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

MATH152 April 2017

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Question A 16
For which value or values of θ in the interval $(-\pi /2,\pi /2]$ does the matrix ${\text{Ref}}_{\theta }$ have ${\begin{pmatrix}{\sqrt {3}}\\3\end{pmatrix}}$ as an eigenvector? Recall: ${\text{Ref}}_{\theta }$ is reflection through a line that makes an angle $\theta$ with the $x$-axis.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The eigenvectors of this matrix will correspond to eigenvalues $\pm 1$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $\color {blue}\theta ={\frac {\pi }{3}},-{\frac {\pi }{6}}$.