Spectrum (functional analysis)

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 ${\displaystyle T-\lambda I}$ 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 ${\displaystyle T}$)

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 converse 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 ${\textstyle \sum _{i=-\infty }^{+\infty }v_{i}^{2}}$. 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 nonzero solution in this space, since it implies that all the values ${\displaystyle v_{i}}$ have the same absolute value (if ${\displaystyle \vert \lambda \vert =1}$) or are a geometric progression (if ${\displaystyle \vert \lambda \vert \neq 1}$); 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 ${\displaystyle i}$).

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[2] that for any element ${\displaystyle T}$ of a Banach algebra,

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

Let X be a Banach space and ${\displaystyle T:\,D(T)\to X}$ be a linear operator defined on domain ${\displaystyle D(T)\subseteq X}$. A complex number λ is said to be in the resolvent set (also called regular set) of ${\displaystyle T}$ if the operator

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

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

${\displaystyle S:\,X\rightarrow D(T)}$

such that

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

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, ${\displaystyle T-\lambda I}$ 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 ${\displaystyle (T-\lambda I)^{-1}}$ 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 ${\displaystyle T-\lambda I}$ 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 ${\displaystyle \sigma (T)=\mathbb {C} }$.

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 σ_p(T).

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

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

The set of approximate eigenvalues is known as the approximate point spectrum, denoted by ${\displaystyle \sigma _{\mathrm {ap} }(T)}$.

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

For example, consider the right shift R on ${\displaystyle l^{2}(\mathbb {Z} )}$ defined by

${\displaystyle R:\,e_{j}\mapsto e_{j+1},\quad j\in \mathbb {Z} ,}$

where ${\displaystyle {\big (}e_{j}{\big )}_{j\in \mathbb {N} }}$ is the standard orthonormal basis in ${\displaystyle l^{2}(\mathbb {Z} )}$. Direct calculation shows R has no eigenvalues, but every λ with |λ| = 1 is an approximate eigenvalue; letting x_n be the vector

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

one can see that ||x_n|| = 1 for all n, but

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

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 ${\displaystyle L^{2}}$ 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 ${\displaystyle T-\lambda I}$ is injective and has dense range, but is not surjective, is called the continuous spectrum of T, denoted by ${\displaystyle \sigma _{\mathbb {c} }(T)}$. The continuous spectrum therefore consists of those approximate eigenvalues which are not eigenvalues and do not lie in the residual spectrum. That is,

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

For example, ${\displaystyle A:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle e_{j}\mapsto e_{j}/j}$, ${\displaystyle j\in \mathbb {N} }$, is injective and has a dense range, yet ${\displaystyle \mathrm {Ran} (A)\subsetneq l^{2}(\mathbb {N} )}$. Indeed, if ${\textstyle x=\sum _{j\in \mathbb {N} }c_{j}e_{j}\in l^{2}(\mathbb {N} )}$ with ${\displaystyle c_{j}\in \mathbb {C} }$ such that ${\textstyle \sum _{j\in \mathbb {N} }|c_{j}|^{2}<\infty }$, one does not necessarily have ${\textstyle \sum _{j\in \mathbb {N} }\left|jc_{j}\right|^{2}<\infty }$, and then ${\textstyle \sum _{j\in \mathbb {N} }jc_{j}e_{j}\notin l^{2}(\mathbb {N} )}$.

The set of ${\displaystyle \lambda \in \mathbb {C} }$ for which ${\displaystyle T-\lambda I}$ does not have dense range is known as the compression spectrum of T and is denoted by ${\displaystyle \sigma _{\mathrm {cp} }(T)}$.

The set of ${\displaystyle \lambda \in \mathbb {C} }$ for which ${\displaystyle T-\lambda I}$ is injective but does not have dense range is known as the residual spectrum of T and is denoted by ${\displaystyle \sigma _{\mathrm {r} }(T)}$:

${\displaystyle \sigma _{\mathrm {r} }(T)=\sigma _{\mathrm {cp} }(T)\setminus \sigma _{\mathrm {p} }(T).}$

An operator may be injective, even bounded below, but still not invertible. The right shift on ${\displaystyle l^{2}(\mathbb {N} )}$, ${\displaystyle R:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle R:\,e_{j}\mapsto e_{j+1},\,j\in \mathbb {N} }$, 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 (${\displaystyle e_{1}\not \in \mathrm {Ran} (R)}$), and moreover ${\displaystyle \mathrm {Ran} (R)}$ is not dense in ${\displaystyle l^{2}(\mathbb {N} )}$ (${\displaystyle e_{1}\notin {\overline {\mathrm {Ran} (R)}}}$).

