 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-1/2a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

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

Solving the three nullspace systems gives the eigenvalues and eigenspaces

E_{- 1} = span {(1, 1, 1)^{T}}, E_{1} = span {(1, 1, 0)^{T}}, E_{2} = span {(0, 1, - 1)^{T}} .

For example, at

λ = 1

z = 0

and

x = y

; at

λ = 2

x = 0

and

y = - z

; at

λ = - 1

x = y = z

Putting these eigenvectors in columns proves their linear independence:

P = 111110 01 - 1, det P = 1, P^{- 1} A P = diag (- 1, 1, 2) .

Hence $A$ is a diagonalizable matrix over the reals.

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