Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / P / 1 / a

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- Solution a

Solution

The Gershgorin circle theorem states that every eigenvalue of

A

lies in at least one disc

D_{i} = {z \in C : ∣ z - a_{ii} ∣ \leq \sum_{j \neq = i} ∣ a_{ij} ∣} .

For a nonzero complex eigenvector

v

, choose an index

i

with

∣ v_{i} ∣ = max_{j} ∣ v_{j} ∣ > 0

. The

i

th row of

A v = λ v

gives

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

Dividing by

∣ v_{i} ∣

proves membership in

D_{i}

. This is valid for real matrices even when the eigenvalues are complex.

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