Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 28

MATH152 April 2015

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Question A 28
Let $A$ be a $2\times 2$ matrix which represents a reflection across a line in $\mathbb {R} ^{2}$. Suppose that ${\begin{bmatrix}2\\3\end{bmatrix}}$ is an eigenvector with eigenvalue $1$ and ${\begin{bmatrix}\alpha \\6\end{bmatrix}}$ is an eigenvector with eigenvalue $\beta \neq 1$. What are the values of $\alpha$ and $\beta$?

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
Observe that the points fixed by a reflection across a line are exactly the points on the line of reflection. What happens to vectors perpendicular to this line, and what does this tell you about the eigenvalues and eigenvectors of $A$?

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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. Recall that the fixed points (i.e., eigenvectors of eigenvalue 1) of a reflection are the points on the line of reflection. Since we are given that $A{\begin{bmatrix}2\\3\end{bmatrix}}={\begin{bmatrix}2\\3\end{bmatrix}},$ the vector ${\begin{bmatrix}2\\3\end{bmatrix}}$ must lie on the line of reflection. Any vector perpendicular to ${\begin{bmatrix}2\\3\end{bmatrix}}$ must therefore point in the opposite direction after reflection (i.e., must be an eigenvector of eigenvalue -1). For instance, $A{\begin{bmatrix}-3\\2\end{bmatrix}}=-{\begin{bmatrix}-3\\2\end{bmatrix}}={\begin{bmatrix}3\\-2\end{bmatrix}}.$ We have thus found two eigenvalues of $A$: -1 and 1. These are all of its eigenvalues, since a 2-by-2 matrix can have at most 2 distinct eigenvalues. Since the eigenvector we seek does not correspond to the eigenvalue 1, it must correspond to the eigenvalue $\beta =-1.$ Now, the eigenspace of each of these eigenvalues is one-dimensional (since each has dimension at least 1, and the sum of their dimensions is at most 2), so it follows that ${\begin{bmatrix}\alpha \\6\end{bmatrix}}=t{\begin{bmatrix}-3\\2\end{bmatrix}}$ for some $t\in \mathbb {R} ,$ whence $\alpha =-9.$ Hence $\color {blue}\alpha =-9,\beta =-1.$