# Spectrum (functional analysis)

(Redirected from Point spectrum)

In mathematics, particularly in functional analysis, the spectrum of a bounded linear operator (or, more generally, an unbounded linear operator) is a generalisation of the set of eigenvalues of a matrix. Specifically, a complex number λ is said to be in the spectrum of a bounded linear operator T if ${\displaystyle T-\lambda I}$ is not invertible, where I is the identity operator. The study of spectra and related properties is known as spectral theory, which has numerous applications, most notably the mathematical formulation of quantum mechanics.

The spectrum of an operator on a finite-dimensional vector space is precisely the set of eigenvalues. However an operator on an infinite-dimensional space may have additional elements in its spectrum, and may have no eigenvalues. For example, consider the right shift operator R on the Hilbert space 2,

${\displaystyle (x_{1},x_{2},\dots )\mapsto (0,x_{1},x_{2},\dots ).}$

This has no eigenvalues, since if Rxx then by expanding this expression we see that x1=0, x2=0, etc. On the other hand, 0 is in the spectrum because the operator R − 0 (i.e. R itself) is not invertible: it is not surjective since any vector with non-zero first component is not in its range. In fact every bounded linear operator on a complex Banach space must have a non-empty spectrum.

The notion of spectrum extends to densely defined unbounded operators. In this case a complex number λ is said to be in the spectrum of such an operator T:DX (where D is dense in X) if there is no bounded inverse ${\displaystyle (T-\lambda I)^{-1}:\,X\to D}$. If T is a closed operator (which includes the case that T is a bounded operator), boundedness of such inverses follows automatically if the inverse exists at all.

The space of bounded linear operators B(X) on a Banach space X is an example of a unital Banach algebra. Since the definition of the spectrum does not mention any properties of B(X) except those that any such algebra has, the notion of a spectrum may be generalised to this context by using the same definition verbatim.

## Spectrum of a bounded operator

### Definition

Let ${\displaystyle T}$  be a bounded linear operator acting on a Banach space ${\displaystyle X}$  over the complex scalar field ${\displaystyle \mathbb {C} }$ , and ${\displaystyle I}$  be the identity operator on ${\displaystyle X}$ . The spectrum of ${\displaystyle T}$  is the set of all ${\displaystyle \lambda \in \mathbb {C} }$  for which the operator ${\displaystyle T-\lambda I}$  does not have an inverse that is a bounded linear operator.

Since ${\displaystyle T-\lambda }$  is a linear operator, the inverse is linear if it exists; and, by the bounded inverse theorem, it is bounded. Therefore, the spectrum consists precisely of those scalars ${\displaystyle \lambda }$  for which ${\displaystyle T-\lambda I}$  is not bijective.

The spectrum of a given operator ${\displaystyle T}$  is often denoted ${\displaystyle \sigma (T)}$ , and its complement, the resolvent set, is denoted ${\displaystyle \rho (T)=\mathbb {C} \setminus \sigma (T)}$ . (${\displaystyle \rho (T)}$  is sometimes used to denote the spectral radius of ${\displaystyle T}$ )

### Spectrum and eigenvalues

If ${\displaystyle \lambda }$  is an eigenvalue of ${\displaystyle T}$ , then the operator ${\displaystyle T-\lambda I}$  is not one-to-one, and therefore its inverse ${\displaystyle (T-\lambda I)^{-1}}$  is not defined. However, the inverse statement is not true: the operator ${\displaystyle T-\lambda I}$  may not have an inverse, even if ${\displaystyle \lambda }$  is not an eigenvalue. Thus the spectrum of an operator always contains all its eigenvalues, but is not limited to them.

