Exercise 5.1.2

Compute ${[T]}_{\beta }$ as what we did in previous chapters. If that matrix is diagonal, then $\beta$ is a basis consisting of eigenvectors.

1.

No. ${[T]}_{\beta }=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right)$.

2.

Yes. ${[T]}_{\beta }=\left(\begin{array}{cc}\hfill -2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -3\hfill \end{array}\right)$.

3.

Yes. ${[T]}_{\beta }=\left(\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$.

4.

No. $\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill -4\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$.

5.

No. ${[T]}_{\beta }=\left(\begin{array}{cccc}\hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$.

6.

Yes. $\left(\begin{array}{cccc}\hfill -3\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$.