 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/1f/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

A

λ

for which

A v = λ v

for some nonzero eigenvector

v

. Its corresponding eigenspace is

E_{λ} = ker (A - λ I) .

Write

v = (a, b, c, d)^{T}

. The displayed matrix is the outer product

A = v v^{T}

, so

A x = v (v^{T} x) = v (v \cdot x) .

Its column space is contained in

span {v}

and is nonzero because

A v = ∥ v ∥^{2} v \neq = 0

. Hence

A

v

∥ v ∥^{2} = a^{2} + b^{2} + c^{2} + d^{2}

v^{⊥}

0

. Thus

E_{∥ v ∥^{2}} = span {v}, E_{0} = v^{⊥} .

Since

R^{4} = span {v} \oplus v^{⊥}

A

has an eigenbasis and is therefore a diagonalizable.

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