Let $A$ be an element of $SO_{3}$,show that if it is defined,the vector $$((a_{23}+a_{32})^{-1},(a_{13}+a_{31})^{-1},(a_{12}+a_{21})^{-1})^T$$ is an eigenvector with eigenvalue $1$
I found this Closed-form for eigenvectors of rotation matrix
I kown this: ifIf $A \in SO_{3}(R)$, then $A$ has an eigenvalue $\lambda=1$, see page 22 the proof:https://www.math.ucdavis.edu/~mathclub/cheat_sheets/notes.150B.pdf
But I can't prove my problem ,Thank you
To elaborate on user1551’s comment : since $A$ is an element of $SO_3$, it is a rotation around an axis. Let $\overrightarrow{k}=(k_1,k_2,k_3)$ be a directing vector for the axis and call $\theta$ the angle.
Then, Rodrigues’ formula (proved on Wikipedia) tells us that for any vector $\overrightarrow{x}$ ,
$$ A\overrightarrow{x}=(\cos(\theta))\overrightarrow{x}+ (\sin(\theta))(\overrightarrow{k} \wedge \overrightarrow{x})+ (1-\cos(\theta))(\overrightarrow{k}.\overrightarrow{x}) \overrightarrow{k} \tag{1} $$
And hence
$$ A=\left( \begin{array}{ccc} \cos(\theta) & 0 & 0 \\ 0 & \cos(\theta) & 0 \\ 0 & 0 & \cos(\theta) \\ \end{array} \right) + \left( \begin{array}{ccc} & -k_3\sin(\theta) & k_2\sin(\theta) \\ k_3\sin(\theta) & 0 & -k_1\sin(\theta) \\ -k_2\sin(\theta) & k_1\sin(\theta) & 0 \\ \end{array} \right) + \left( \begin{array}{ccc} k_1^2(1-\cos(\theta)) & k_1k_2(1-\cos(\theta)) & k_1k_3(1-\cos(\theta)) \\ k_1k_2(1-\cos(\theta)) & k_2^2(1-\cos(\theta)) & k_2k_3(1-\cos(\theta)) \\ k_1k_3(1-\cos(\theta)) & k_2k_3(1-\cos(\theta)) & k_3^2(1-\cos(\theta)) \\ \end{array} \right) \tag{2} $$
We deduce that for $i\neq j$,
$$ a_{ij}+a_{ji}=2k_ik_j(1-\cos(\theta)), \text{so} \ \frac{1}{a_{ij}+a_{ji}}=\lambda k_{6-(i+j)} \tag{3} $$
where $\lambda$ is the constant $\frac{k_1k_2k_3}{2(1-\cos(\theta))}$. The result is now clear.
