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# Spectrum of an operator

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 $\ell^2(\mathbf Z)$ 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.

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## Definition

Let X be a complex Banach space, and B(X) the Banach algebra of bounded linear operators on X. Then if I denotes the identity operator, and TB(X) then the spectrum of T (normally written as σ(T) ) consists of λ such that λ I - T is not invertible in the algebra of bounded linear operators on X. Note that by the closed graph theorem, this condition is equivalent to asserting λ I - T fails to be bijective.

## Basic properties

Theorem: The spectrum is non-empty, bounded, and closed.

Proof: Suppose 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 C$\to$C. It is not hard to see that

$\lim_{\mu \to \lambda} \frac{F(\lambda) - F(\mu)}{\lambda - \mu} = -\Phi( R(\lambda)^2 )$

so F is an analytic function. However, F(λ) is O-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 Neumann series expansion (named after the German mathematician Carl Neumann),

$(I - A)^{-1} = \sum_{n = 0}^\infty A^n$,

which is valid for any AB(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

$r(T) = \sup \{|\lambda| : \lambda \in \sigma(T)\}$

is 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. It follows that the set of invertible operators is open, and hence, since the function CB(X) defined by λ → λ I - T 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 spectrum

Loosely 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

If 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

More 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 xX. 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

$\lim_{n \to \infty} \|Tx_n - \lambda x_n\| = 0$.

The 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 $\ell^2(\mathbf{Z})$, there are no eigenvectors, but every λ with |λ| = 1 is an approximate eigenvector; letting xn be the vector

$\frac{1}\sqrt{n}(\dots, 0, 1, \lambda, \lambda^2, \dots, \lambda^{n-1}, 0, \dots)$

then ||xn|| = 1 for all n, but

$Tx_n - \lambda x_n = \frac{2}\sqrt{n} \to 0$.

### Compression spectrum

The unilateral shift on $\ell^2(\mathbf{N})$ 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.

## Further results

The spectral radius formula states that

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

This 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).