# Bases and wedge product

Let $\{v_1,\dots, v_r\}$ and $\{w_1,\dots, w_r\}$ be linearly independent sets of a vector space $V$. If $v_1 \wedge \dots \wedge v_r =c w_1 \wedge \dots \wedge w_r$ for some nonzero number $c$, then $\{v_1,\dots, v_r\}$ and $\{w_1,\dots, w_r\}$ are two bases of the same $r-$dimensional subspace.

To approach this problem, I first extend two sets to bases of $V$, say $\{v_1,\dots, v_d\}$ and $\{w_1,\dots, w_d\}$ where $d$ is the dimension of the vector space $V$. Then each $v_i$ can be expressed into a linear combination of $w_1,\dots,w_d$. But how to show that each $v_i$ can be expressed into a linear combination of $w_1,\dots,w_r$? Thanks in advance!

Before continuing, recall the fact that a set of vectors $\{u_1,\dotsc,u_k\}$ in $V$ is linearly independent if and only if $u_1 \wedge \dotsc \wedge u_k \neq 0$ in $\bigwedge^k V$.
Let $S_1 = \operatorname{Span}\{v_1,\dotsc,v_r\}$ and $S_2 = \operatorname{Span}\{w_1,\dotsc,w_r\}$, and suppose that $S_1 \neq S_2$, which means that $S_1 \setminus S_2 \neq \emptyset$ or $S_2 \setminus S_1 \neq \emptyset$. Without loss of generality, we can assume that $S_1 \setminus S_2 \neq \emptyset$, so that there exists some non-zero vector $w_{r+1} \in S_1 \setminus S_2$. Now, given your hypothesis on $\{v_1,\dotsc,v_r\}$ and $\{w_1,\dotsc,w_r\}$ and the fact I recalled above, what is the relationship between $v_1 \wedge \dotsc \wedge v_r \wedge w_{r+1}$ and $w_1 \wedge \dotsc \wedge w_r \wedge w_{r+1}$, and why does this give you a contradiction?