Science:Math Exam Resources/Courses/MATH221/April 2013/Question 07 (a)/Solution 1

Figure for reflection A

• For any non-trivial reflection, the eigenvalues are always 1 and -1.

• For the eigenvalue 1, we can pick any vector on the line itself. Such a vector will not be moved by the linear transformation. So, for us, we pick $\left[{\begin{array}{c}4\\3\end{array}}\right]$ as an eigenvector to the eigenvalue 1. Note that (x,y) = (4,3) is a point on the line 3x=4y since 3(4)=4(3).

• The eigenvalue of -1 comes from picking a vector in the line that is perpendicular (i.e. orthogonal) to the line $4y=3x\$. Any such vector will simply change its sign, thus will correspond to the eigenvalue -1. We choose $\left[{\begin{array}{c}-3\\4\end{array}}\right]$ as an eigenvector to the eigenvalue -1.