Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 28/Solution 1

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 $\color {blue}\alpha =-9,\beta =-1.$