# Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 16/Solution 1

First we will find the eigenvector corresponding to the eigenvalue 1. Remember that the only fixed points (i.e. eigenvectors corresponding to the eigenvalue 1) of a reflection are those which are on the line used for the reflection. Therefore ${\displaystyle {\begin{pmatrix}{\sqrt {3}}\\3\end{pmatrix}}}$ is on the line whose angle we are looking for. Let us denote our line in the generic form ${\displaystyle y=Ax+b}$. Since the line must pass through the origin, thus ${\displaystyle b=0}$ and since ${\displaystyle {\begin{pmatrix}{\sqrt {3}}\\3\end{pmatrix}}}$ is on the line, ${\displaystyle A}$ satisfies ${\displaystyle A{\sqrt {3}}=3}$. Therefore, the line we are looking for is given by the equation ${\displaystyle y={\sqrt {3}}x}$.
We know that the slope of the line is the same as the tangent of the angle: ${\displaystyle \tan \theta ={\sqrt {3\,}}}$ therefore ${\displaystyle \theta =\arctan({\sqrt {3}})=\pi /3}$.
The other eigenvector will be orthogonal to the previous one: this vectors on this line will be inverted under the reflection and thus will correspond to the eigenvalue ${\displaystyle -1}$. The angle orthogonal to ${\displaystyle \pi /3}$ which lies in the given interval is ${\displaystyle -\pi /6}$.
Answer: ${\displaystyle \color {blue}\theta ={\frac {\pi }{3}},-{\frac {\pi }{6}}}$.