What are the Complex (Non-Real) Eigenvectors of $3\times 3$ Rotation Matrices? A $3\times 3$ rotation matrix $R$ that rotates $\mathbb{R}^3$ around the 
unit vector $v\in\mathbb{R}^3$ by angle $\theta$ (as defined by 
Rodrigues' 
rotation formula) satisfies the eigendecomposition 
$$
R=W\Sigma W^\mathrm{*} \enspace,
$$
where
$$
W=\left(\begin{matrix}v \;|\; x \;|\; y \end{matrix}\right) 
$$
is a unitary matrix of eigenvectors, and
$$
\Sigma=\mathrm{diag}\left(1,e^{i\theta},e^{-i\theta}\right)
$$
is the matrix of the corresponding eigenvalues.
What expressions define the non-real eigenvectors $x$ and $y$?
 A: Theorem
$x$ and $y$ are non-real eigenvectors of $R$ if and only if 
\begin{align}
x &= a+ib\\
y &= \pm(a-ib)\\
b &= a\times v \enspace,
\end{align}
where $a\in\mathbb{R}^3$ is an arbitrary vector satisfying
$$
  ||a||^2=\frac{1}{2} \;,\; v^\mathrm{T}a=0 \enspace,   
$$
which implies that $W$ has 1 d.o.f.
Proof of Necessity
We will first of all show that if $w=c+id$ and $u$ are non-real eigenvectors
of $R$ with eigenvalues $e^{i\theta}$ and $e^{-i\theta}$ respectively, for
some $c$ and $d$ in $\mathbb{R}^3$, then 
$$
v^\mathrm{T}c=0 \;,\; \; d = c\times v \;,\; ||c||^2=\frac{1}{2} 
  \;,\quad \mbox{and}\quad u=\pm\overline{w}=\pm(c-id)
$$
We will only prove that $\overline{w}$ is a possible value of eigenvector $u$. Clearly if this is the case then $u=\pm\overline{w}$.
Proof that $v^\mathrm{T}c=0$
Since $W$ is unitary, $w$ must satisfy
$$ v^\mathrm{T}w \;=\; v^\mathrm{T}c+iv^\mathrm{T}d \;=\; 0 \enspace.$$
As $v$ is real, this condition is only satisfied when $v$ is orthogonal to
both $c$ and $d$.
Proof that $d = c\times v$
By the eigenvector equation,
\begin{align}
Rw &= e^{i\theta}w \\
  &= (\cos\theta+i\sin\theta)w \\
  &= -d\sin\theta+c\cos\theta+i(c\sin\theta+d\cos\theta) \enspace.
\end{align}
Furthermore, by Rodrigues' rotation formula,
\begin{align}
Rw &= w\cos\theta + (v\times w)\sin\theta+v(v^\mathrm{T}w)(1-\cos\theta) \\
  &= w\cos\theta + (v\times w)\sin\theta \quad,\quad\mbox{as $v^\mathrm{T}w=0$}\\
  &= (v\times c)\sin\theta+c\cos\theta+
    i\left((v\times d)\sin\theta+d\cos\theta\right)
\end{align}
The result follows by comparing similar terms in the real and imaginary parts of
these two expressions for $Rw$.
Proof that $||c||^2=\frac{1}{2}$
The fact that the eigenvectors have unit norms implies that
\begin{align}
1 &= w^*w \\
  &= (c^\mathrm{T}-id^\mathrm{T})(c+id) \\
  &= c^\mathrm{T}c+d^\mathrm{T}d \\
  &= c^\mathrm{T}c+(c\times v)^\mathrm{T}(c\times v) \\
  &= c^\mathrm{T}c+(c^\mathrm{T}c)v^\mathrm{T}v-(c^\mathrm{T}v)(v^\mathrm{T}c)
    \quad,\quad \mbox{by cross-product properties}\\
  &= 2||c||^2 \\
\Rightarrow ||c||^2 &= \frac{1}{2} \enspace.
\end{align}
Proof that $\overline{w}$ is an Eigenvector
As $W$ is unitary, $\overline{w}$ must satisfy 
$$ v^\mathrm{T}\overline{w} \;=\; \overline{w}^*w \;=\; 0 \quad\mbox{and}\quad ||\overline{w}||^2=1 \enspace. $$
The first 3 lines of the proof that $||c||^2=\frac{1}{2}$ show that
$||\overline{w}||^2=1$, so it suffices to show that $\overline{w}$ has the
two orthogonality properties and that $e^{-i\theta}$ is $\overline{w}$'s
eigenvalue.
