 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4/16h/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

Let

A

A v = λ v

with

v \neq = 0

. Choose a vertex

i

maximizing

∣ v_{i} ∣ > 0

. Then

∣ λ ∣∣ v_{i} ∣ = \sum_{j \sim i} v_{j} \leq de g (i) ∣ v_{i} ∣ \leq Δ∣ v_{i} ∣.

Dividing gives

∣ λ ∣ \leq Δ.

This argument works even with a complex eigenvector; for a finite undirected graph the matrix is real symmetric and all eigenvalues are real.

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