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 ${\begin{pmatrix}{\sqrt {3}}\\3\end{pmatrix}}$ is on the line whose angle we are looking for. Let us denote our line in the generic form $y=Ax+b$. Since the line must pass through the origin, thus $b=0$ and since ${\begin{pmatrix}{\sqrt {3}}\\3\end{pmatrix}}$ is on the line, $A$ satisfies $A{\sqrt {3}}=3$. Therefore, the line we are looking for is given by the equation $y={\sqrt {3}}x$.

We know that the slope of the line is the same as the tangent of the angle: $\tan \theta ={\sqrt {3\,}}$ therefore $\theta =\arctan({\sqrt {3}})=\pi /3$.

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 $-1$. The angle orthogonal to $\pi /3$ which lies in the given interval is $-\pi /6$.

Answer: $\color {blue}\theta ={\frac {\pi }{3}},-{\frac {\pi }{6}}$.