 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/6/b/solution

U

U^{†} U = U U^{†} = I

. Equivalently, it preserves the complex inner product and hence the norm. If

Uv = μv

with

v \neq = 0

, then

v^{†} v = (Uv)^{†} (Uv) = ∣ μ ∣^{2} v^{†} v .

Division by the positive squared norm gives

∣ μ ∣ = 1

. Thus every eigenvalue lies on the unit circle.

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