$ is not decomposable, and so not all $2$-forms are pullbacks of top forms. (In fact, the $GL(V)$-orbit of $\gamma$ is open, so in this sense nearly all $2$-forms in $V$ are pullbacks of top forms.)
