$$R(θ) = \begin{pmatrix} \cosθ & -\sinθ \\ \sinθ & \cosθ \end{pmatrix}$$
$0 < θ < π$
Now, I understand that there are not any eigenvectors/values over $\mathbb R$ (but do has over the Complex field) for this Matrix, but how I show this by geometry?
And one more question - is this matrix an Orthogonal Matrix?
Thank you.
First of all, the matrix does have eigenvalues, but they are not real numbers. If you are comfortable with complex numbers you can use the usual methods to show that the eigenvalues are $e^{\pm i\theta}$.
The matrix $R(\theta)$ represents the rotation of a vector by an angle $\theta$ with $0<\theta<\pi$. If it has a real eigenvalue $\lambda$ with eigenvector $\bf v$, then by definition $R(\theta)\bf v=\lambda \bf v$. That is, rotating $\bf v$ through the angle $\theta$ gives the same result as multiplying $\bf v$ by a real scalar. If you draw a diagram, it is geometrically obvious that this is impossible.
The matrix is orthogonal because if you multiply it by its transpose you get the identity matrix.