# linear algebra - eigenvalues/vectors & diagnalization

$$R(θ) = \begin{pmatrix} \cosθ & -\sinθ \\ \sinθ & \cosθ \end{pmatrix}$$

$0 < θ < π$

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.

• A skew-symmetric matrix has purely imaginary eigenvalues. Except in the case $\theta = n\pi$, this matrix is skew-symmetric. – M. Vinay Jun 9 '14 at 13:21
• Has no eigenvalues over what field? What's your definition of orthogonal matrix? – Git Gud Jun 9 '14 at 13:21
• And yes, it is an orthogonal matrix because its transpose is its inverse. You can easily check this by multiplying it by its transpose and verifying that it is the identity matrix. – M. Vinay Jun 9 '14 at 13:22
• Do you really mean $\cos(0)$ in the last entry? – Git Gud Jun 9 '14 at 13:23
• Notice: I have edited the post, sorry. And still don't understand about the geometry. – New_Math_ Jun 9 '14 at 13:25

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.