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

Use the convention that the inner product is conjugate-linear in its first argument and linear in its second, as in part (b). The adjoint operator is characterized byExistence and uniqueness follow from the finite-dimensional representation of linear functionals by the inner product. More concretely, choose an orthonormal basis $e_i$ and let $a_{ij}=⟨e_i,T{e}_j⟩$. Thenwhich defines $T^{\ast }$ uniquely and verifies the required identity by expansion in that basis. Thus its matrix is the conjugate transpose, proved here rather than assumed. A self-adjoint operator satisfies $T^{\ast }=T$, equivalently $⟨Tv,w⟩=⟨v,Tw⟩$ for every pair of vectors.

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 $\lambda$. Normalize it to have norm one. Self-adjointness givesso $\lambda$ is real.

The orthogonal complement $v^{\perp }$ is an invariant subspace: if $w\perp v$, thenThe restriction of $T$ to $v^{\perp }$ is again self-adjoint, since the same inner product identity holds on that subspace. By induction it has an orthonormal basis of eigenvectors. Adjoining $v$ givesThe existence of the first eigenvector is the essential use of the complex numbers in this proof of the finite-dimensional spectral theorem.

Let $e_1,\dots ,e_d$ be the given orthonormal basis, with $T{e}_j={\lambda }_j{e}_j$. From the defining identity of the adjoint operator,These coordinates prove $T^{\ast }{e}_j={\overline{\lambda }}_j{e}_j$ without assuming a matrix formula. Thereforefor every basis vector. Thus $T$ is a normal operator, andThe 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.