Exercise 5.4.6

Follow the prove of Theorem 5.22. And we know that the dimension is the maximum number $k$ such that

is independent and the set is a basis of the subspace.

1.

Calculate that

$$z=(1,0,0,0),T(z)=(1,0,1,1),$$

$$T^2(z)=(1,-1,2,2),T^3(z)=(0,-3,3,3).$$

So we know the dimension is $3$ and the set

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

is a basis.

2.

Calculate that

$$z=x^3,T(z)=6x,T^2(z)=0.$$

So we know that the dimension is $2$ and the set

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

is a basis.

3.

Calculate that $T(z)=z$. So we know that the dimension is $1$ and $\{z\}$ is a basis.

4.

Calculate that

$$z=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),T(z)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 2\hfill \end{array}\right),$$

$$T^2(z)=\left(\begin{array}{cc}\hfill 3\hfill & \hfill 3\hfill \\ \hfill 6\hfill & \hfill 6\hfill \end{array}\right).$$

So we know that the dimension is $2$ and the set

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

is a basis.