$v^\mathrm{T}\overline{w}=0$ follows from $v$'s orthogonality to $c$ and
$d$, which we established above. To prove that $\overline{w}^*w=0$: 
\begin{align}
\overline{w}^*w &= w^\mathrm{T}w \\
  &= (c+id)^\mathrm{T}(c+id) \\
  &= c^\mathrm{T}c+2ic^\mathrm{T}d-d^\mathrm{T}d \\
  &= ||c||^2+0-||c\times v||^2 \\
  &= ||c||^2-||c||^2 \quad,\quad\mbox{as $v^\mathrm{T}c=0$ and $||v||=1$}\\
  &= 0 \enspace. \\
\end{align}
To show that $e^{-i\theta}$ is $\overline{w}$'s eigenvalue, note that our
proof that $d = c\times v$ shows that
\begin{align}
Rc &= -d\sin\theta+c\cos\theta \\
Rd &= c\sin\theta+d\cos\theta \enspace.
\end{align}
Substituting these results into $R\overline{w}$ and rearranging gives
\begin{align}
R\overline{w} &= Rc-iRd \\
  &= (c-id)\cos\theta-(ic+d)\sin\theta \\
  &= (c-id)\cos\theta-i(c-id)\sin\theta \\
  &= (\cos\theta-i\sin\theta)(c-id) \\
  &= e^{-i\theta}\overline{w} \enspace.
\end{align}
Proof of Sufficiency
We will now prove that $x=a+ib$ and $y=a-ib$ are a valid pair of non-real
eigenvectors of $R$. Again, it is obvious that if $y$ is an eigenvector, then $-y$ is also an eigenvector corresponding to the same eigenvalue.
Proof that $W^*W=I$
For $W$ to be unitary, $x$ and $y$ must satisfy
$$
v^\mathrm{T}x \;=\; v^\mathrm{T}y \;=\; y^*x \;=\; 0
\enspace,
$$
and
$$
x^*x=y^*y=1\enspace.
$$
It is obviously true that 
$v^\mathrm{T}x \;=\; v^\mathrm{T}y \;=\; 0$, since 
$v^\mathrm{T}a=0$ by definition and $v^\mathrm{T}(a\times v)=0$.
The third orthogonality condition follows from our proof above that
$\overline{w}$ is an eigenvector. Also, as $y=\overline{x}$, the unit norm conditions follow
from the first 3 lines of our proof above that $||c||^2=\frac{1}{2}$ —
just substitute $x$ for $w$.
Proof that $x$ and $y$ are Eigenvectors
Next we must prove that $x$ satisfies the eigenvector equation
$$
Rx \;=\; e^{i\theta}x \enspace.
$$
Combining Rodrigues' rotation formula with the fact that $v^\mathrm{T}x=0$ 
gives
$$ Rx = x\cos\theta + (v\times x)\sin\theta \enspace. $$
As
\begin{align}
v\times x &= v\times(a+i(a\times v)) \\
  &= v\times a + iv\times(a\times v) \\
  &= v\times a + ia \quad,\quad\mbox{as $v^\mathrm{T}a=0$ and $||v||=1$}\\
  &= i(i(a\times v) + a) \\
  &= ix \enspace,
\end{align}
it follows by substitution that
\begin{align}
Rx &= x\cos\theta + ix\sin\theta \\
  &= (\cos\theta + i\sin\theta)x \\
  &= e^{i\theta}x \enspace.
\end{align}
To show that $y$ is also an eigenvector with eigenvalue $e^{-i\theta}$, we
can once again use Rodrigues' rotation formula and the fact that
$v^\mathrm{T}y=0$ to obtain
\begin{align}
Ry &= y\cos\theta + (v\times y)\sin\theta \\
  &= y\cos\theta-iy\sin\theta \enspace,
\end{align}
where the result
$$ v\times y = -iy $$
follows from a slight modification of the proof that $v\times x = ix$. Thus
\begin{align}
Ry &= (\cos\theta-i\sin\theta)y \\
  &= e^{-i\theta}y \enspace.
\end{align}
QED.
