 Solution

ID: past-exam-of-the-numerical-analysis-course-of-the-university-of-oxford/2014/3/a/solution

A = (a_{ij}) \in R^{n \times n}

belongs to at least one of the closed Gershgorin discs

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

The eigenvalue may be complex even though the matrix is real. To prove the theorem, choose a nonzero complex eigenvector

v

with

A v = λ v

and an index

i

at which

∣ v_{i} ∣

is maximal. Then

∣ v_{i} ∣ > 0

, and the

i

th row gives

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

∣ λ - a_{ii} ∣ \leq \sum_{j \neq = i} ∣ a_{ij} ∣ \frac{∣ v _{j} ∣}{∣ v _{i} ∣} \leq \sum_{j \neq = i} ∣ a_{ij} ∣ = r_{i} .

Thus

λ \in D_{i}

, proving the assertion without any symmetry assumption.

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