Chapter 6

Spectrum of a Bounded Operator

The spectrum generalizes the notion of eigenvalues from linear algebra to infinite dimensions, where an operator can fail to be invertible without having an eigenvector.

Resolvent set and spectrum

Definition For $T \in B(H)$, the resolvent set $\rho(T)$ is the set of $\lambda \in \mathbb{C}$ such that $\lambda I - T$ is invertible in $B(H)$ (i.e. bijective with bounded inverse). The spectrum is its complement, $$ \sigma(T) = \mathbb{C} \setminus \rho(T). $$

The spectrum decomposes into three disjoint parts:

Point spectrum $\sigma_p(T)$: $\lambda$ such that $\lambda I - T$ is not injective — these are the eigenvalues, with eigenvector any nonzero $x$ satisfying $Tx = \lambda x$.
Continuous spectrum $\sigma_c(T)$: $\lambda I - T$ is injective with dense but not closed (equivalently, not all of $H$) range.
Residual spectrum $\sigma_r(T)$: $\lambda I - T$ is injective but its range is not dense.

In infinite dimensions, unlike finite dimensions, an operator can fail to be invertible because it fails to be surjective even while remaining injective — this is exactly what produces continuous and residual spectrum with no accompanying eigenvector.

Basic facts

Theorem For any $T \in B(H)$:

$\sigma(T)$ is a nonempty, compact subset of $\mathbb{C}$;
$\sigma(T) \subseteq \{ \lambda \in \mathbb{C} : |\lambda| \le \|T\| \}$.

Compactness (closedness + boundedness) follows because $\rho(T)$ is open (invertibility is an open condition) and because $\lambda I - T$ is invertible whenever $|\lambda| > \|T\|$, via the Neumann series $$ (\lambda I - T)^{-1} = \frac{1}{\lambda}\sum_{n=0}^{\infty} \left(\frac{T}{\lambda}\right)^n, $$ which converges in operator norm when $|\lambda| > \|T\|$. Non-emptiness requires a complex-analytic argument (the resolvent $\lambda \mapsto (\lambda I - T)^{-1}$ is an analytic $B(H)$-valued function on $\rho(T)$; if $\sigma(T)$ were empty this function would be entire and bounded, forcing it to be $0$ by Liouville's theorem — contradiction).

Spectral radius

Definition The spectral radius of $T$ is $$ r(T) = \sup_{\lambda \in \sigma(T)} |\lambda|. $$

Spectral radius formula $$ r(T) = \lim_{n \to \infty} \|T^n\|^{1/n}. $$

In general $r(T) \le \|T\|$, and the inequality can be strict — for example, a nonzero nilpotent operator ($T^n = 0$ for some $n$) has $r(T) = 0$ but $\|T\| > 0$.

Spectra of special operators

Theorem

If $T$ is self-adjoint, then $\sigma(T) \subseteq \mathbb{R}$, and $r(T) = \|T\|$.
If $T$ is unitary, then $\sigma(T) \subseteq \{ \lambda \in \mathbb{C} : |\lambda| = 1 \}$.
If $T$ is normal, then $r(T) = \|T\|$.

Why self-adjoint spectra are real (sketch). For $T = T^*$ and $\lambda = a + bi$ with $b \ne 0$, one computes $$ \|(\lambda I - T)x\|^2 = \|(a I - T)x\|^2 + b^2 \|x\|^2 \ge b^2 \|x\|^2, $$ so $\lambda I - T$ is bounded below, hence injective with closed range; a similar bound for $\overline{\lambda} I - T = (\lambda I - T)^*$ shows the range is also dense, so $\lambda I - T$ is invertible and $\lambda \notin \sigma(T)$.

Looking ahead For compact self-adjoint operators, the spectral picture becomes completely explicit: the spectrum consists of real eigenvalues accumulating only at $0$, with an orthonormal basis of eigenvectors. This is the subject of Chapter 7.

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