 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/40c/a/solution

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

λ

of a complex matrix

A = (a_{ij})

lies in at least one disk

∣ λ - a_{ii} ∣ \leq j \neq = i \sum ∣ a_{ij} ∣ .

To prove it, choose a nonzero eigenvector

x

for

λ

and an index

i

such that

∣ x_{i} ∣ = max_{j} ∣ x_{j} ∣ > 0

. The

i

th component of

A x = λ x

gives

(λ - a_{ii}) x_{i} = \sum_{j \neq = i} a_{ij} x_{j} .

Taking absolute values and using

∣ x_{j} ∣ \leq ∣ x_{i} ∣

yields

∣ λ - a_{ii} ∣ ∣ x_{i} ∣ \leq \sum_{j \neq = i} ∣ a_{ij} ∣ ∣ x_{j} ∣ \leq ∣ x_{i} ∣ \sum_{j \neq = i} ∣ a_{ij} ∣.

∣ x_{i} ∣

proves the theorem.

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