Expressing wedge product as a multiple of a determinant 
Let $V \subset \mathbb{R}^n$ be a $k$-dimensional subspace. Let $(a_1, \dots , a_n)$ and $(b_1 , \dots , b_n)$ be bases of $\mathbb{R}^n$ such that $V = \text{span}\{a_1, \dots , a_k\} = \text{span}\{b_1 , \dots , b_k \}$, and let $(\alpha_1, \dots , \alpha_n)$ and $(\beta_1 , \dots , \beta_n )$ be dual bases respectively. Show that there exists $c \ne 0$ satisfying $$\alpha_1 \wedge \dots \wedge \alpha_k = c\beta_1 \wedge \dots \wedge \beta_k.$$

Since the spans are equal we can write $b_i=\sum_{j=1}^k v^i_ja_j$ and  we get that \begin{align*} \alpha_1 \wedge \dots \wedge \alpha_k(b_{i_1}, \dots, b_{i_k}) &= \det \begin{pmatrix} \alpha_1(b_{i_1}) & \dots  & \alpha_1(b_{i_k})\\ \vdots & \ddots & \vdots\\ \alpha_{k}(b_{i_1}) & \dots  & \alpha_{k}(b_{i_k}) \end{pmatrix}\\ &= \det \begin{pmatrix} v^{i_1}_1 & \dots  & v^{i_k}_1\\ \vdots & \ddots & \vdots\\ v^{i_1}_k & \dots  & v^{i_k}_k \end{pmatrix} \end{align*} for an arbitary input $b_{i_1}, \dots, b_{i_k}$ where $(i_1, \dots, i_k) \in \{1, \dots, k\}^n$. But we are missing the $\beta_1 \wedge \dots \wedge \beta_k$ term completely. I'm starting to think that this is not a true result. I can show this if we have equality between the spans in the dual, but with this formulation I don't think that this can be saved. Could I have a hint on where I'm miscalculating or whether this is not even doable in the first place?
 A: $
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$The actual question is as follows: for any $k$, if $\mathrm{span}\{a_i\}_{i=1}^k = \mathrm{span}\{b_i\}_{i=1}^k$ does that imply that $\mathrm{span}\{\alpha_i\}_{i=1}^k = \mathrm{span}\{\beta_i\}_{i=1}^k$?
The answer is no. Consider $n = 2$ and $k = 1$. Geometrically, it suffices to think about the lines spanned by $a_1, a_2, b_1, b_2$ individually. Then $\alpha_1$ is simply $a_2$ in the sense that $\ker(\alpha_1) = \mathrm{span}\{a_2\}$, and similarly $\alpha_2$ is $a_1$, $\beta_1$ is $b_2$, and $\beta_2$ is $b_1$.
But if $a_1$ and $b_1$ define the same line, this puts absolutely no constraints on $a_2$ and $b_2$, and hence neither $\alpha_1$ nor $\beta_1$. That is to say, any concrete example where $a_2, b_2$ point in different directions is a counterexample.
However, it will be true that $\alpha_2$ and $\beta_2$ have the same span! This fact generalizes. Let $\idx n = (1,2,\dotsc,n)$ be a multi-index, and let $\idx k \subseteq \idx n$. Then $a_{\idx k}$ has the same span as $b_{\idx k}$ if and only if $\alpha_{\idx n\setminus\idx k}$ has the same span as $\beta_{\idx n\setminus\idx k}$. This is a manifestation of what are called the Poincaré isomorphisms. This is best understood through a proper understanding the exterior algebra, in particular the exterior algebra of $V^*\oplus V$. A possible reference is Multilinear Algebra by Werner Greub.
