Q:

Invariant subspace of an orthogonal operator. 

I was trying to understand a proof in Prasolov's book Problems and Theorem's in linear algebra.  To prove a result on orthogonal matrices  Prokasov uses a geometric argument which is not clear to me.  So I attempted to convert his geometric argument into an algebraic one. 
This result is proved prior to proving spectral decomposition for normal operators, so I would prefer proofs that avoid spectral decomposition. 
This is what I think Prasalov proves via his geometric arguments, and I need a hint to see why it is true (if it is true):
Let $A$ be a real orthogonal matrix of size $n > 1$ which does not have $1$ or $-1$ as an eigenvalue. 
Let $x_0 \in \mathbb{R}^n$ with $\|x_0\|=1$ be such that
$$
x_0^TAx_0 =  \sup_{ \{x : \|x\| = 1\}} x^TAx. 
$$
Let $W$ denote the subspace spanned by $x_0$ and $Ax_0$, then $AW \subset W.$ (Note : It is sufficient to prove $A^2x_0 \in W.$)
Added later. 
Original argument paraphrased by me( from page 64, initial pages torn off from my copy so no edition information available):

Since $A$ does not have $-1,1$ as eigenvalues $Ax$ is never parallel to $x$. The function $\phi(x) = \angle(x,Ax)$ is a continuous function on $\{x : \|x\|=1\}$ and attains its minimum at, say, $x_0$. Let $\phi_0 = \phi(x_0)$, and let $e$ denote the bisector of the angle between $x_0$ and $Ax_0$, then
$$
\phi_0 \leq \angle(e,Ae) \leq \angle(e,Ax_0) + \angle(Ax_0,Ae) \leq \phi_0/2 + \phi_0/2 = \phi_0, \tag{1}
$$
and therefore (according to Prasolov) $Ae$ belongs to $\text{span}(x_0,e)$ and this plane is invariant to $A$ as $Ax_0,Ae \in \text{span}(x_0,e).$ 

In (1) Prasolov is using $\angle(Ax,Ay) = \angle(x,y)$ which follows from the orthgonality of $A$.  
To come up with my version of the question, I used the fact that angle between $x$ and $Ax$ is smallest on $\{x:\|x\|=1\}$ when $x^T Ax$ takes the  largest value (since $A$ is orthogonal if $\|x\|=1$ then $\|Ax\| = \|x\| = 1$).  
Also $x_0 + Ax_0$ bisects the angle between $x_0$ and $Ax_0$ and $\text{span}(x_0,x_0+Ax_0) = \text{span}(x_0,Ax_0).$

A:

$\sup_{||x||=1}x^TAx=1/2\sup_{||x||=1}x^T(A+A^T)x$. Thus $x_0$ is an eigenvector of $A+A^T$, with norm $1$, associated to the eigenvalue $\lambda=\sup(spectrum(A+A^T))$. Thus $Ax_0=\lambda x_0-A^Tx_0$ and $A^2 x_0=\lambda Ax_0-AA^Tx_0=\lambda Ax_0-x_0$ and we are done.