Exercise 4.6.9

Pick a $A=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$.

$G=A^{T}A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$, solve for eigenvalues, we find its

${\lambda }_1={\lambda }_2=3,{\lambda }_3=0$.

The eigenvectors are not completely determined because $G$ has a repeated eigenvalue $\lambda =3$.

Solve for eigenvectors using eigenvalue 3, so we have the eigenvector $v_1=\frac{1}{\sqrt{6}}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill -2\hfill \end{array}\right]$, $v_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right]$, note we need to make $v_2$ orthogonal to $v_1$

Compute the left singular vectors as : $u_1=\frac{1}{\sqrt{2}}\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -1\hfill \\ \hfill -1\hfill \end{array}\right]$, $u_2=\frac{1}{\sqrt{6}}\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$

so we have $A=U\mathrm{\Sigma }{V}^T=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -\frac{2}{\sqrt{6}}\hfill \\ \hfill -\frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{6}}\hfill \\ \hfill -\frac{1}{\sqrt{2}}\hfill & \hfill -\frac{1}{\sqrt{6}}\hfill \end{array}\right]\left[\begin{array}{cc}\hfill \sqrt{3}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \sqrt{3}\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill \frac{1}{\sqrt{6}}\hfill & \hfill \frac{1}{\sqrt{6}}\hfill & \hfill -\frac{2}{\sqrt{6}}\hfill \\ \hfill \frac{1}{\sqrt{2}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill & \hfill 0\hfill \end{array}\right]$