Show that a given matrix always has an eigenvector in $\mathbb{R}$ Can somebody give a hint? The given exercise is, for all $\theta$ in $\mathbb{R}$, show that the matrix always has an eigenvector in $\mathbb{R^2}$ $$
A = \begin{pmatrix}
\cos(\theta) & \sin(\theta) \\
\sin(\theta) & -\cos(\theta)
\end{pmatrix}
$$
The use of determinants isn't allowed, it is just allowed that I should use de definition of eigenvector and eigenvalue, and the fact that the eigenvectors are linearly independent if they are distinct (if needed). 
I don't fully understand the problem because I can't see how is possible that this matrix can give me a multiple scalar of some vector.

the matrix $\pmatrix{\cos \theta & \sin \theta\\ \sin \theta & -\cos \theta}$ represents  the reflection on a mirror along the line $y = \tan (\frac{\theta}{2})\  x.$ therefore $\pmatrix{\cos (\theta/2) \\\sin(\theta/2)}$ is an eigenvector corresponding to the eigenvalue $1$  and $\pmatrix{\sin (\theta/2) \\-\cos(\theta/2)}$ is an eigenvector corresponding to the eigenvalue $-1$.
