The spectrum of an operator on a finite-dimensionalvector 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 spaceℓ2,
This has no eigenvalues, since if Rx=λx then by expanding this expression we see that x1=0, x2=0, etc. On the other hand, 0 is in the spectrum because although the operator R − 0 (i.e. R itself) is invertible, the inverse is defined on a set which is not dense in ℓ2. In fact every bounded linear operator on a complexBanach space must have a non-empty spectrum.
The notion of spectrum extends to unbounded (i.e. not necessarily bounded) operators. A complex numberλ is said to be in the spectrum of an unbounded operator defined on domain if there is no bounded inverse defined on the whole of If T is closed (which includes the case when T is bounded), boundedness of follows automatically from its existence.
The space of bounded linear operators B(X) on a Banach space X is an example of a unitalBanach 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
Let be a bounded linear operator acting on a Banach space over the complex scalar field , and be the identity operator on . The spectrum of is the set of all for which the operator does not have an inverse that is a bounded linear operator.
Since 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 for which is not bijective.
The spectrum of a given operator is often denoted , and its complement, the resolvent set, is denoted . ( is sometimes used to denote the spectral radius of )
Relation to eigenvalues
If is an eigenvalue of , then the operator is not one-to-one, and therefore its inverse is not defined. However, the converse statement is not true: the operator may not have an inverse, even if is not an eigenvalue. Thus the spectrum of an operator always contains all its eigenvalues, but is not limited to them.
that have a finite sum of squares . The bilateral shift operator simply displaces every element of the sequence by one position; namely if then for every integer . The eigenvalue equation has no nonzero solution in this space, since it implies that all the values have the same absolute value (if ) or are a geometric progression (if ); either way, the sum of their squares would not be finite. However, the operator is not invertible if . For example, the sequence such that is in ; but there is no sequence in such that (that is, for all ).
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.
One can extend the definition of spectrum to unbounded operators on a Banach spaceX. These operators which are no longer elements in the Banach algebra B(X).
Let X be a Banach space and be a linear operator defined on domain .
A complex number λ is said to be in the resolvent set (also called regular set) of if the operator
has a bounded everywhere-defined inverse, i.e. if there exists a bounded operator
A complex number λ is then in the spectrum if λ is not in the resolvent set.
For λ to be in the resolvent (i.e. not in the spectrum), just like in the bounded case, 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.
By the closed graph theorem, boundedness of does follow directly from its existence when T is closed. Then, just as in the bounded case, a complex number λ lies in the spectrum of a closed operator T if and only if is not bijective. Note that the class of closed operators includes all bounded operators.
The spectrum of an unbounded operator is in general a closed, possibly empty, subset of the complex plane.
If the operator T is not closed, then .
A bounded operator T on a Banach space is invertible, i.e. has a bounded inverse, if and only if T is bounded below, i.e. for some and has dense range. Accordingly, the spectrum of T can be divided into the following parts:
if is not bounded below. In particular, this is the case if 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, could be one-to-one but still not 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).
if does not have dense range. The set of such λ is called the compression spectrum of T, denoted by . If does not have dense range but is injective, λ is said to be in the residual spectrum of T, denoted by .
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.
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 x ∈ X. So the spectrum includes the set of approximate eigenvalues, which are those λ such that T - λI is not bounded below; equivalently, it is the set of λ for which there is a sequence of unit vectors x1, x2, ... for which
The set of approximate eigenvalues is known as the approximate point spectrum, denoted by .
It is easy to see that the eigenvalues lie in the approximate point spectrum.
For example, consider the right shift R on defined by
where is the standard orthonormal basis in . Direct calculation shows R has no eigenvalues, but every λ with |λ| = 1 is an approximate eigenvalue; letting xn be the vector
one can see that ||xn|| = 1 for all n, but
Since R is a unitary operator, its spectrum lies on the unit circle. Therefore, the approximate point spectrum of R is its entire spectrum.
This conclusion is also true for a more general class of operators.
A unitary operator is normal. By the 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 space) to a multiplication operator. It can be shown that the approximate point spectrum of a bounded multiplication operator equals its spectrum.
The set of all λ for which is injective and has dense range, but is not surjective, is called the continuous spectrum of T, denoted by . The continuous spectrum therefore consists of those approximate eigenvalues which are not eigenvalues and do not lie in the residual spectrum. That is,
For example, , , , is injective and has a dense range, yet .
