find all invariant subspaces of $T_A$:$ \mathbb{R^3} \to \mathbb{R^3}$ 
Let $T_A$ :$ \mathbb{R^3} \to \mathbb{R^3}$ be a linear transformation,
$$A=\left(
    \begin{matrix}
    0 & -1 & 0 \\
    1 & 0 & 0 \\
    0 & 0 & 2 \\
    \end{matrix} \right)
$$
find all invariant subspaces of $T_A$.

*0 dimension:* {0}
*3 dimension:* $\mathbb{R^3}$ itself.
*1 dimension:* The characteristic polynomial is $(x-2)(x^2 +1)$ therefore we have only 1 eigenvalue $\lambda= 2$ so the invariant subspace of dimension 1 must be the eigenspace of $\lambda= 2$ which is V=span{$(0,0,1)$}.
*2 dimension:* Here is where I need you help. How can I know how much invariant subspaces of  dimension 2 there is? and how can I find them?
 A: From the block structure of the matrix you can see that $U:=\langle (0,0,1)^T\rangle$ is a one-dimensional invariant subspace, and that $W:=\langle (1,0,0)^T, (0,1,0)^T\rangle$ is a two-dimensional invariant subspace.
Can there be another one-dimensional invariant subspace, $\langle u \rangle$, say? This would mean that $Au\in\langle u\rangle$, so that $u$ would be an eigenvector. But you know that the only eigenvalue is $2$ and a simple calculation (solve the equations $Au=2u$) gives that $u$ must be a multiple of $(0,0,1)^T$, and so $\langle u\rangle=U$ after all.
Can there be another two-dimensional invariant subspace, $W_1$ say, with $W\not=W$? If so then $W\cap W_1$ is an invariant subspace. ( If you have doubts, note that $x\in W\cap W_1$ implies $x\in W$ and so $Ax\in W$, and similarly that $Ax\in W_1$.) As $W\not= W_1$ we have that $W+W_1>W$ and so $\dim (W+W_1)=3$. Hence $\dim W\cap W_1=\dim W +\dim W_1 -\dim (W+W_1)=1$. That is, $W\cap W_1$ is a one-dimensional invariant subspace, and so is $U$. But clearly $U$ is not a subspace of $W$, so we have a contradiction. Hence the only two-dimensional invariant subspace is $W$.
