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

Table of contents
- Solution ii

Solution

Proceed by induction on

dim V

. The result is immediate for dimension zero. In positive dimension, the characteristic polynomial has a root over the complex numbers, so there is a nonzero eigenvector

v

with eigenvalue

λ

. Normalize it to have norm one. Self-adjointness gives

\overline{λ} ⟨ v, v ⟩ = ⟨ T v, v ⟩ = ⟨ v, T v ⟩ = λ ⟨ v, v ⟩,

so

λ

is real.

The orthogonal complement

v^{⊥}

is an invariant subspace: if

w ⊥ v

, then

⟨ v, Tw ⟩ = ⟨ T v, w ⟩ = \overline{λ} ⟨ v, w ⟩ = 0.

The restriction of

T

to

v^{⊥}

is again self-adjoint, since the same inner product identity holds on that subspace. By induction it has an orthonormal basis of eigenvectors. Adjoining

v

gives

an orthonormal basis of eigenvectors of T .

The existence of the first eigenvector is the essential use of the complex numbers in this proof of the finite-dimensional spectral theorem.

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