# Is “being decomposable” preserved under taking a subspace?

Let $$V$$ be a vector space over some field, and $$W \le V$$ a vector subspace. Let $$1 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 $$\bigwedge^k W$$?

The answer is positive. Let $$w_i$$ be a basis for $$W$$. Write $$\omega=\sum_{1 \le i_1 <\dots. Let $$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.
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 $$\bigwedge^k W$$.
$$\omega$$ 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 $$\bigwedge^k W$$.