 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2019/6/solution

Let

A v = λ v

with

v \neq = 0

v^{†} A v = λ v^{†} v .

Taking the complex conjugate and using

A^{†} = - A

yields

\overline{λ} v^{†} v = v^{†} A^{†} v = - λ v^{†} v

. Since

v^{†} v > 0

\overline{λ} = - λ, Re λ = 0.

Thus every eigenvalue of an anti-Hermitian matrix lies on the imaginary axis, including the possible zero eigenvalue; each nonzero eigenvalue is purely imaginary.

Now take

A u = λ u

and

A v = μv

. On one hand

u^{†} A v = μ u^{†} v

. On the other hand

u^{†} A v = - (A u)^{†} v = - \overline{λ} u^{†} v = λ u^{†} v .

Consequently

(μ - λ) u^{†} v = 0

. Distinct eigenvalues therefore imply

u^{†} v = 0,

which is the required orthogonality of the eigenvectors.

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