Exercise 1.8.6

Here $A$ is the transpose of the original matrix $A$ in example 1, if $A=U\mathrm{\Sigma }{V}^T$, then we have $A^T=V\mathrm{\Sigma }{U}^T$, so the singular values of $A$ are the same as in the example 1, i.e. $\mathrm{\Sigma }=\left[\begin{array}{cc}\hfill \sqrt{45}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \sqrt{5}\hfill \end{array}\right]$, whilethe left singular vectors are $\frac{1}{\sqrt{2}}\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ and the right singular vectors are $\frac{1}{\sqrt{10}}\left[\begin{array}{cc}\hfill 1\hfill & \hfill -3\hfill \\ \hfill 3\hfill & \hfill 1\hfill \end{array}\right]$