I'm learning about the Gram-Schmidt process, and there's this question:
Are those list just the list $(e_1,...,e_m)$ produced by the Gram-Schmidt process, with the signs flipped? And if it's true, how do I know that there are not more?
By the way here's the formulation of the process in the textbook:
You are right: the $2^m$ possibilities are exactly those given by $(\pm e_1,\pm e_2,\dots, \pm e_m)$ where the $e_j$ are results of Gram-Schmidt.
Let us write $V_j=\operatorname{span}(v_1,\dots,v_j)$. Then $V_1$ is one-dimensional and there are the two unit vectors $\pm \frac{v_1}{\|v_1\|}$ in it. These are your two choices for $e_1$. No other unit vectors span $V_1$. The orthogonal complement $W_2$ of $V_1$ in $V_2$ is also one-dimensional. It is spanned by $v_2-\left<e_1,v_2\right>e_1$, and there are exactly two unit vectors in $W_2$, which are your choices for $e_2$. You can proceed inductively.
The statement is essentially the (almost) uniqueness of orthonormal bases for complete flags, and mentioned in their Wikipedia article.