 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/40c/e/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

r = A v - λ v, ∥ r ∥_{2} = ϵ, ∥ v ∥_{2} = 1.

E = - r v^{T} .

Then

(A + E) v = A v - r (v^{T} v) = λ v,

(λ, v)

is an exact eigenpair of the perturbed matrix

A + E

. Moreover,

∥ E ∥_{2} = ∥ r ∥_{2} ∥ v ∥_{2} = ϵ,

Follow the simple eigenvalue branch from

λ (0) = λ

for

A (t) = A + tE

∣ λ - λ ∣ = ∣ λ (1) - λ (0) ∣ \approx ∣ λ^{'} (0) ∣ \leq ∥ E ∥_{2} s (λ) .

Since the residual is at the machine precision scale, this proves

∣ λ - λ ∣ ≲ ϵs (λ) .

 New to topics? Read the docs here!