 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/13/a/solution

det (A - λ I) = (2 - λ) (4 - λ) + 2 = (λ - 3)^{2} + 1.

Thus the eigenvalues and corresponding eigenvectors over

C

are

λ_{+} = 3 + i, v_{+} = (- 1 + i, 1)^{T}; λ_{-} = 3 - i, v_{-} = (- 1 - i, 1)^{T} .

All nonzero scalar multiples of these vectors are also eigenvectors. Distinct eigenvalues make them linearly independent. Put them in the columns of

B = (- 1 + i 1 - 1 - i 1), B^{- 1} = \frac{1}{2 i} (1 - 1 1 + i - 1 + i) .

A v_{\pm} = λ_{\pm} v_{\pm}

are exactly

A B = BC

C = B^{- 1} A B = (3 + i 0 0 3 - i) .

There is no diagonalization over

R

, since the matrix has no real eigenvalues.

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