Show there exists an invariant subspace $W\subseteq \mathbb R^n$.. Let $T$ be an orthogonal operator in $\mathbb R^n$. How can I show there exists in invariant subspace $W\subset \mathbb R^n$ such that $\textrm{dim}(W)=1$ or $\textrm{dim}(W)=2$? There was a hint for this problem: Consider $T+T^*$.. Any help will be welcome, thanks..
 A: It is not necessary to even use the assumption that $T$ is orthogonal. In fact, $T$ being orthogonal allows us to conclude a much stronger conclusion:
There exists a collection of mutually orthogonal $T$-subspaces $\{W_1,\ W_2,\ \cdots,\ W_k\}$ each of which has dimension $1$ or $2$ and together
$$\mathbb{R}^n = W_1 \oplus W_2 \oplus \cdots \oplus W_k$$
This essentially says that orthogonal operators are generalized (improper) rotations.
Anyways, back to the question at hand, we can prove the following.
Theorem: Let $A$ be a linear operator on $\mathbb{R}^n$. Then $A$ has an invariant subspace of dimension $1$ or $2$.
Proof: Suppose $A$ has a real eigenvalue. Then $A$ has a real eigenvector which trivially spans an invariant subspace of dimension $1$. In particular, we see that this holds for odd $n$.
Now suppose that $\lambda = \lambda_1 + i\lambda_2$ is a complex eigenvalue of $A$ with $\lambda_1,\ \lambda_2\in\mathbb{R}$. Let $\mathbf{v} = \mathbf{v}_1 + i\mathbf{v}_2$ be the corresponding eigenvector.
Then we have
$$\begin{align}\lambda\mathbf{v} = (\lambda_1 + i\lambda_2)(\mathbf{v}_1 + i\mathbf{v}_2) &= (\lambda_1\mathbf{v}_1 - \lambda_2\mathbf{v}_2) + i(\lambda_1\mathbf{v}_2 + \lambda_2\mathbf{v}_1)
\\&=A\mathbf{v}_1 + iA\mathbf{v}_2 = A\mathbf{v}\end{align}$$
Equating real and imaginary parts, we have
$$A\mathbf{v}_1 = \lambda_1\mathbf{v}_1 - \lambda_2\mathbf{v}_2$$
$$A\mathbf{v}_2 = \lambda_1\mathbf{v}_2 + \lambda_2\mathbf{v}_1$$
Let $W$ be the subspace spanned by $\{\mathbf{v}_1,\ \mathbf{v}_2\}$. Then it follows that 
$$A(c_1\mathbf{v}_1 + c_2\mathbf{v}_2) = (c_1\lambda_1 + c_2\lambda_2)\mathbf{v}_1 + (c_2\lambda_1 - c_1\lambda_2)\mathbf{v}_2$$
Therefore $W$ is $A$ invariant. $\square$
