Below is a problem from Spivak's Calculus on Manifolds:
- If $v\in \mathbf R^2$, what is $v\times$?
- If $v_1\dots,v_{n-1}\in \mathbf R^n$ are linearly independent, show that $[v_1,\dots,v_{n-1},v_1\times\dots\times v_{n-1}]$ is the usual orientation of $\mathbf R^n$.
See this question for Spivak's definition of cross product.
For 1, I don't even understand what this question asks.
For 2, I think this amounts to showing that the matrix whose columns are $v_1,\dots,v_{n-1},v_1\times\dots\times v_{n-1}$ has positive determinant. But I don't know how to proceed using Spivak's confusing definition.
Prof @Ted suggestion is correct. To show that the vector $z:= v_1\times\cdots\times v_{n-1} \neq 0$, do the following. Since $v_1,\dots,v_{n-1}$ is linearly independent, there exists a nonzero vector $v_n \in \mathbb{R}^n$ orthogonal to $v_1,\dots,v_{n-1}$. So $v_1,\dots,v_n$ is a basis for $\mathbb{R}^n$. Then $z = \sum_{i=1}^n z^i v_i$ is nonzero iff we can show that at least one of the components is nonzero. By definition of $\varphi : \mathbb{R}^n \to \mathbb{R}$, that is $\varphi(w) = \langle w,z \rangle = \det (v_1,\dots,v_{n-1},w)$, we have $$ \det (v_1,\dots,v_n) = \langle v_n ,z \rangle = \langle v_n,z^iv_i \rangle = \sum_{i=1}^{n-1} z^i \langle v_n,v_i \rangle + z^n \langle v_n,v_n \rangle = 0 + z^n |v_n|^2. $$ Since $v_1,\dots,v_n$ are linearly independent ($\det (v_1,\dots,v_n) \neq 0$) and $v_n \neq 0$, then $z^n \neq 0$.