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

MATH152 April 2015

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[hide] 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 $\beta$?

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[show] 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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[show] Solution

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