Exercise 5.2.3

For these question, we may choose arbitrary matrix representation, usually use the standard basis, and do the same as what we did in the previous exercises. So here we’ll have ${[T]}_{\beta }=D$ and the set of column vectors of $Q$ is the ordered basis $\beta$.

1.

It’s not diagonalizable since $\dim <!--\; FUNCTION\; APPLICATION\; -->(E_0)$ is $1$ but not $4$.

2.

It’s diagonalizable with $D=\left(\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$ and $Q=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$.

3.

It’s not diagonalizable since its characteristic polynomial does not split.

4.

It’s diagonalizable with $D=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ and $Q=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$.

5.

It’s diagonalizable with $D=\left(\begin{array}{cc}\hfill 1-i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill i+1\hfill \end{array}\right)$ and $Q=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 1\hfill \end{array}\right)$.

6.

It’s diagonalizable with $D=\left(\begin{array}{cccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$ and $Q=\left(\begin{array}{cccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$.