spectrum (infinite-dim)

In finite dimensions an eigenvalue is any lambda where T - lambda I fails to be invertible, and failure of invertibility means exactly the existence of an eigenvector. In infinite dimensions those two things come apart. The spectrum keeps the right definition — it is the set of lambda where T - lambda I is not invertible — but invertibility can now fail for reasons that have nothing to do with an eigenvector existing. The spectrum is genuinely richer than the set of eigenvalues.

Precisely: for a bounded operator T on a Banach space, the spectrum sigma(T) is the set of complex lambda such that T - lambda I has no bounded inverse. Its complement, the resolvent set, is where (T - lambda I)^-1 exists and is bounded. The spectrum is always a nonempty, compact subset of the complex plane, and it sits inside the disc |lambda| <= ||T||. Unlike a matrix, an operator need not have a single eigenvalue, yet its spectrum is never empty.

Why invertibility can fail without eigenvectors: T - lambda I might be injective (no eigenvector) yet not surjective, or have a range that is dense but not closed, so its inverse exists only on a proper or non-closed subset and is unbounded. These failures are invisible in finite dimensions, where injective forces surjective. They are the reason the spectrum splits into point, continuous, and residual parts.

Why it matters: the spectrum is the true infinite-dimensional replacement for the list of eigenvalues. It controls stability, the convergence of operator power series and the Neumann series for inverses, and the functional calculus f(T). The spectral radius sup{|lambda| : lambda in sigma(T)} equals lim ||T^n||^(1/n), tying the geometry of the spectrum to the growth of the operator's powers.