Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-2/16/b/i/solution

With , the characteristic polynomial factors into , up to the overall sign convention. The eigenvalues are and . Integer values require an odd integer satisfying . The positive integer factorization forces and
The resulting eigenvalues and corresponding nonzero eigenvectors are
Any nonzero scalar multiple is equally valid. The factorization also shows that no other positive integer works.

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