 Solution

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

Take

v \in U^{⊥}

and

u \in U

. Because

U

T u \in U

. The adjoint operator relation yields

⟨ T v, u ⟩ = ⟨ v, T^{*} u ⟩ = - ⟨ v, T u ⟩ = 0.

This holds for every

u \in U

, so

T v \in U^{⊥}

. Therefore the orthogonal complement is also an invariant subspace.

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