 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2009/p/p1/solution

λ

A = (a_{ij})

belongs to at least one closed disc

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

To prove it, take an eigenvector

v \neq = 0

and choose

i

with

∣ v_{i} ∣ = max_{j} ∣ v_{j} ∣ > 0

. The

i

th row of

A v = λ v

gives

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

∣ v_{i} ∣

proves the assertion. The proof works for real matrices even when their eigenvalues or eigenvectors are complex.

For

B + μ I

, the three discs have centres

2 + μ, 3 + μ, 4 + μ

2, 3, 4

. Every point in their union has real part at least

μ

. Thus for

μ > 0

no disc contains zero, and zero cannot be an eigenvalue. Hence $B + μ C$ is nonsingular.

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