Spectrum of an operatorCategory:Operator theoryCategory:Functional analysisCategory:Linear algebra In functional analysis, the concept of the spectrum of an operator is a generalisation of the concept of eigenvalues, which is much more useful in the case of operators on infinite-dimensional spaces. For example, the bilateral shift operator on the Hilbert space has no eigenvalues at all; but we shall see below that any bounded linear operator on a complex Banach space must have non-empty spectrum. The study of the properties of spectra is known as spectral theory.
Basic propertiesTheorem: The spectrum is non-empty, bounded, and closed. Proof:\nSuppose the spectrum is empty; then the function R(λ) = (λI - T)-1 is defined everywhere on the complex plane. So if Φ is any linear functional on B(X), F(λ) = Φ(R(λ)) is a continuous function . It is not hard to see that \n:\nso F is an analytic function. However, F(λ) is O(λ-1) for large λ so F is a bounded analytic function, and hence constant by Liouville's theorem, and thus everywhere zero as it is zero at infinity. However, by the Hahn-Banach theorem this implies that R(λ) is zero for all λ, which is obviously a contradiction. The boundedness of the spectrum is immediate from the von Neumann series expansion,\n:,\nwhich is valid for any A ∈ B(X) with ||A|| < 1. This implies that if |λ| > ||T||, (λ I - T) is invertible (taking A = T/λ). So σ(T) is bounded, and the spectral radius \n:\nis bounded above by ||T||. Furthermore, the Neumann series implies that for any two operators A, B with A invertible and ||A - B|| < ||A-1||-1, B must also be invertible. So the set of invertible operators is open, and hence, since the function defined by is continuous, the set of λ for which λ I - T is invertible is open, so its complement is closed; but this complement is exactly σ(T).
Classification of points in the spectrumLoosely speaking, there are a variety of ways in which an operator S can fail to be invertible, and this allows us to classify the points of the spectrum into various types.Point spectrum\nIf an operator is not injective (so there is some nonzero x with S(x) = 0), then it is clearly not invertible. So if λ is an eigenvalue of T, we necessarily have λ ∈ σ(T). The set of eigenvalues of T is sometimes called the point spectrum of T.Approximate point spectrum\nMore generally, S is not invertible if it is not bounded below; that is, if there is no 'c' > 0 such that ||Sx|| > c||x|| for all x ∈ X. So the spectrum includes the set of approximate eigenvalues, which are those λ such that T - λ I is not bounded below; equivalently, it is the set of λ for which there is a sequence of unit vectors x1, x2, ... for which\n:. \nThe set of approximate eigenvalues is known as the approximate point spectrum. For example, in the example in the first paragraph of the bilateral shift on , there are no eigenvectors, but every λ with |λ| = 1 is an approximate eigenvector; letting xn be the vector\n:\nthen ||xn|| = 1 for all n, but \n:.Compression spectrum\nThe unilateral shift on gives an example of yet another way in which an operator can fail to be invertible; this shift operator is bounded below (by 1; it is obviously norm-preserving) but it is not invertible as it is not surjective. The set of λ for which λ I - T is not surjective is known as the compression spectrum of T. This exhausts the possibilities, since if T is surjective and bounded below, T is invertible.\n \nFurther resultsThe spectral radius formula states that \n:.\nThis can be proved using similar methods to the above theorem, considering the power series expansion of F(1/λ); this must converge for all λ > r(T), and applying the uniform boundedness principle to the series coefficients gives the result. If T is a compact operator, then it can be shown that any nonzero approximate eigenvalue is in fact an eigenvalue. If X is a Hilbert space and T is a normal operator, then a remarkable result known as the spectral theorem gives an analogue of the diagonalisation theorem for normal finite-dimensional operators (Hermitian matrices, for example).External linkAn account of the spectral theorem |
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is continuous, the set of λ for which λ I - T is invertible is open, so its complement is closed; but this complement is exactly σ(T).
