b'}$. This ends the proof in the case where ${\bf a}$ and ${\bf b}$ are colinear.

Let us now consider the general case where ${\bf a}$ and ${\bf b}$ are not colinear. It is clear that ${\bf a'}$ and ${\bf b'}$ cannot be colinear either. We take two bases $({\bf a}, {\bf b}, {\bf f_3}, \cdots, {\bf f_n})$ and $({\bf a'}, {\bf b'}, {\bf f_3'}, \cdots, {\bf f_n'})$ of $\mathbbm{C}^n$ and use the Gram-Schmidt process to obtain two orthonormal bases $\mathcal{B}=({\bf e_1}, \cdots, {\bf e_n})$ and $\mathcal{B}'=({\bf e_1'}, \cdots, {\bf e_n'})$. Note that vectors ${\bf e_1}, {\bf e_2}, {\bf e_1'}$ and $e{\bf _2'}$ are given by: $$\begin{aligned}
{\bf e_1} = \frac{{\bf a}}{\sqrt{A}} &,& {\bf e_2} = \frac{{\bf b} - \langle {\bf e_1}|{\bf b}\rangle {\bf e_1}}{||{\bf b} - \langle {\bf e_1}|{\bf b}\rangle {\bf e_1} ||}\\
{\bf e_1'} = \frac{{\bf a'}}{\sqrt{A}} &,& {\bf e_2'} = \frac{{\bf b'} - \langle {\bf e_1'}|{\bf b'}\rangle {\bf e_1'}}{||{\bf b'} - \langle {\bf e_1'}|{\bf b}\rangle {\bf e_1'} ||}.\end{aligned}$$ Let us call $U$ the unitary operator mapping $\mathcal{B}$ to $\mathcal{B}'$. It is easy to see that $U$ maps ${\bf a}$ and ${\bf b}$ to ${\bf a'}$ and ${\bf b'}$, respectively. This concludes our proof.

Distance between the symmetrized 