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Let $V$ be a vector space over some field, and $W \le V$ a vector subspace. Let $1<k<\dim V$ be an integer.

Suppose $\omega \in \bigwedge^k W$ is decomposable as an element in $\bigwedge^k V$. Is it decomposable as an element in $\bigwedge^k W$?

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The answer is positive. Let $w_i$ be a basis for $W$. Write $\omega=\sum_{1 \le i_1 <\dots<i_k \le \dim W)} w_{i_1} \wedge \ldots \wedge w_{i_k}$. Let $v \in V \setminus W$. Since $w_i,v$ are linearly independent, we can complete them into a basis of $V$. This implies that $\omega \wedge v=\sum_{1 \le i_1 <\dots<i_k \le \dim W)} w_{i_1} \wedge \ldots \wedge w_{i_k} \wedge v \neq 0$.

By assumption, $\omega \in \bigwedge^k W$ is decomposable in $\bigwedge^k V$, i.e. $\omega=v_1 \wedge \dots \wedge v_k$, for some $v_i \in V$. Since $v_1 \wedge \dots \wedge v_k \wedge v=\omega \wedge v \neq 0$, we deduce that $v \notin \{v_1,\dots,v_k\}$. Since $v \in V \setminus W$ was arbitrary, we have proved that $V \setminus W=W^c \subseteq \{v_1,\dots,v_k\}^c$, so $ \{v_1,\dots,v_k\} \subseteq W$, so $\omega=v_1 \wedge \dots \wedge v_k$ is decomposable in $\bigwedge^k W$.

An alternative proof:

$\omega$ is decomposable in $\bigwedge^k V$, if and only if $V_{\omega}=\{ v \in V \, | \, v \wedge \omega=0\}$ is $k$-dimensioal.

Now, $v \in V \setminus W \Rightarrow v \wedge \omega \neq 0 \Rightarrow v \notin V_{\omega}$, i.e. $W^c \subseteq V_{\omega}^c$. Thus, so $V_{\omega} \subseteq W$, hence $$ W_{\omega}=V_{\omega} \cap W=V_{\omega}$$ is also $k$-dimensional, so $\omega$ is decomposable in $\bigwedge^k W$.

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