Exercise 6.B.10

Proof. Just apply the Gram-Schmidt Procedure once on $v_1,\dots ,v_m$ to get the orthonormal an $e_1,\dots ,e_m$. Note that, if we switch the order of this list without relabeling the vectors, we still have $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_j)=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(e_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,e_j)$ for all $j\in \{1,\dots ,m\}$, because the vectors themselves remain the same. There $2^m$ such permutations. □