Improper rotation matrix in $2D$ The following is the related problem:
Improper Rotations in Even Dimensions
I want the simpler explanation.   

An improper rotation is rotation, followed by reflection in the plane perpendicular to the axis of rotation.    

Consider an improper rotation in $2D$:    
$$R_I=\begin{bmatrix} \cos\theta & \sin\theta  \\ \sin\theta   & -\cos\theta  \end{bmatrix} $$  
Obviously, $\text {det}( R_I)=-1$.   
Consider the following simple case:   


*

*$R_I\begin{bmatrix} 1\\ 0  \end{bmatrix}=\begin{bmatrix} \cos\theta \\ \sin\theta   \end{bmatrix}$. However, this is the same as $R\begin{bmatrix} 1\\ 0  \end{bmatrix}$, where $R$ is the proper rotation matrix.
$$R=\begin{bmatrix} \cos\theta & -\sin\theta  \\ \sin\theta   & \cos\theta  \end{bmatrix}$$     

*$R_I\begin{bmatrix} 0\\ 1  \end{bmatrix}=\begin{bmatrix} \sin\theta \\ -\cos\theta    \end{bmatrix}$. and $R\begin{bmatrix} 0\\ 1  \end{bmatrix}=\begin{bmatrix} -\sin\theta  \\ \cos\theta   \end{bmatrix}$. Both are $180^o$ out of phase.     


I am confused about how to explain the definition of improper rotation by just two examples.   
 A: 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

A 2-dimensional improper rotation is just the orthogonal reflection w.r.t. a line through the origin.

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 $\vec{u}$.
