Exercise 1.8.12

$A^{T}A=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 10\hfill \\ \hfill 10\hfill & \hfill 20\hfill \end{array}\right]$, it has eigenvalues of ${\lambda }_1=0,{\lambda }_2=25$. The corresponding eigenvectors (which are also right singular vectors) are $v_1=\frac{1}{\sqrt{5}}\left[\begin{array}{c}\hfill 2\hfill \\ \hfill -1\hfill \end{array}\right]$and $v_2=\frac{1}{\sqrt{5}}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$.

So $A^{T}A=\mathit{Q\Lambda }{Q}^T=\frac{1}{5}\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 25\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 5\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

So we can compute the left singular vectors: $u_i=\frac{A{v}_i}{{\sigma }_i}$. Note we only need to compute where eigenvalue is larger than 0, i.e. ${\sigma }_2=5$. $u_2=\frac{1}{\sqrt{5}}\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]$Select $u_1$ to be orthogonal to $u_2$, we have $u_1=\frac{1}{\sqrt{5}}\left[\begin{array}{c}\hfill -1\hfill \\ \hfill 2\hfill \end{array}\right]$

So we have $U=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$ and $\mathrm{\Sigma }=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$, $V=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill -1\hfill \end{array}\right]$, then

$A=U\mathrm{\Sigma }{V}^T=\frac{1}{5}\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 5\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill -1\hfill \end{array}\right]$