 Solution

ID: past-exam-of-the-linear-algebra-course-of-the-university-of-oxford/2023/3/a/ii/solution

R

and

S

both satisfy the defining identity of an adjoint operator, then

⟨ u, (R - S) w ⟩ = ⟨ T u, w ⟩ - ⟨ T u, w ⟩ = 0

for every

u, w \in V

. Choosing

u = (R - S) w

gives

∥ (R - S) w ∥^{2} = 0

(R - S) w = 0

for every

w

, proving

R = S .

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