Finding all cyclic subspaces of $\mathbf{C}^2$ and $\mathbf{R}^2$ w.r.t to rotational linear transformation

Question

I have been studying linear algebra and came across this question that I am not quite sure how to solve.

The question begins by letting $T:\mathbf{R}^2 \rightarrow \mathbf{R}^2 $ be a rotation counter-clockwise 90 degrees and let $S: \mathbf{C}^2 \rightarrow \mathbf{C}^2$ be defined by $S(x,y) = (-y,x)$. The question asks to find all cyclic subspaces of $\mathbf{R}^2$ with respect to $T$ and all cyclic subspaces of $\mathbf{C}^2$ with respect to S.

The hint says to consider the independence or dependence of the list $x, Tx$ for $x \in \mathbf{R}^2$ and the set $y,Sy$ for $y\in\mathbf{C}^2$ but I'm not sure how to solve this problem

Any help would be greatly appreciated!

Answer 1

For both types of linear transformations it is true that $$T^2=-I, T^4=I$$ where $I$ is the identity transformation. This means that we only need to check whether $x,Tx$ are linearly independent, since all the other vectors generated by powers of $T$ are linearly dependent on these two.

Now to find all possible cyclic subspaces, consider the vector subspaces of $\mathbb{R^2}$ and of $\mathbb{C^2}$ one by one ranked by their dimensionality.

The zero-dimensional subspace is a cyclic subspace for both vector spaces. When $x=(0~ 0)^T$ it is obvious that $Tx=x=0$ as well and the subspace closes.

• Consider a general vector in $\mathbb{R}^2$, $x=\begin{pmatrix}x_1\\x_2\end{pmatrix}$. Then $$Tx=\begin{pmatrix}-x_2\\x_1\end{pmatrix}$$

Unless $x_1,x_2=0$ there exists no $\lambda\in \mathbb{R}$ such that $Tx+\lambda x=0$. Thus the two vectors are always linearly independent of one another and therefore they form a basis of $\mathbb{R^2}$. We conclude that for $T:\mathbb{R}^2\to\mathbb{R}^2$, the two T-cyclic subspaces are the origin and $\mathbb{R^2}$ itself.

• Consider a general vector in $\mathbb{C}^2$, $x=\begin{pmatrix}z_1\\z_2\end{pmatrix}$. Then $$Tx=\begin{pmatrix}-z_2\\z_1\end{pmatrix}$$ Suppose there exists $\lambda\in \mathbb{C}$ such that $Tx+\lambda x=0$. This equation can be satisfied if $$z_1=\lambda z_2~,~ z_2=-\lambda z_1$$ substituting the second equation into the 1st one and discarding the case when $z_1=z_2=0$ this leads to $$\lambda^2+1=0$$ which indicates that $\lambda=\pm i~, ~z_1=\pm i z_2$ can satisfy the equation! We conclude that there are two additional separate one-dimensional T-cyclic subspaces in $\mathbb{C}^2$ spanned by the vectors $\begin{pmatrix}1\\i\end{pmatrix}, \begin{pmatrix}1\\-i\end{pmatrix}$, that $T$ leaves invariant. $\mathbb{C}^2$ is another obvious cyclic subspace generated by any vector other than the ones mentioned above.