 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/3/b/i/solution

Use the convention of part (a): the inner product is linear in its first argument. If

T v = λ v

with

v \neq = 0

, the adjoint operator identity and

T^{*} = - T

give

λ ⟨ v, v ⟩ = ⟨ T v, v ⟩ = ⟨ v, T^{*} v ⟩ = - ⟨ v, T v ⟩ = - \overline{λ} ⟨ v, v ⟩ .

⟨ v, v ⟩ > 0

, so

λ = - \overline{λ}, Re λ = 0.

Thus every eigenvalue of a skew-adjoint operator is purely imaginary, including zero.

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