Let $V$ be an $n$-dimensional vector space and $\{e_i\}_{i=1}^n \in V$ be a basis and $v_1,\cdots, v_k \in V$ be linearly independent vectors. Furthermore let $V^*$ be the dual space of $V$ with $\{e^i\}_{i=1}^n$ as the dual basis. Then if I take a vector $v \in V$ then its components wrt $\{e_i\}_{i=1}^n$ can be found by acting the dual basis $\{e^i\}_{i=1}^n$ on it, that is $$ v^i = e^i(v) $$ Now if I take the vector constructed by the wedge product $$ w = v_1\wedge\cdots\wedge v_k $$ which lives in a $\Lambda^k V$ and which represents the oriented $k$-dimensional volume in $V$ spanned by $v_1,\cdots, v_k \in V$. Then...
Question: does it make sense to say that the components of $w$ in $\Lambda^k V$ wrt $\{e_{i_1}\wedge\cdots\wedge e_{i_k}\}_{1 \le i_1<\cdots<i_k\le n}$ can be found by acting $\{e^{i_1}\wedge\cdots\wedge e^{i_k}\}_{1\le i_1<\cdots<i_k\le n}$, which is the (dual) basis of the dual space $(\Lambda^k V)^* = \Lambda^k V^*$, on $w$ ? that is \begin{align} w^{i_1\cdots i_k} &= (e^{i_1}\wedge\cdots\wedge e^{i_k})(v_1\wedge\cdots\wedge v_k) \\ &= (e^{i_1}\wedge\cdots\wedge e^{i_k})(v_1,\cdots,v_k) \\ &=\sum_{\sigma\in S_k}\text{sgn}(\sigma)\ e^{i_1}(v_{\sigma(1)})\cdots e^{i_k}(v_{\sigma(k)})\\ &= \det\left( e^{i}(v_j) \right), \qquad\qquad i \in \{i_1,\dots,i_k\}, j \in \{1,\dots,k \}. \end{align}
The last expression of the component is the oriented volume of the "shadow" of the $k$-dimensional volume in $V$ on the subspace spanned by $e_{i_1},\dots, e_{i_k} \in V$.