Exercise 6.5.5

For these problems, try to diagonalize the matrix which is not diagonalized yes. And check whether it can be diagonalized on an orthonormal basis.

1.

No. They have different eigenvalues.

2.

No. Their determinant is different.

3.

No. They have different eigenvalues.

4.

Yes. We have

$${\left(\begin{array}{ccc}\hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{2}}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)}^{\text{∗}}\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill \frac{1}{\sqrt{2}}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{\sqrt{2}}\hfill & \hfill -\frac{1}{\sqrt{2}}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -i\hfill \end{array}\right).$$

5.

No. One is symmetric but the other is not.