For example, consider the Hilbert space ${\displaystyle \ell ^{2}(\mathbb {Z} )}$ , that consists of all bi-infinite sequences of real numbers

${\displaystyle v=(\ldots ,v_{-2},v_{-1},v_{0},v_{1},v_{2},\ldots )}$

that have a finite sum of squares ${\displaystyle \sum _{i=-\infty }^{+\infty }v_{i}^{2}}$ . The bilateral shift operator ${\displaystyle T}$  simply displaces every element of the sequence by one position; namely if ${\displaystyle u=T(v)}$  then ${\displaystyle u_{i}=v_{i-1}}$  for every integer ${\displaystyle i}$ . The eigenvalue equation ${\displaystyle T(v)=\lambda v}$  has no solution in this space, since it implies that all the values ${\displaystyle v_{i}}$  have the same absolute value (if ${\displaystyle \lambda =1}$ ) or are a geometric progression (if ${\displaystyle \lambda \neq 1}$ ); either way, the sum of their squares would not be finite. However, the operator ${\displaystyle T-\lambda I}$  is not invertible if ${\displaystyle |\lambda |=1}$ . For example, the sequence ${\displaystyle u}$  such that ${\displaystyle u_{i}=1/(|i|+1)}$  is in ${\displaystyle \ell ^{2}(\mathbb {Z} )}$ ; but there is no sequence ${\displaystyle v}$  in ${\displaystyle \ell ^{2}(\mathbb {Z} )}$  such that ${\displaystyle (T-I)v=u}$  (that is, ${\displaystyle v_{i-1}=u_{i}+v_{i}}$  for all ${\displaystyle i}$ ).

### Basic properties

The spectrum of a bounded operator T is always a closed, bounded and non-empty subset of the complex plane.

If the spectrum were empty, then the resolvent function

${\displaystyle R(\lambda )=(T-\lambda I)^{-1},\qquad \lambda \in \mathbb {C} ,}$

would be defined everywhere on the complex plane and bounded. But it can be shown that the resolvent function R is holomorphic on its domain. By the vector-valued version of Liouville's theorem, this function is constant, thus everywhere zero as it is zero at infinity. This would be a contradiction.

The boundedness of the spectrum follows from the Neumann series expansion in λ; the spectrum σ(T) is bounded by ||T||. A similar result shows the closedness of the spectrum.

The bound ||T|| on the spectrum can be refined somewhat. The spectral radius, r(T), of T is the radius of the smallest circle in the complex plane which is centered at the origin and contains the spectrum σ(T) inside of it, i.e.

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

The spectral radius formula says[1] that for any element ${\displaystyle T}$  of a Banach algebra,

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

## Classification of points in the spectrum of an operator

A bounded operator T on a Banach space is invertible, i.e. has a bounded inverse, if and only if T is bounded below and has dense range. Accordingly, the spectrum of T can be divided into the following parts:

1. λ ∈ σ(T), if ${\displaystyle T-\lambda I}$  is not bounded below. In particular, this is the case if ${\displaystyle T-\lambda I}$  is not injective, that is, λ is an eigenvalue. The set of eigenvalues is called the point spectrum of T and denoted by σp(T). Alternatively, ${\displaystyle T-\lambda I}$  could be one-to-one but still not be bounded below. Such λ is not an eigenvalue but still an approximate eigenvalue of T (eigenvalues themselves are also approximate eigenvalues). The set of approximate eigenvalues (which includes the point spectrum) is called the approximate point spectrum of T, denoted by σap(T).
2. λ ∈ σ(T), if ${\displaystyle T-\lambda I}$  does not have dense range. The set of such λ is called the compression spectrum of T, denoted by σcp(T). For a subset: If ${\displaystyle T-\lambda I}$  does not have dense range but is injective, λ is said to be in the residual spectrum of T, denoted by σr(T).

Note that the approximate point spectrum and residual spectrum are not necessarily disjoint (however, the point spectrum and the residual spectrum are).

The following subsections provide more details on the three parts of σ(T) sketched above.

