 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-1/7b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

det (M - t I) = (2 - t) [(1 - t) (3 - t) + 2] - 2 (2 - t) = (2 - t) (1 - t) (3 - t) .

Thus the eigenvalues are

1, 2, 3

. Solving the corresponding homogeneous linear systems gives

E_{1} = span {(- 1, 0, 1)^{T}}, E_{2} = span {(1, 1, 0)^{T}}, E_{3} = span {(1, 1, 1)^{T}} .

Every nonzero vector in the displayed eigenspace is an eigenvector. Direct multiplication verifies the three eigenvalue equations. The distinct eigenvalues imply these eigenvectors are linearly independent and form a basis of

R^{3}

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