Exercise 1.9.4

We have $A-A_1=\left[\begin{array}{cc}\hfill \frac{3}{2}\hfill & \hfill -\frac{3}{2}\hfill \\ \hfill -\frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill \end{array}\right]$, so $\left|\right|A-A_1|{\text{|}}_{\infty }=\mathit{max}(3,1)=3$.

Let rank-1 matrix be $A_2=\left[\begin{array}{cc}\hfill \frac{3}{2}\hfill & \hfill \frac{15}{11}\hfill \\ \hfill \frac{9}{2}\hfill & \hfill \frac{45}{11}\hfill \end{array}\right]$, we have $A-A_2=\left[\begin{array}{cc}\hfill \frac{3}{2}\hfill & \hfill -\frac{15}{11}\hfill \\ \hfill -\frac{1}{2}\hfill & \hfill \frac{10}{11}\hfill \end{array}\right]$, $\left|\right|A-A_2|{\text{|}}_{\infty }=\mathit{max}(\frac{3}{2}+\frac{15}{11},\frac{1}{2}+\frac{10}{11})=\frac{63}{22}$, which is closer to $A$ than $A_1$.