explicit description of eigenvector of a rotation

Question

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?

Answer 1

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.

Answer 2

Let me use letter $R$ instead of $A$ for the considered matrix.

Generally for a rotation matrix $R(v,\theta)$ you can calculate the axis unit vector from the formula: $v= {\dfrac {1}{2sin(\theta)}}\begin{bmatrix} r_{32}-r_{23} \\ r_{13}-r_{31} \\ r_{21} -r_{12} \end{bmatrix}$ where $r_{ij}$ are appropriate entries of $R$ matrix.

Now the problem is equivalent to the checking whether cross product
$c=\begin{bmatrix} r_{32}-r_{23} \\ r_{13}-r_{31} \\ r_{21} -r_{12} \end{bmatrix} \times \begin{bmatrix} (r_{32}+r_{23})^{-1} \\ (r_{13}+r_{31})^{-1} \\ (r_{21} +r_{12})^{-1}) \end{bmatrix}$ is really a zero vector what is equivalent to the condition for the vectors to be parallel.

Calculations can be checked with WolphramAlpha.
Use the command

[r_32-r_23,r_13-r_31,r_21-r_12] x [1/(r_32+r_23),1/(r_31+r_13),1/(r_21+r_12)]

$c=[\dfrac{r_{12}^2}{(r_{12} + r_{21}) (r_{13} + r_{31})} + \dfrac {r_{13}^2}{ (r_{12} + r_{21}) (r_{13} + r_{31})} - \dfrac {r_{21}^2}{ (r_{12} + r_{21}) (r_{13} + r_{31})} - \dfrac {r_{31}^2}{ (r_{12} + r_{21}) (r_{13} + r_{31}) } \ , - \dfrac {r_{12}^2}{ (r_{12} + r_{21}) (r_{23} + r_{32}) } + \dfrac{ r_{21}^2 }{ (r_{12} + r_{21}) (r_{23} + r_{32}) } + \dfrac {r_{23}^2 }{ (r_{12} + r_{21}) (r_{23} + r_{32} ) } - \dfrac{ r_{32}^2 }{ (r_{12} + r_{21}) (r_{23} + r_{32}) } \ ,- \dfrac{r_{13}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32}) } - \dfrac{ r_{23}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32}) } + \dfrac {r_{31}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32}) } + \dfrac {r_{32}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32})) }]^T$

what really will give the searched zero vector if we additionally take into consideration that norms of column and row vectors for rotation matrix are equal to $1$.

(for example:
$ \ r_{11}^2+r_{12}^2+r_{13}^2=1$, $ \ r_{11}^2+r_{21}^2+r_{31}^2=1$ from this $ \ r_{12}^2+r_{13}^2-(r_{21}^2+r_{31}^2)=0$, etc..)

Answer 3

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.