 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/11/a/solution

Solving

(A - 4 I) v = 0

gives

v_{2} = 0

and

v_{1} = v_{3}

. A convenient normalized eigenvector is

e_{1} = \frac{1}{2} (1, 0, 1), A e_{1} = 4 e_{1} .

The displayed nonzero vector proves that four is an eigenvalue. Every nonzero multiple represents the same eigendirection.

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