Sometimes its easier to work by inspection, noting any obvious properties a matrix may have, than to proceed with the full formalism of the characteristic polynomial, etc.; I think such is the case with the present matrix. Setting
$A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}, \tag{1}$
it is very easy to see that, with
$\mathbf j = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \tag{2}$
we have
$A \mathbf j = \mathbf j = 1 \mathbf j, \tag{3}$
which shows us that $1$ is an eigenvalue of $A$ with eigenvector $\mathbf j$. Next we observe that, setting
$\mathbf i = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \tag{4}$
and
$\mathbf k = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \tag{5}$
we have
$A \mathbf i= \cos \theta \mathbf i + \sin \theta \mathbf k \tag{6}$
and
$A\mathbf k = -\sin \theta \mathbf i + \cos \theta \mathbf k. \tag{7}$
(6) and (7) together show that the subspace $\mathcal S = \text{span}\{ \mathbf i, \mathbf k \}$ is invariant under the action of $A$; that is $A \mathcal S \subset \mathcal S$. Therefore any eigenvalue of the restriction of $A$ to $\mathcal S$, $A_{\mathcal S}$, will necessarily be an eigenvalue of $A$. It is easy to see that the matrix $A_{\mathcal S}$ is
$A_{\mathcal S} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \tag{8}$
and the characteristic polynomial $p_{A_{\mathcal S}}(\lambda)$ of $A_{\mathcal S}$ is given by the simple quadratic
$p_{A_{\mathcal S}}(\lambda) = \det(A_{\mathcal S} - \lambda I) = (\cos \theta - \lambda)^2 + \sin^2 \theta$
$= \lambda^2 - 2(\cos \theta) \lambda + \cos^2 \theta + \sin^2 \theta = \lambda^2 - 2(\cos \theta)\lambda + 1; \tag{9}$
it is easy to see from the quadratic formula that the roots of $p_{A_{\mathcal S}}(\lambda)$ are
$\lambda = \dfrac{1}{2}(2\cos \theta \pm \sqrt{4\cos^2 \theta - 4}) = \cos \theta \pm \sqrt{-\sin^2 \theta} = \cos \theta \pm i \sin \theta = e^{\pm i \theta}; \tag{10}$
we thus see these eigenvalues are in agreement with those given by Victor Liu in his answer.
In the above calculations, we have effectively replaced factoring the cubic characteristic equation of $A$ with the decomposition of $\Bbb R^3$ (or $\Bbb C^3$) into two separate invariant subspaces of $A$, $\text{span}\{\mathbf j \}$ and $\text{span} \{ \mathbf i, \mathbf k \}$.
Of course one can if one chooses exploit the cubic $p_A(\lambda)$ directly:
$p_A(\lambda) = \det (A - \lambda I) = \det(\begin{bmatrix} \cos \theta - \lambda & 0 & -\sin \theta \\ 0 & 1 - \lambda & 0 \\ \sin \theta & 0 & \cos \theta - \lambda \end{bmatrix})$
$ = (\cos \theta - \lambda)^2(1 - \lambda) + \sin^2 \theta (1 - \lambda) = ((\cos \theta - \lambda)^2 + \sin^2 \theta)(1 - \lambda)$
$= (\lambda^2 - 2(\cos \theta)\lambda + 1)(1 - \lambda), \tag{11}$
where evaluation of the determinant is made easy using Sarrus' rule. It is easy to see from (11) that the roots of $p_A(\lambda)$ are the same as the eigenvalues discovered above, basically by "inspection".
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
Best Answer
I assume that an improper rotation means an element of the orthogonal group with determinant $=-1$. In other words, an $n$-dimensional improper rotation is represented by a matrix $R$ such that $RR^T=I_n$ and $\det R=-1$.
All such matrices have $\lambda=-1$ as an eigenvalue. This is because $$\det(R+I)=\det(R+RR^T)=\det R \det (I+R^T)=-\det(I+R^T)=-\det(R+I),$$ which implies that $\det(R+I)=0$.
This has the following corollary
Proof. We saw above that $\lambda_1=-1$ is an eigenvalue of $R$. If $\lambda_2$ is the other eigenvalue, then $\lambda_1\lambda_2=\det R=-1$, so we can conclude that $\lambda_2=1$. If $\vec{u}$ is an eigenvector belonging to $\lambda_2$, and $\vec{v}\perp\vec{u}$ is another unit vector, then (because $R$ preserves lengths and angles) we can conclude that $R\vec{v}\perp R\vec{u}$. But here $R\vec{u}=\vec{u}$, and in 2D the only unit vectors $\perp\vec{u}$ are $\pm\vec{v}$. So we can conclude that $R\vec{v}=\pm\vec{v}$. The plus sign cannot occur, for then we would have $R=I_2$. Therefore $R\vec{v}=-\vec{v}$. This implies that $R$ is the orthogonal reflection w.r.t. the line spanned by $\vec{u}$.