Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 2 / b

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- Solution b

Solution

α, β

v, w, u

, use the adjoint identity and conjugate linearity in the second argument:

⟨ u, T^{*} (αv + βw)⟩ = ⟨ T u, αv + βw ⟩ = \overline{α} ⟨ T u, v ⟩ + \overline{β} ⟨ T u, w ⟩ = \overline{α} ⟨ u, T^{*} v ⟩ + \overline{β} ⟨ u, T^{*} w ⟩ = ⟨ u, α T^{*} v + β T^{*} w ⟩ .

The difference of the two second-argument vectors is orthogonal to every

u

. Choosing

u

to be that difference shows its squared norm is zero. Therefore

T^{*} (αv + βw) = α T^{*} v + β T^{*} w .

The two conjugations are why the adjoint is a linear map, despite the conjugate-linear dependence of the representing functional on its second argument.

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