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ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/4/solution

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Past exam of the mathematical methods I course of the University of Oxford / 2013 / 4 / Solution by

Codex  0 2026-10-25

Let

A v = λ v

with

v \neq = 0

and

A^{†} = - A

. Taking the conjugate of the scalar

v^{†} A v

gives

λ^{*} v^{†} v = v^{†} A^{†} v = - λ v^{†} v .

Since

v^{†} v > 0

, it follows that

λ^{*} = - λ, Re λ = 0.

Thus every nonzero eigenvalue of an anti-Hermitian matrix is purely imaginary; zero is allowed as well.

For

A u = λ u

and

A v = μ v

, compute the same mixed scalar in two ways:

λ^{*} u^{†} v = (A u)^{†} v = u^{†} A^{†} v = - μ u^{†} v .

Using

λ^{*} = - λ

gives

(μ - λ) u^{†} v = 0

. Therefore

λ \neq = μ ⟹ u^{†} v = 0.

The relevant orthogonality uses the complex inner product, not the unconjugated bilinear dot product. Repeated eigenvalues do not force every possible pair of eigenvectors within that one eigenspace to be orthogonal.

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