Exercise 3.4.8

Note, we have $A^{T}A=\left[\begin{array}{cc}\hfill 16\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$ and $A^{T}b=\left[\begin{array}{c}\hfill 4\hfill \\ \hfill 2\hfill \end{array}\right]$

Following Example 4, and use $u_0=(1,0)=z_0$ with $\rho =2$, we have

• Now compute $z_1$:

  $$\begin{array}{llll}\hfill z_1& =S_{\frac{\lambda }{\rho }}(x_1+u_0)& \hfill & \\ \hfill & =\mathit{sign}(x_1+u_0)\max <!--\; FUNCTION\; APPLICATION\; -->\left(|x_1+u_0|-\frac{\lambda }{\rho },0\right)& \hfill & \\ \hfill & =\mathit{sign}(\left[\begin{array}{c}\hfill \frac{8}{7}\hfill \\ \hfill \frac{5}{14}\hfill \end{array}\right])\max <!--\; FUNCTION\; APPLICATION\; -->\left(\left|\left[\begin{array}{c}\hfill \frac{8}{7}\hfill \\ \hfill \frac{5}{14}\hfill \end{array}\right]\right|-1,0\right)& \hfill & \\ \hfill & =\max <!--\; FUNCTION\; APPLICATION\; -->\left(\left[\begin{array}{c}\hfill \frac{8}{7}\hfill \\ \hfill \frac{5}{14}\hfill \end{array}\right]-1,0\right)& \hfill & \\ \hfill & =\max <!--\; FUNCTION\; APPLICATION\; -->\left(\left[\begin{array}{c}\hfill \frac{1}{7}\hfill \\ \hfill -\frac{9}{14}\hfill \end{array}\right],0\right)& \hfill & \\ \hfill & =\left[\begin{array}{c}\hfill \frac{1}{7}\hfill \\ \hfill 0\hfill \end{array}\right]& \hfill & \end{array}$$
• And we have $u_1=u_0+x_1-z_1=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill \frac{5}{14}\hfill \end{array}\right]$