 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2010/algebra/2/c/i/solution

In fact the adjoint invariance follows from the orthogonal decomposition and the invariance under

T

, without using the normal operator property. For

x \in V_{i}

and

y \in V_{j}

with

j \neq = i

⟨ T^{*} x, y ⟩ = ⟨ x, T y ⟩ = 0,

since

T y \in V_{j}

. Hence

T^{*} x

is orthogonal to every other summand and therefore lies in

V_{i}

. Moreover, for

x, y \in V_{i}

, the adjoint identity restricted to that subspace shows

T_{i}^{*} = T^{*} ∣_{V_{i}} .

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