Exercise 7.4.2

Find the factorization of the characteristic polynomial. Find the basis of $K_{\varphi }$ for each some monic irreducible polynomial factor consisting of $T$-cyclic bases through the proof of Theorem 7.22. Write down the basics with some appropriate order as the columns of $Q$. Then compute $C=Q^{-1}\mathit{AQ}$ or find $C$ by the dot diagram. And I want to emphasize that I compute these answers in Exercises 7.4.2 and 7.4.3 by HAND!

1.

It is a Jordan canonical form. So

$$Q=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 9\hfill \end{array}\right)$$

and

$$C=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 27\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill -27\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 9\hfill \end{array}\right).$$

2.

It has been already the rational canonical form since the characteristic polynomial $t^2+t+1$ is irreducible in $ℝ$. So $C=A$ and $Q=I$.

3.

It is diagonalizable in $ℂ$. So

$$Q=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt{3}i+1}{2}\hfill & \hfill \frac{1-\sqrt{3}i}{2}\hfill \end{array}\right)$$

and

$$C=\left(\begin{array}{cc}\hfill -\frac{\sqrt{3}i+1}{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{\sqrt{3}i-1}{2}\hfill \end{array}\right).$$

4.

Try the generating vector $(1,0,0,0)$. So

$$Q=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill -7\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -4\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -4\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -4\hfill & \hfill -8\hfill \end{array}\right)$$

and

$$C=\left(\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right).$$

5.

Use $(0,-1,0,1)$ and $(3,1,1,0)$ as generating vectors. So

$$Q=\left(\begin{array}{cccc}\hfill 0\hfill & \hfill -3\hfill & \hfill 3\hfill & \hfill 8\hfill \\ \hfill -1\hfill & \hfill -2\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill -3\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -4\hfill & \hfill 0\hfill & \hfill 7\hfill \end{array}\right)$$

and

$$C=\left(\begin{array}{cccc}\hfill 0\hfill & \hfill -2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right).$$