Let $k \in \mathbb{N}$ and let $V$ be a finite-dimensional vector space. Consider the intersection map \begin{eqnarray*} \mathcal{I} : &L(\mathbb{R}^k, V) \times L(V, \mathbb{R^k}) &\longrightarrow \,\,\,\,\,ℝ \\ &(B, C) &\mapsto \det (C \circ B) \end{eqnarray*} Now let $\text{Span }L(ℝ^k, V)$ be the (formal) vector space of all (finite) linear combinations of elements of $L(ℝ^k, V)$, and let $\tilde{\mathcal{I}}$ be the linear extension of $\mathcal{I}$ to $\text{Span }L(ℝ^k, V)$. Now for $X, \tilde{X} ∈ \text{Span }L(ℝ^k, V)$, consider the equivalence relation ~ given by
$$ X \sim \tilde{X} \iff (\forall C \in L(ℝ^k, V))\left(\tilde{\mathcal{I}}(X, C) = \tilde{\mathcal{I}}(\tilde{X}, C)\right). $$ Now let $\bigwedge^k V := \text{Span }L(ℝ^k, V)/\sim$.
Suppose $\dim V =: l ∈ ℕ_{<k}$.
To demsonstrate: $\bigwedge^k V = 0$.
For this is want to show that all elements are equivalent, i.e. that for all $C ∈ L(ℝ^k, V)$, we have, for $X = \sum_{i=0}^n b_i B_i, \, \tilde{X} = \sum_{i=0}^n \tilde{b_i} \tilde{B_i}$:
$$\sum_{i=0}^n b_i\det(C \circ B_i) = \sum_{j=0}^m\tilde{b_i} \det(C \circ \tilde{B_i}).$$
Can anyone argue why this would be the case?
It's enough to prove that any $X\in\mathrm{Span} L(\Bbb R^k, V)$ is equivalent to $0$, which follows immediately from the fact that $\det(C\circ B) =0$ for all $B\in L(\Bbb R^k, V)$, since the rank of both $C$ and $B$, hence also of $C\circ B$, is at most $\dim V$ which is $<k$ by assumption.