Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 3 / b / ii

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- Solution ii

Solution

Take

v \in U^{⊥}

and

u \in U

. Because

U

is an invariant subspace,

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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