An exterior $k$-tensor $\Theta \in \Lambda^kV^*$ is ${\bf decomposable}$ if there exist $\theta^1, \dots, \theta^k$ such that $\Theta = \theta^1\wedge\cdots\wedge\theta^k. $
Assume $\dim V = 4$.
Question: Prove that any $\Theta \in \Lambda^3V^*$ is decomposable.
Suppose the basis of $V^*$ is $l_1,l_2,l_3,l_4$.
I know that for each $\Theta \in \Lambda^kV^*$
I need to find the specific $\theta^{i}=a_{i1}l_1+a_{i2}l_2+a_{i3}l_3+a_{i4}l_4$, $i=1,2,3$. such that $\Theta = \theta^1\wedge\theta^2\wedge\theta^3$.
But I do not know how to find the coefficient $a_{ij}$.
Could you please give me a hint？ Thanks a lot!
Hint: More generally, every $\Theta \in [V^*]^{\wedge (n-1)}$ is decomposable when $\dim V = n$. Fixing a basis $(e_i)_{i=1}^n$ of $V$ and its dual basis $(e^i)_{i=1}^n$, let $J\colon V\to V^*$ be the non-canonical isomorphism given by $J(e_i)=e^i$. Writing $$\Theta =\sum_{i=1}^n (-1)^{i-1}a_i e^1\wedge\cdots\wedge \widehat{e^i}\wedge\cdots\wedge e^n$$with $a_1,\ldots,a_n\in \Bbb R$ and hat denoting omission, consider the linear functional $\varphi = \sum_{i=1}^n a_ie^i$. It's not hard to see that $$\Theta \wedge J(v) = \varphi(v)\, e^1\wedge\cdots \wedge e^n$$ for every $v\in V$. If $\varphi =0$, then $\Theta=0$ and we're done. Else, $\ker\varphi$ is a codimension-one subspace of $V$. Pick a basis $(v_1,\ldots, v_{n-1})$ of $\ker \varphi$. How are $\Theta$ and $J(v_1)\wedge\cdots\wedge J(v_{n-1})$. related?
Alternatively, try induction on $n$. The base case is trivial. And if $\Theta\wedge v = 0$ with $v\in \ker \varphi$, then $\Theta = v\wedge \Theta'$ with $\Theta'\in [\ker\varphi]^{\wedge (n-2)}$.