Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 6 (a)/Solution 2

Following the second hint, Since matrix $A$ is a orthogonal rotation matrix, therefore it has determinant $\pm 1$. For this question, the determinant is $1$.

Out of three eigenvalues, one of them must be 1 from the property of an orthogonal rotation matrix. Now suppose the left two eigenvalues are $z_{1}$ and $z_{2}$. Since the sum of the eigenvalues is the trace of the matrix, and since their product is the determinant, we have $1+z_{1}+z_{2}=1$ and $z_{1}\cdot z_{2}=1$.

$z_{1}=i,z_{2}=-i$

answer: $\color {blue}1,i,-i$