<p>Let A be a 3x3 matrix such that A<sup>2</sup> = I. If the determinant of A is -1 and the trace of A is 0, then the eigenvalues of A are:</p>
Step-by-Step Solution
Key Concept: Since A^2 = I, the eigenvalues lambda must satisfy lambda^2 = 1, so lambda = 1 or -1. Let the eigenvalues be lambda1, lambda2, lambda3. Given det(A) = lambda1*lambda2*lambda3 = -1 and trace(A) = lambda1 + lambda2 + lambda3 = 0. Solving these with lambda_i in {1, -1} gives 1, -1, -1.
<p>Given A<sup>2</sup> = I, the minimal polynomial divides x<sup>2</sup> - 1 = 0, so eigenvalues are 1 or -1. Let eigenvalues be λ<sub>1</sub>, λ<sub>2</sub>, λ<sub>3</sub>. We have λ<sub>1</sub>λ<sub>2</sub>λ<sub>3</sub> = det(A) = -1 and λ<sub>1</sub> + λ<sub>2</sub> + λ<sub>3</sub> = trace(A) = 0. The only combination of 1 and -1 satisfying these is 1, -1, -1.</p>
Correct Answer: 2