In a first linear-algebra course you learned to diagonalise a matrix: find the eigenvalues \lambda and eigenvectors of an n \times n matrix A, and — when there are enough of them — rewrite A as a stretch by \lambda_i along each eigen-direction. For a symmetric real matrix this always works perfectly: the spectral theorem for matrices hands you an orthonormal basis of eigenvectors and real eigenvalues. The whole subject in front of you is the answer to one question: what survives when the matrix becomes an infinite-dimensional operator?

The stakes are not academic. In quantum mechanics an observable — energy, position, momentum — is a self-adjoint operator on a Hilbert space of states, and the numbers a laboratory can actually measure are exactly the points of that operator's spectrum. The discrete energy levels of a hydrogen atom, the allowed frequencies of a vibrating string, the resonances of a drum — each is a spectrum. So "find the spectrum" is the infinite-dimensional cousin of "find the eigenvalues", and getting it right is the difference between predicting a spectral line and not.

Here is the first surprise, and the organising idea of the whole page. In finite dimensions the set of eigenvalues is the whole story: A - \lambda I fails to be invertible precisely when it has a non-trivial kernel, i.e. precisely when \lambda is an eigenvalue. In infinite dimensions invertibility can fail without any kernel at all — an operator can be injective yet still not invertible, because its inverse is unbounded or its range is too small. The spectrum is therefore bigger than the eigenvalues, and learning to see that gap is the heart of the matter.

The definition: invertibility, not eigenvalues

Fix a complex Banach space X \ne \{0\} and a bounded linear operator T : X \to X. Throughout, I is the identity and we abbreviate T - \lambda I to T - \lambda. We ask, for each complex number \lambda, a single yes/no question: is T - \lambda invertible as a bounded operator? By the bounded-inverse theorem, "invertible" here means bijective — a bounded inverse then comes for free.

In finite dimensions a linear map is injective iff surjective iff invertible, so those three lines collapse into one and \sigma(A) is just the eigenvalue list. The whole richness of spectral theory comes from these implications breaking apart once X is infinite-dimensional: an operator can be injective but not surjective, or have dense-but-not-closed range, and each failure is a different way to land in the spectrum.

Anatomy of the spectrum: three ways to fail

Because T - \lambda can miss invertibility for genuinely different reasons, the spectrum splits into three disjoint pieces according to how the map fails. Ask two questions: is T - \lambda injective, and is its range dense in X?

• Point spectrum \sigma_p(T). T - \lambda is not injective. Then \lambda is an honest eigenvalue with an eigenvector in the kernel — the finite-dimensional notion, still present here.
• Continuous spectrum \sigma_c(T). T - \lambda is injective and has dense range, but that range is not all of X. The inverse exists on a dense set but is unbounded, so there is no bounded inverse: invertibility fails "at infinity", not by a kernel.
• Residual spectrum \sigma_r(T). T - \lambda is injective but its range is not even dense. There is a whole direction the operator's image never approaches.

The single most important sentence on this page: in infinite dimensions \sigma_c and \sigma_r can be non-empty even when \sigma_p is empty. An operator may have no eigenvalues whatsoever and still have a large spectrum. Keep this in view — the shift operator below is exactly such a beast, and the misconception it kills is the classic exam trap.

What the spectrum always looks like

Before computing any example, three structural facts pin down where the spectrum can live. They are proved with the Neumann series and a little complex analysis, and they hold for every bounded operator on a complex Banach space.

Compactness plus non-emptiness means the maximum in the next definition is genuinely attained.

The bound above says r(T) \le \lVert T \rVert. Remarkably, the spectral radius is computable from the norms of powers of T alone, with no reference to any \lambda — this is Gelfand's formula:

A stark example: a nonzero nilpotent operator N with N^2 = 0 has \lVert N \rVert > 0 but \lVert N^n \rVert = 0 for n \ge 2, so r(N) = 0 and \sigma(N) = \{0\}: a "big" operator with a one-point spectrum. Gelfand's formula is what tells norm and spectral radius apart.

Worked example 1: a multiplication operator — spectrum = range

Let X = L^2[0,1] and let \varphi \in C[0,1] be continuous. Define the multiplication operator M_\varphi by (M_\varphi f)(x) = \varphi(x)\,f(x). It is bounded with \lVert M_\varphi \rVert = \max_x |\varphi(x)|. Claim:

Why. The operator M_\varphi - \lambda = M_{\varphi - \lambda} is again multiplication, now by \varphi - \lambda. If \lambda \notin \varphi([0,1]) then \varphi - \lambda is bounded away from 0, so 1/(\varphi - \lambda) is continuous and bounded and multiplication by it is a bounded inverse — thus \lambda \in \rho(M_\varphi). Conversely if \lambda = \varphi(x_0), then \varphi - \lambda is tiny on a small interval around x_0; multiplying by its "inverse" would blow functions up without bound, so no bounded inverse exists and \lambda \in \sigma(M_\varphi).