Indeed, if with such that , one does not necessarily have , and then .
The set of for which does not have dense range is known as the compression spectrum of T and is denoted by .
The set of for which is injective but does not have dense range is known as the residual spectrum of T and is denoted by :
An operator may be injective, even bounded below, but still not invertible. The right shift on , , , 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 (), and moreover is not dense in
The peripheral spectrum of an operator is defined as the set of points in its spectrum which have modulus equal to its spectral radius.
There are five similar definitions of the essential spectrum of closed densely defined linear operator which satisfy
All these spectra , coincide in the case of self-adjoint operators.
The essential spectrum is defined as the set of points of the spectrum such that is not semi-Fredholm. (The operator is semi-Fredholm if its range is closed and either its kernel or cokernel (or both) is finite-dimensional.) Example 1: for the operator , (because the range of this operator is not closed: the range does not include all of although its closure does). Example 2: for , for any (because both kernel and cokernel of this operator are infinite-dimensional).
The essential spectrum is defined as the set of points of the spectrum such that the operator either has infinite-dimensional kernel or has a range which is not closed. It can also be characterized in terms of Weyl's criterion: there exists a sequence in the space X such that , and such that contains no convergent subsequence. Such a sequence is called a singular sequence (or a singular Weyl sequence). Example: for the operator , if j is even and when j is odd (kernel is infinite-dimensional; cokernel is zero-dimensional). Note that .
The essential spectrum is defined as the set of points of the spectrum such that is not Fredholm. (The operator is Fredholm if its range is closed and both its kernel and cokernel are finite-dimensional.) Example: for the operator , (kernel is zero-dimensional, cokernel is infinite-dimensional). Note that .
The essential spectrum is defined as the set of points of the spectrum such that is not Fredholm of index zero. It could also be characterized as the largest part of the spectrum of A which is preserved by compact perturbations. In other words, ; here denotes the set of all compact operators on X. Example: where is the right shift operator, , for (its kernel is zero, its cokernel is one-dimensional). Note that .
The essential spectrum is the union of with all components of that do not intersect with the resolvent set . It can also be characterized as . Example: consider the operator , for , . Since , one has . For any with , the range of is dense but not closed, hence the boundary of the unit disc is in the first type of the essential spectrum: . For any with , has a closed range, one-dimensional kernel, and one-dimensional cokernel, so although for ; thus, for . There are two components of : and . The component has no intersection with the resolvent set; by definition, .
Example: Hydrogen atom
The hydrogen atom provides an example of different types of the spectra. The hydrogen atom Hamiltonian operator, , with domain has a discrete set of eigenvalues (the discrete spectrum , which in this case coincides with the point spectrum since there are no eigenvalues embedded into the continuous spectrum) that can be computed by the Rydberg formula. Their corresponding eigenfunctions are called eigenstates, or the bound states. The result of the ionization process is described by the continuous part of the spectrum (the energy of the collision/ionization is not "quantized"), represented by (it also coincides with the essential spectrum, ).
Theorem — For a bounded (or, more generally, closed and densely defined) operator T,
In particular, .
Suppose that is not dense in X.
By the Hahn–Banach theorem, there exists a non-zero that vanishes on .
For all x ∈ X,
Therefore, and is an eigenvalue of T*.
Conversely, suppose that is an eigenvalue of T*. Then there exists a non-zero such that , i.e.
If is dense in X, then φ must be the zero functional, a contradiction.
The claim is proved.
We also get by the following argument: X embeds isometrically into X**.
Therefore, for every non-zero element in the kernel of there exists a non-zero element in X** which vanishes on .
Thus can not be dense.
The definitions of the resolvent and spectrum can be extended to any continuous linear operator acting on a Banach space over the real field (instead of the complex field ) via its complexification. In this case we define the resolvent set as the set of all such that is invertible as an operator acting on the complexified space ; then we define .
The real spectrum of a continuous linear operator acting on a real Banach space , denoted , is defined as the set of all for which fails to be invertible in the real algebra of bounded linear operators acting on . In this case we have . 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
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Let B be a complex Banach algebra containing a unite. 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 numbersλ 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 unital Banach algebra.