### Point spectrum

If an operator is not injective (so there is some nonzero x with T(x) = 0), then it is clearly not invertible. So if λ is an eigenvalue of T, one necessarily has λ ∈ σ(T). The set of eigenvalues of T is also called the point spectrum of T, denoted by σp(T).

### Approximate point spectrum

More generally, by the bounded inverse theorem, T is not invertible if it is not bounded below; that is, if there is no c > 0 such that ||Tx|| ≥ c||x|| for all xX. So the spectrum includes the set of approximate eigenvalues, which are those λ such that TI is not bounded below; equivalently, it is the set of λ for which there is a sequence of unit vectors x1, x2, ... for which

${\displaystyle \lim _{n\to \infty }\|Tx_{n}-\lambda x_{n}\|=0}$ .

The set of approximate eigenvalues is known as the approximate point spectrum, denoted by σap(T).

It is easy to see that the eigenvalues lie in the approximate point spectrum.

Example Consider the bilateral shift T on l2(Z) defined by

${\displaystyle T(\cdots ,a_{-1},{\hat {a}}_{0},a_{1},\cdots )=(\cdots ,{\hat {a}}_{-1},a_{0},a_{1},\cdots )}$

where the ˆ denotes the zero-th position. Direct calculation shows T has no eigenvalues, but every λ with |λ| = 1 is an approximate eigenvalue; letting xn be the vector

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

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

${\displaystyle \|Tx_{n}-\lambda x_{n}\|={\sqrt {\frac {2}{n}}}\to 0.}$

Since T is a unitary operator, its spectrum lies on the unit circle. Therefore, the approximate point spectrum of T is its entire spectrum. This is true for a more general class of operators.

A unitary operator is normal. By spectral theorem, a bounded operator on a Hilbert space H is normal if and only if it is equivalent (after identification of H with an L^2 space) to a multiplication operator. It can be shown that the approximate point spectrum of a bounded multiplication operator equals its spectrum.

### Continuous spectrum

The set of all λ for which ${\displaystyle T-\lambda I}$  is injective and has dense range, but is not surjective, is called the continuous spectrum of T, denoted by σc(T) . The continuous spectrum therefore consists of those approximate eigenvalues which are not eigenvalues and do not lie in the residual spectrum. That is,

${\displaystyle \sigma _{c}(T)=\sigma _{ap}(T)\setminus (\sigma _{r}(T)\cup \sigma _{p}(T))}$ .

#### Example

The hydrogen atom provides an example of this decomposition. The hydrogen atom Hamiltonian operator has a discrete set of eigenvalues that can be computed by the Rydberg formula. Their corresponding eigenfunctions are called eigenstates. The bound states of the hydrogen atom correspond to the eigenvalues, whereas the end result of the ionization process is described by the continuous part of the spectrum (the energy of the collision/ionization is not "quantized").[citation needed]

### Residual spectrum

An operator may be injective, even bounded below, but not invertible. The unilateral shift on ${\displaystyle l^{2}(\mathbb {N} )}$  is such an example. This shift operator is an isometry, therefore bounded below by 1. But it is not invertible as it is not surjective. The set of λ for which ${\displaystyle T-\lambda I}$  is injective but does not have dense range is known as the residual spectrum or compression spectrum of T and is denoted by σr(T).

### Essential spectrum

There are several similar definitions of the essential spectrum which all coincide for self-adjoint operators but differ in a general case. In particular, the essential spectrum ${\displaystyle \sigma _{\mathrm {ess} ,3}(A)}$  of a closed densely defined linear operator ${\displaystyle A:\,{\mathfrak {B}}\to {\mathfrak {B}}}$  can be characterized as the set of points ${\displaystyle \lambda }$  of the spectrum such that ${\displaystyle A-\lambda I_{\mathfrak {B}}}$  is not Fredholm. Here ${\displaystyle I_{\mathfrak {B}}}$  is the identity operator in the Banach space ${\displaystyle {\mathfrak {B}}}$ .