Now the subtle part. Is \lambda = \varphi(x_0) an eigenvalue? We would need (\varphi - \lambda) f = 0 a.e. with f \ne 0 in L^2 — i.e. f supported where \varphi = \lambda. If \varphi hits the value \lambda only on a set of measure zero (e.g. \varphi(x) = x, each value taken at a single point), then no such f exists: there are no eigenvalues at all, and the entire spectrum [0,1] is continuous spectrum. So M_x on L^2[0,1] has \sigma = [0,1] but \sigma_p = \varnothing — our first operator whose spectrum is all "continuous". This is precisely the position operator of a particle on a segment: its spectrum is the continuum of positions it can be found at.

Worked example 2: the shift — a full disc with no eigenvalues

On \ell^2 = \ell^2(\mathbb{N}) the unilateral (right) shift is

It is an isometry: \lVert Sx \rVert = \lVert x \rVert, so \lVert S \rVert = 1 and every point of \sigma(S) satisfies |\lambda| \le 1.

No eigenvalues. Suppose Sx = \lambda x. Comparing coordinates gives 0 = \lambda x_1 and x_{k} = \lambda x_{k+1} for all k. If \lambda = 0 then all x_k = 0; if \lambda \ne 0 the first equation forces x_1 = 0 and the recursion propagates 0 through every coordinate. Either way x = 0, so \sigma_p(S) = \varnothing: the shift has not a single eigenvalue.

Yet the spectrum is the entire closed unit disc. Since r(S) \le \lVert S \rVert = 1 we have \sigma(S) \subseteq \overline{\mathbb{D}} = \{ |\lambda| \le 1 \}. The reverse inclusion comes from the adjoint, the backward shift S^*(x_1, x_2, \dots) = (x_2, x_3, \dots). For every |\lambda| < 1 the vector (1, \lambda, \lambda^2, \dots) \in \ell^2 is an eigenvector of S^*, so the open disc lies in \sigma_p(S^*); a general fact links the two, \sigma(S^*) = \overline{\sigma(S)}, and the spectrum is closed, so \overline{\mathbb{D}} \subseteq \sigma(S). Putting the pieces together:

A two-dimensional continuum of spectrum, and every last point of it is continuous or residual — not one eigenvalue among them.

The crown jewel: the spectral theorem for compact self-adjoint operators

Now we return home. Restrict to a Hilbert space H and to the friendliest operators of all: those that are both compact and self-adjoint (T = T^*). Self-adjointness is the infinite-dimensional analogue of a symmetric matrix; compactness is the analogue of being "finite-rank up to a small error", and it is what forces the pathologies of the shift to disappear. For this class the finite-dimensional dream comes true verbatim.

First, two facts that hold for any bounded self-adjoint T on H — the same computations you saw for symmetric matrices, now with the Hilbert inner product playing the role of the dot product.

• Real spectrum. If T = T^* then \sigma(T) \subseteq \mathbb{R}. For an eigenvalue, \lambda \langle x, x \rangle = \langle Tx, x \rangle = \langle x, Tx \rangle = \overline{\lambda}\langle x, x\rangle, forcing \lambda = \overline{\lambda}.
• Orthogonal eigenspaces. Eigenvectors for distinct eigenvalues are orthogonal: if Tx = \lambda x and Ty = \mu y with \lambda \ne \mu, then \lambda \langle x, y \rangle = \langle Tx, y\rangle = \langle x, Ty\rangle = \mu \langle x, y \rangle, so \langle x, y \rangle = 0.
• Norm = spectral radius. For self-adjoint (indeed any normal) operators Gelfand's formula tightens to an equality: r(T) = \lVert T \rVert, and moreover at least one of \pm \lVert T \rVert is in the spectrum.

Read that last line beside the matrix case. Diagonalising a symmetric matrix A = \sum_i \lambda_i\, e_i e_i^{\mathsf T} says Ax = \sum_i \lambda_i (e_i^{\mathsf T} x)\, e_i — decompose x along the orthonormal eigenbasis, scale the i-th component by \lambda_i, reassemble. The spectral theorem is that same sentence with the finite sum replaced by a convergent series and the dot product replaced by \langle x, e_n \rangle. Compactness is the hypothesis that makes the eigenvalues pile up only at 0 and lets the sum converge; without it (the shift again) there may be no eigenbasis to expand in.

The spectrum itself is then completely transparent: \sigma(T) = \{0\} \cup \{\lambda_n\} — the eigenvalues together with their limit point 0 (which sits in the spectrum whenever H is infinite-dimensional, since a compact operator can never be invertible there). A discrete constellation of real eigenvalues marching in toward the origin: that is the picture below.

Seeing the spectrum in the complex plane

Every spectrum lives inside the closed disc |\lambda| \le \lVert T \rVert (normalise \lVert T \rVert = 1, so it is the unit disc), is compact, and is non-empty. Within that frame the three examples above look utterly different. Flip between them:

• Compact self-adjoint — a discrete string of real eigenvalues (points on the horizontal axis) accumulating at the origin, exactly as the spectral theorem promises. The largest one sits on the boundary because r(T) = \lVert T \rVert.
• Unilateral shift — the entire filled disc, though not one point of it is an eigenvalue. This is the "watch out!" picture: maximal spectrum, empty point spectrum.
• Multiplication by e^{2\pi i x} — the spectrum is the range of the multiplier, the unit circle: a one-dimensional curve of spectrum, no interior, no eigenvalues.

Where this pays off