Exercise 1.8.8

For $A=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$, we have $A^TA =

$\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 9\hfill \end{array}\right]$

$ and $A{A}^T=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 9\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$.

We compute the eigen values as ${\lambda }_1=0,{\lambda }_2=4,{\lambda }_3=9$, so we have singular values as ${\sigma }_1=0,{\sigma }_2=2,{\sigma }_3=3$.

Let’s now find the eigenvectors of $A^{T}A$ with eigenvalues $4,9$: $x_1=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]$and $x_2=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]$

Choose $v_1=x_2=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]$ and ${\sigma }_1=\sqrt{{\lambda }_2}=3$, we have $u_1=\frac{A{v}_1}{{\sigma }_1}=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]$.

Choose $v_2=x_1=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]$ and ${\sigma }_1=\sqrt{{\lambda }_1}=2$, we have $u_2=\frac{A{v}_2}{{\sigma }_2}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]$.

So we have right singular matrix: $V=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \hfill \end{array}\right]$, and left singular matrix: $U=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right]$ note we picked the third singular vector to be perpendicular to other two singular vectors.

And we have $\mathrm{\Sigma }=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \hfill \end{array}\right]$

It can be shown that $A=U\mathrm{\Sigma }{V}^T$