Exercise 1.8.7

$A-{\sigma }_1{u}_1{v}_1^T=U\mathrm{\Sigma }{V}^T-\left[\begin{array}{cccc}\hfill u_1\hfill & \hfill u_2\hfill & \hfill \dots \hfill & \hfill u_m\hfill \end{array}\right]\left[\begin{array}{cccc}\hfill {\sigma }_1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{c}\hfill v_1^T\hfill \\ \hfill v_2^T\hfill \\ \hfill \dots \hfill \\ \hfill v_n^T\hfill \end{array}\right]=U\mathrm{\Sigma }{V}^T-U\left[\begin{array}{cccc}\hfill {\sigma }_1\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill & \hfill \dots \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \end{array}\right]V^T=U\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\sigma }_2\hfill & \hfill \dots \hfill & \hfill 0\hfill \\ \hfill \dots \hfill & \hfill \dots \hfill & \hfill {\sigma }_r\hfill & \hfill \dots \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill 0\hfill \end{array}\right]V^T$

So the norm of $\left|\right|A-{\sigma }_1{u}_1{v}_1^T\left|\right|=\sqrt{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=2}^r{\sigma }_i^2}$ This can also be seen by : $A-{\sigma }_1{u}_1{v}_1^T={\sigma }_2{u}_2{v}_2^T+\cdots +{\sigma }_r{u}_r{v}_r^T$.

The rank of the reduced matrix is: $r-1$.