Exercise 1.5.7

• ${\overline{x}}_1^T{x}_1=4$
• ${\overline{x}}_2^T{x}_2=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -i\hfill & \hfill -1\hfill & \hfill i\hfill \end{array}\right]\left[\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \\ \hfill -1\hfill \\ \hfill -i\hfill \end{array}\right]=4$
• ${\overline{x}}_3^T{x}_3=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]=4$
• ${\overline{x}}_4^T{x}_4=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill i\hfill & \hfill -1\hfill & \hfill i\hfill \end{array}\right]\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -i\hfill \\ \hfill -1\hfill \\ \hfill -i\hfill \end{array}\right]=4$
• ${\overline{x}}_1^T{x}_2=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \\ \hfill -1\hfill \\ \hfill -i\hfill \end{array}\right]=0$

It’s easy to check that for $i\ne j$, ${\overline{x}}_i^T{x}_j=0$, so ${\overline{Q}}^{T}Q=I$.