Geometric means: Eigenvalue, eigenvector 
Find the eigenpairs for the matrix $M=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}$ in terms of $\theta$.

It seems that the eigenvalues are $e^{±i\theta}$ and the eigenvectors are $(1, -i)^T$ and $(1, i)^T$.
Then it comes to the second part of the question.

Explain by geometric means why a real eigenvector cannot be expected for $M$.

I have no idea what should I do in order to explain there is no real eigenvector for $M$ by geometric means. Please give me advice.
 A: Hint: Interpret the matrix as a geometric transformation of the plane - it should be a familiar one. Does this transformation fix any lines through the origin? (The answer will depend on the value of $\theta$).
A: The eigenvector $(1,i)$ can be seen as a left-polarized light propagating along z-axis, $$(1,i) \sim \left(e^{- i \omega t}, e^{- i \left(\omega t + \frac{\pi}{2}\right)} \right)$$
whose initial position, by taking the real parts at $t=0$, is 
$$\left( \cos\left(0\right), \cos\left(\frac{\pi}{2}\right) \right) = (1,0).$$
$M$ rotates the coordinates from $xyz$ to $x'y'z'$, with $z'=z$, by an angle $\theta$ in the $xy$-plane. The polarized light in the $x'y'z'$ system is
$$
\left(e^{- i \omega t - i \theta}, e^{- i \left(\omega t + \frac{\pi}{2}\right) - i\theta} \right)
= e^{-i\theta }\left(e^{- i \omega t}, e^{- i \left(\omega t + \frac{\pi}{2}\right)} \right)
$$
Thus, the eigenvalue $e^{-i\theta}$. The initial position in $x'y'z'$ system is 
$$
\left( \cos\left( \theta \right), \cos\left(\frac{\pi}{2}+\theta \right) \right) =
\left( \cos\left( \theta \right),-\sin\left(\theta \right)\right).
$$
The interpretation for $(1,-i)$ is similar, using a right-polarized light.
