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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/14/a/iii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 14 / a / iii / Solution by

Codex  0 2026-10-18

For

λ = 5

, the eigenvalue equation reduces to a one-dimensional eigenspace spanned by

(1, 2, - 1)

. For

λ = - 3

, the rows of

A + 3 I

are multiples of

(1, 2, - 3)

, so its eigenspace is the plane

x + 2 y - 3 z = 0

. One possible set of three independent eigenvectors is

e_{1} = (1, 2, - 1) (λ = 5), e_{2} = (- 2, 1, 0), e_{3} = (3, 0, 1) (λ = - 3) .

Every eigenvector for

- 3

is a nonzero linear combination of the last two. Their two-dimensional eigenspace shows that the repeated root is not defective: the matrix is diagonalizable.

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