Exercise 2.2.9

$q_1=\frac{a}{\left|a\right|}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$,then $A_2=b-(b^T{q}_1)q_1=\left[\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \end{array}\right]-2\sqrt{2}\frac{1}{\sqrt{2}}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill -2\hfill \end{array}\right]$, and $q_2=\frac{A_2}{\left|A_2\right|}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$. In the end we have