Past exam of the linear algebra course of the University of Oxford / 2017 / 3 / a / iii

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- Solution iii

Solution

Let

e_{1}, \dots, e_{d}

be the given orthonormal basis, with

T e_{j} = λ_{j} e_{j}

. From the defining identity of the adjoint operator,

⟨ e_{i}, T^{*} e_{j} ⟩ = ⟨ T e_{i}, e_{j} ⟩ = \overline{λ}_{i} δ_{ij} .

These coordinates prove

T^{*} e_{j} = \overline{λ}_{j} e_{j}

without assuming a matrix formula. Therefore

T^{*} T e_{j} = ∣ λ_{j} ∣^{2} e_{j} = T T^{*} e_{j}

for every basis vector. Thus

T

is a normal operator, and

T^{*} T = T T^{*}, T^{*} = T ⟺ λ_{j} \in R for every j .

The condition of possessing an orthonormal basis of eigenvectors is already part of the hypothesis; real eigenvalues alone would not imply self-adjointness for an arbitrary linear transformation.

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