Exercise 5.4.9

See Example 5.4.6.

1.

For the first method, we may calculate $T^3(z)=(3,-3,3,3)$ and represent it as a linear combination of the basis

$$T^3(z)=0z-3T(z)+3T^2(z).$$

So the characteristic polynomial is $-t^3+3t^2-3t$. For the second method, denote $\beta$ to be the ordered basis

$$\{z,T(z),T^2(z),T^3(z)\}.$$

And we may calculate the matrix representation

$${[T_W]}_{\beta }=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right)$$

and directly find the characteristic polynomial of it to get the same result.

2.

For the first method, we may calculate $T^3(z)=0$ and represent it as a linear combination of the basis

$$T^2(z)=0z+0T(z).$$

So the characteristic polynomial is $t^2$. For the second method, denote $\beta$ to be the ordered basis

$$\{z,T(z)\}.$$

And we may calculate the matrix representation

$${[T_W]}_{\beta }=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$$

and directly find the characteristic polynomial of it to get the same result.

3.

For the first method, we may calculate $T(z)=z$. So the characteristic polynomial is $-t+1$. For the second method, denote $\beta$ to be the ordered basis $\{z\}$. And we may calculate the matrix representation

$${[T_W]}_{\beta }=\left(\begin{array}{c}\hfill 1\hfill \end{array}\right)$$

and directly find the characteristic polynomial of it to get the same result.

4.

For the first method, we may calculate $T^2(z)=3T(z)$. So the characteristic polynomial is $t^2-3t$. For the second method, denote $\beta$ to be the ordered basis

$$\{z,T(z)\}.$$

And we may calculate the matrix representation

$${[T_W]}_{\beta }=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right)$$

and directly find the characteristic polynomial of it to get the same result.