Exercise 2.3.10

By looking at equation (4) on page 140, we see that $A=Y\mathit{BZ}=\left[\begin{array}{c}\hfill I_r\hfill \\ \hfill C_{m-r}{B}^{-1}\hfill \end{array}\right]B\left[\begin{array}{cc}\hfill I_r\hfill & \hfill B^{-1}{Z}_{n-r}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill B\hfill \\ \hfill C_{m-r}\hfill \end{array}\right]\left[\begin{array}{cc}\hfill I_r\hfill & \hfill B^{-1}{Z}_{n-r}\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill B\hfill & \hfill Z_{n-r}\hfill \\ \hfill C_{m-r}\hfill & \hfill C_{m-r}{B}^{-1}{Z}_{n-r}\hfill \end{array}\right]$

This helps us select $B,C,Z$ matrices from the original $A$. Notice here $A$ has a rank-3 while $B$ has a rank-2, so we are trying to approximate the matrix $A$ by $Y\mathit{BZ}$ with rank-2.

The submatrix $B_{\mathit{max}}=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]$.

We have $B^{-1}=\frac{1}{22}\left[\begin{array}{cc}\hfill 5\hfill & \hfill -1\hfill \\ \hfill -3\hfill & \hfill 5\hfill \end{array}\right]$ and $C_{m-r}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ Solve for $C_{m-r}{B}^{-1}$ we have $C_{m-r}{B}^{-1}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \end{array}\right]\frac{1}{22}\left[\begin{array}{cc}\hfill 5\hfill & \hfill -1\hfill \\ \hfill -3\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \frac{1}{11}\hfill & \hfill \frac{2}{11}\hfill \end{array}\right]$

We also have $Z_{n-r}=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]$, solve for $B^{-1}{Z}_{n-r}=\left[\begin{array}{cc}\hfill 5\hfill & \hfill -1\hfill \\ \hfill -3\hfill & \hfill 5\hfill \end{array}\right]\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \end{array}\right]=\frac{1}{22}\left[\begin{array}{c}\hfill 9\hfill \\ \hfill -1\hfill \end{array}\right]$

So $A\approx \left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \\ \hfill \frac{1}{11}\hfill & \hfill \frac{2}{11}\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 5\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill \frac{9}{22}\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -\frac{1}{22}\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 5\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 5\hfill \\ \hfill \frac{4}{11}\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$

Note the element at $(3,1)$ is $\frac{4}{11}$ instead of the 3 in the original matrix $A$.