### Discrete spectrum

The discrete spectrum is defined as the set of normal eigenvalues. Equivalently, it can be characterized as the set of isolated points of the spectrum such that the corresponding Riesz projector is of finite rank.

### Peripheral spectrum

The peripheral spectrum of an operator is defined as the set of points in its spectrum which have modulus equal to its spectral radius.[2]

## Further results

If T is a compact operator, or, more generally, an inessential operator, then it can be shown that the spectrum is countable, that zero is the only possible limit point, and that any nonzero λ in the spectrum is 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).

## Spectrum of an unbounded operator

One can extend the definition of spectrum for unbounded operators on a Banach space X, operators which are no longer elements in the Banach algebra B(X). One proceeds in a manner similar to the bounded case. A complex number λ is said to be in the resolvent set, that is, the complement of the spectrum of a linear operator

${\displaystyle T:D\subset X\to X}$

if the operator

${\displaystyle T-\lambda I:D\to X}$

has a bounded inverse, i.e. if there exists a bounded operator

${\displaystyle S:X\rightarrow D}$

such that

${\displaystyle S(T-\lambda I)=I_{D},\,(T-\lambda I)S=I_{X}.}$

A complex number λ is then in the spectrum if this property fails to hold. One can classify the spectrum in exactly the same way as in the bounded case.

The spectrum of an unbounded operator is in general a closed, possibly empty, subset of the complex plane.

For λ to be in the resolvent (i.e. not in the spectrum), as in the bounded case A − λI must be bijective, since it must have a two-sided inverse. As before if an inverse exists then its linearity is immediate, but in general it may not be bounded, so this condition must be checked separately.

However, boundedness of the inverse does follow directly from its existence if one introduces the additional assumption that T is closed; this follows from the closed graph theorem. Therefore, as in the bounded case, a complex number λ lies in the spectrum of a closed operator T if and only if T − λI is not bijective. Note that the class of closed operators includes all bounded operators.

Via its spectral measures, one can define a decomposition of the spectrum of any self adjoint operator, bounded or otherwise into absolutely continuous, pure point, and singular parts.

## Spectrum of a real operator

The definitions of the resolvent and spectrum can be extended to any continuous linear operator ${\displaystyle T}$  acting on a Banach space ${\displaystyle X}$  over the real field ${\displaystyle \mathbb {R} }$  (instead of the complex field ${\displaystyle \mathbb {C} }$ ) via its complexification ${\displaystyle T_{\mathbb {C} }}$ . In this case we define the resolvent set ${\displaystyle \rho (T)}$  as the set of all ${\displaystyle \lambda \in \mathbb {C} }$  such that ${\displaystyle T_{\mathbb {C} }-\lambda I}$  is invertible as an operator acting on the complexified space ${\displaystyle X_{\mathbb {C} }}$ ; then we define ${\displaystyle \sigma (T)=\mathbb {C} \setminus \rho (T)}$ .

### Real spectrum

The real spectrum of a continuous linear operator ${\displaystyle T}$  acting on a real Banach space ${\displaystyle X}$ , denoted ${\displaystyle \sigma _{\mathbb {R} }(T)}$ , is defined as the set of all ${\displaystyle \lambda \in \mathbb {R} }$  for which ${\displaystyle T-\lambda I}$  fails to be invertible in the real algebra of bounded linear operators acting on ${\displaystyle X}$ . In this case we have ${\displaystyle \sigma (T)\cap \mathbb {R} =\sigma _{\mathbb {R} }(T)}$ . Note that the real spectrum may or may not coincide with the complex spectrum. In particular, the real spectrum could be empty.

## Spectrum of a unital Banach algebra

Let B be a complex Banach algebra containing a unit e. Then we define the spectrum σ(x) (or more explicitly σB(x)) of an element x of B to be the set of those complex numbers λ for which λe − x is not invertible in B. This extends the definition for bounded linear operators B(X) on a Banach space X, since B(X) is a Banach algebra.