Proposition

If $c:[0, 1] \rightarrow (\mathbb{R}^n)^n$ is continuous and each $(c^1(t), \cdots, c^n(t))$ is a basis for $\mathbb{R}^n$, show that $[c^1(0), \cdots, c^n(0)] = [c^1(1), \cdots, c^n(1)]$.

Solution

Since $\{ e_1, \cdots, e_n \}$ is a basis of $\mathbb{R}^n$, there must exist $a_{11}, \cdots, a_{1n} \in \mathbb{R}$ such that $c_1(0) = a_{11}e_1 + \cdots + a_{1n}e_n$. Similarly, for each $i = 2, \cdots, n$, $c_i(0) = a_{i1}e_1 + \cdots + a_{in}e_n$ for some $a_{ij}$. We are given that $(c^1(0), \cdots, c^n(0))$ is a basis, so we can conclude that $[e_1, \cdots, e_n] = [c_1(0), \cdots, c_n(0)]$ if and only if $\det(A) > 0$ where $A = (a_{ij})$. (P.82[Spivak]) Moreover,

so $A = (\det \circ c)(0)$. Therefore, $[c_1(0), \cdots, c_n(0)]$ is the usual orientation if and only if $(\det \circ c)(0) > 0$.

Using the exact same argument, we conclude that $[c_1(1), \cdots, c_n(1)]$ is the usual orientation if and only if $(\det \circ c)(1) > 0$.

If $[c_1(0), \cdots, c_n(0)] \ne [c_1(1), \cdots, c_n(1)]$, then $\det \circ c$ achieves a positive value at one end of $[0, 1]$ and a negative value at the other end of $[0, 1]$. By the intermediate value theorem, $(\det \circ c)(t_0) = 0$ for some $t_0 \in (0, 1)$. However, this implies that $(c^1(t_0), \cdots, c^n(t_0))$ is linearly dependent, and thus is not a basis. This is a contradiction, so $[c_1(0), \cdots, c_n(0)] = [c_1(1), \cdots, c_n(1)]$.