Past exam of the linear algebra course of the University of Oxford / 2023 / 3 / b / i

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- Solution i

Solution

For the normal operator

T

, the adjoint operator identity and

T^{*} T = T T^{*}

give

∥ T v ∥^{2} = ⟨ v, T^{*} T v ⟩ = ⟨ v, T T^{*} v ⟩ = ∥ T^{*} v ∥^{2} .

For the last equality,

⟨ T T^{*} v, v ⟩ = ⟨ T^{*} v, T^{*} v ⟩

and complex conjugation allow reversal of the two arguments. Taking the nonnegative square roots proves

∥ T v ∥ = ∥ T^{*} v ∥ (v \in V) .

In particular the two kernels coincide.

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