Exercise 6.7.6

Use Theorem 6.29. So we compute $A^{\text{†}}=V{\mathrm{\Sigma }}^{\text{†}}{U}^{\text{∗}}$.

1.

$$A^{\text{†}}=\frac{1}{6}\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right).$$

2.

$$A^{\text{†}}=\left(\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{1}{2}\hfill & \hfill -\frac{1}{2}\hfill \end{array}\right).$$

3.

$$A^{\text{†}}=\frac{1}{5}\left(\begin{array}{cccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \end{array}\right).$$

4.

$$A^{\text{†}}=A^{-1}=\left(\begin{array}{ccc}\hfill \frac{1}{3}\hfill & \hfill \frac{1}{3}\hfill & \hfill \frac{1}{3}\hfill \\ \hfill \frac{1}{3}\hfill & \hfill -\frac{2}{3}\hfill & \hfill \frac{1}{3}\hfill \\ \hfill \frac{1}{3}\hfill & \hfill \frac{1}{3}\hfill & \hfill -\frac{2}{3}\hfill \end{array}\right).$$

5.

$$A^{\text{†}}=\frac{1}{6}\left(\begin{array}{cc}\hfill 1-i\hfill & \hfill 1+i\hfill \\ \hfill 1\hfill & \hfill i\hfill \end{array}\right).$$

6.

$$A^{\text{†}}=\left(\begin{array}{ccc}\hfill \frac{1}{6}\hfill & \hfill \frac{1}{6}\hfill & \hfill \frac{1}{6}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 0\hfill & \hfill -\frac{1}{2}\hfill \\ \hfill \frac{1}{6}\hfill & \hfill -\frac{1}{3}\hfill & \hfill \frac{1}{6}\hfill \\ \hfill \frac{1}{6}\hfill & \hfill \frac{1}{6}\hfill & \hfill \frac{1}{6}\hfill \end{array}\right).$$