 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017/3/ii/solution

Uv = λ v

with

v \neq = 0

, then

∥ v ∥^{2} = ∥ Uv ∥^{2} = ∣ λ ∣^{2} ∥ v ∥^{2} .

Dividing by the positive squared norm gives

∣ λ ∣ = 1.

Unlike the Hermitian case, this constrains eigenvalues to the unit circle, not necessarily the real axis.

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