Spectral theory studies an operator through the complex numbers for which it fails to be invertible. For Hilbert-space operators, the adjoint links this analytic definition to geometry.
Hilbert adjoints
Adjoints are defined by moving an operator across the inner product. In Hilbert spaces this motion is represented by another operator, not merely by an abstract dual map.
Definition. For \(T\in\mathcal L(H,K)\), the Hilbert adjoint \(T^*\in\mathcal L(K,H)\) is determined by \(\langle Tx,y\rangle_K=\langle x,T^*y\rangle_H\). An operator on \(H\) is self-adjoint if \(T=T^*\), normal if \(TT^*=T^*T\), and unitary if \(T^*T=TT^*=I\).
Proposition. \((S+T)^*=S^*+T^*\), \((\alpha T)^*=\overline\alpha T^*\), \((ST)^*=T^*S^*\), and \(\|T^*T\|=\|T\|^2\).
Proof. The algebraic identities follow by inserting the definitions inside inner products and using uniqueness of representing vectors. For the norm identity, \(\|T^*T\|\le\|T^*\|\|T\|=\|T\|^2\). Conversely, \(\|Tx\|^2=\langle T^*Tx,x\rangle\le\|T^*T\|\|x\|^2\), so \(\|T\|^2\le\|T^*T\|\).
Spectrum and resolvent
The spectrum records where the inverse problem fails. Thinking of \(\lambda I-T\) as a parameter-dependent equation makes resolvent estimates a substitute for eigenvalue computations.
Definition. For \(T\in\mathcal L(X)\), the resolvent set is \(\rho(T)=\{\lambda\in\mathbb C:\lambda I-T\text{ is bijective with bounded inverse}\}\). The spectrum is \(\sigma(T)=\mathbb C\setminus\rho(T)\).
Theorem. The resolvent set is open, and \(R(\lambda,T)=(\lambda I-T)^{-1}\) is analytic on \(\rho(T)\).
Proof. Fix \(\lambda_0\in\rho(T)\). Write \(\lambda I-T=(\lambda_0I-T)[I+(\lambda-\lambda_0)R(\lambda_0,T)]\). If \(|\lambda-\lambda_0|\,\|R(\lambda_0,T)\|<1\), the bracket is invertible by the Neumann series. Thus nearby \(\lambda\) lie in \(\rho(T)\), and the Neumann expansion gives \(R(\lambda,T)=\sum_{n\ge0}(-1)^n(\lambda-\lambda_0)^nR(\lambda_0,T)^{n+1}\), which is analytic in operator norm.
Compactness of the spectrum
The spectrum of a bounded operator is not arbitrary. Compactness follows from a Neumann-series argument at infinity and from stability of the resolvent under small perturbations.
Theorem. For \(T\in\mathcal L(X)\) on a complex Banach space, \(\sigma(T)\) is a nonempty compact subset of \(\{\lambda:|\lambda|\le\|T\|\}\).
Proof. If \(|\lambda|>\|T\|\), then \(\lambda I-T=\lambda(I-\lambda^{-1}T)\) is invertible by the Neumann series, so the spectrum is bounded. It is closed because the resolvent set is open.
To prove nonemptiness, suppose \(\sigma(T)=\varnothing\). Then \(R(\lambda,T)\) is entire and \(\|R(\lambda,T)\|\le (|\lambda|-\|T\|)^{-1}\) for large \(|\lambda|\), so for every \(x\in X\) and \(f\in X^*\), the scalar function \(f(R(\lambda,T)x)\) is entire and tends to \(0\) at infinity. Liouville's theorem gives it identically \(0\). Hahn-Banach then implies \(R(\lambda,T)x=0\) for all \(x\), impossible because \(R(\lambda,T)\) is invertible.
Self-adjoint spectrum
Self-adjointness forces the spectral picture onto the real line. The proof is an energy estimate: non-real shifts have bounded inverses because the imaginary part controls the norm.
Theorem. If \(T=T^*\in\mathcal L(H)\), then \(\sigma(T)\subset\mathbb R\).
Proof. Let \(\lambda=a+ib\) with \(b\ne0\). For \(x\in H\), \(\|(\lambda I-T)x\|\|x\|\ge |\langle(\lambda I-T)x,x\rangle|=|a\|x\|^2-\langle Tx,x\rangle+ib\|x\|^2|\ge |b|\|x\|^2\). Thus \(\|(\lambda I-T)x\|\ge |b|\|x\|\). The range is closed and the kernel is zero.
The adjoint of \(\lambda I-T\) is \(\overline\lambda I-T\), which satisfies the same lower bound, so its kernel is zero. Hence the range of \(\lambda I-T\) is dense because its orthogonal complement is \(\ker(\overline\lambda I-T)\). Being both dense and closed, the range is all of \(H\), and the inverse is bounded by the lower bound. Thus \(\lambda\in\rho(T)\).
Spectral radius
The spectral radius is the analytic size seen by repeated composition. It is often smaller than the operator norm, but it captures the asymptotic growth of powers.
Proposition. For self-adjoint \(T\), \(\|T\|=\sup\{|\lambda|:\lambda\in\sigma(T)\}\).
Proof. The inequality \(r(T)\le\|T\|\) holds for all bounded operators by the spectral containment above. For self-adjoint \(T\), the \(C^*\)-identity gives \(\|T^2\|=\|T\|^2\), hence by induction \(\|T^{2^n}\|=\|T\|^{2^n}\). The spectral radius formula \(r(T)=\lim_n\|T^n\|^{1/n}\), proved from the resolvent expansion and submultiplicativity, yields \(r(T)=\|T\|\).