explicit description of eigenvector of a rotation One of the exercise in Artin's algebra gives an eigenvector of an element of $SO(3)$, in one possible case. Namely, it is asked to show that 

If $A=[a_{ij}]$ is a rotation in $SO(3)$, then the vector 
  $$v=\begin{bmatrix} (a_{23}+a_{32})^{-1}\\ (a_{13}+a_{31})^{-1} \\ (a_{12}+a_{21})^{-1}\end{bmatrix}$$
  is an eigenvector of $A$, whenever these entries of the vector are well-defined in $\mathbb{R}$. 

I couldn't get any way to proceed to prove this. What I tried was, consider $A$ as sum of symmetric and skew-symmetric- $\frac{1}{2}(A+A^t)+\frac{1}{2}(A-A^t)$ and show that $v$ is an eigenvector of both with eigenvalues $1$ and $0$ respectively; but, this was not working beyond long algebraic/symbolic computations. Any hint for this?
 A: I would like to propose another approach to the problem. As it is substantially other that given above it will be presented as a separate answer.
Let $R(u,\theta)$ be rotation matrix with the unit axis vector  $u=[x,y,z]^T$ and rotation angle $\theta$.
In this situation we have    Rodrigues formula for generating the exact form of this matrix.
$R(u,\theta)=I+\sin(\theta)K+(1-\cos(\theta))K^2$
where skew-symetric  matrix $ K=
\begin{bmatrix}   0 & -z & y \\ 
z & 0 &-x \\ -y & x & 0   \end{bmatrix} $
It can be checked additionally that $K^2$ is symmetric matrix and   
$K^2=uu^T-I =  \begin{bmatrix}    x^2 -1 &  xy & xz \\ xy &  y^2-1 &  yz \\  xz  & yz   & z^2-1   \end{bmatrix} $ 
Now we can easily calculate  the given vector $v$
$v=\begin{bmatrix}  { (r_{32}+r_{23})^{-1}  \\(r_{13}+r_{31})^{-1}    \\  (r_{21}+r_{12})^{-1} }\end{bmatrix}=  (1-\cos(\theta))^{-1}\begin{bmatrix}  { (2yz)^{-1}  \\(2xz)^{-1}    \\  (2xy)^{-1} }\end{bmatrix} \ \ \ (*)$.
It is visible that only symmetric part of rotation matrix influences on the form of vector $v$.
The generated vector $v$ is parallel to $u$ because it's sufficient to multiply $v$ by a scalar $c=2(1-\cos(\theta))xyz \ \ $ to obtain $ \ \ u=cv \ \ $ so really the vector $v$ is the eigenvector of $R$ like vector $u$.
The approach is valid for all $\theta$ except of course $\theta=0$ when corresponding matrix becomes identity matrix. Additionally $x,y,z$ must be different from $0$ what is the condition for real components of $v$.
As procedure above seems to be the easiest way of proving that $v$ is an eigenvector of $R$,  the Rodrigues form can be also used for proving that $0$ is eigenvalue for skew-symmetric part of $R$ ( i.e. $\sin(\theta)K$) and $1$ is eigenvalue for symmetric part (i.e. $I+ (1-\cos(\theta))K^2)$ with the given eigenvector $v$. It's sufficient to use form (*) for $v$ in appropriate calculations.
A: If $A\not= A^T$ (that is, if $A$ is not $I_3$ or a U-turn), then $x\in \ker(A-A^T)$ iff $Ax=x$.
It suffices to show that the first component of $(A-A^T)v$ is zero
$\dfrac{a_{1,2}-a_{2,1}}{a_{1,3}+a_{3,1}}+\dfrac{a_{1,3}-a_{3,1}}{a_{1,2}+a_{2,1}}=0$ iff ${a_{1,2}}^2+{a_{1,3}^2}-{a_{2,1}^2}-{a_{3,1}}^2=0$
iff $1-{a_{1,1}^2}-(1-{a_{1,1}^2})=0$.
EDIT. It remains to consider the case when $A\in T$ (the set of U-turns) and, for every $i<j, a_{i,j}\not=0$; $T\subset SO(3)$ is locally isomorphic to a plane included in $\mathbb{R}^3$. Then there is a sequence $(A_n)\subset SO(3)\setminus T$ s.t. $A_n$ tends to $A$. By above proof, for $n$ large enough, we know an explicit $v_n$ s.t. $A_nv_n=v_n$. Conclude by continuity.
