 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/8/c/ii/solution

For the lowest eigenvalue,

- 1

, solving

(H + I) v = 0

gives

v_{2} = 0

and

v_{3} = - v_{1}

. A unit eigenvector is therefore

v = \frac{1}{2} (1, 0, - 1)^{T} .

Any nonzero scalar multiple is an equally valid eigenvector.

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