 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/18/c/ii/solution

For the

C

det (C - μ I) = (- 4 - μ) (4 - μ) - 9 = μ^{2} - 25.

Thus the eigenvalues are

5

and

- 5

. For

μ = 5

(C - 5 I) v = 0

gives

v_{2} = - 3 v_{1}

; for

μ = - 5

, it gives

v_{1} = 3 v_{2}

. Normalizing the eigenvectors yields

μ = 5 : v_{+} = \frac{1}{10} (1 - 3); μ = - 5 : v_{-} = \frac{1}{10} (31) .

Both are unit vectors, their dot product is zero, and either overall sign is allowed for each eigenvector.

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