The peripheral spectrum of an operator is defined as the set of points in its spectrum which have modulus equal to its spectral radius.[3]

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.

There are five similar definitions of the essential spectrum of closed densely defined linear operator ${\displaystyle A:\,X\to X}$ which satisfy

${\displaystyle \sigma _{\mathrm {ess} ,1}(A)\subset \sigma _{\mathrm {ess} ,2}(A)\subset \sigma _{\mathrm {ess} ,3}(A)\subset \sigma _{\mathrm {ess} ,4}(A)\subset \sigma _{\mathrm {ess} ,5}(A)\subset \sigma (A).}$

All these spectra ${\displaystyle \sigma _{\mathrm {ess} ,k}(A),\ 1\leq k\leq 5}$, coincide in the case of self-adjoint operators.

1. The essential spectrum ${\displaystyle \sigma _{\mathrm {ess} ,1}(A)}$ is defined as the set of points ${\displaystyle \lambda }$ of the spectrum such that ${\displaystyle A-\lambda I}$ 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: ${\displaystyle \lambda =0\in \sigma _{\mathrm {ess} ,1}(A)}$ for the operator ${\displaystyle A:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle A:\,e_{j}\mapsto e_{j}/j,~j\in \mathbb {N} }$ (because the range of this operator is not closed: the range does not include all of ${\displaystyle l^{2}(\mathbb {N} )}$ although its closure does).
   Example 2: ${\displaystyle \lambda =0\in \sigma _{\mathrm {ess} ,1}(N)}$ for ${\displaystyle N:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle N:\,v\mapsto 0}$ for any ${\displaystyle v\in l^{2}(\mathbb {N} )}$ (because both kernel and cokernel of this operator are infinite-dimensional).
2. The essential spectrum ${\displaystyle \sigma _{\mathrm {ess} ,2}(A)}$ is defined as the set of points ${\displaystyle \lambda }$ of the spectrum such that the operator either ${\displaystyle A-\lambda I}$ 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 ${\displaystyle (x_{j})_{j\in \mathbb {N} }}$ in the space X such that ${\displaystyle \Vert x_{j}\Vert =1}$, ${\textstyle \lim _{j\to \infty }\left\|(A-\lambda I)x_{j}\right\|=0,}$ and such that ${\displaystyle (x_{j})_{j\in \mathbb {N} }}$ contains no convergent subsequence. Such a sequence is called a singular sequence (or a singular Weyl sequence).
   Example: ${\displaystyle \lambda =0\in \sigma _{\mathrm {ess} ,2}(B)}$ for the operator ${\displaystyle B:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle B:\,e_{j}\mapsto e_{j/2}}$ if j is even and ${\displaystyle e_{j}\mapsto 0}$ when j is odd (kernel is infinite-dimensional; cokernel is zero-dimensional). Note that ${\displaystyle \lambda =0\not \in \sigma _{\mathrm {ess} ,1}(B)}$.
3. The essential spectrum ${\displaystyle \sigma _{\mathrm {ess} ,3}(A)}$ is defined as the set of points ${\displaystyle \lambda }$ of the spectrum such that ${\displaystyle A-\lambda I}$ is not Fredholm. (The operator is Fredholm if its range is closed and both its kernel and cokernel are finite-dimensional.)
   Example: ${\displaystyle \lambda =0\in \sigma _{\mathrm {ess} ,3}(J)}$ for the operator ${\displaystyle J:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle J:\,e_{j}\mapsto e_{2j}}$ (kernel is zero-dimensional, cokernel is infinite-dimensional). Note that ${\displaystyle \lambda =0\not \in \sigma _{\mathrm {ess} ,2}(J)}$.
4. The essential spectrum ${\displaystyle \sigma _{\mathrm {ess} ,4}(A)}$ is defined as the set of points ${\displaystyle \lambda }$ of the spectrum such that ${\displaystyle A-\lambda I}$ 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, ${\textstyle \sigma _{\mathrm {ess} ,4}(A)=\bigcap _{K\in B_{0}(X)}\sigma (A+K)}$; here ${\displaystyle B_{0}(X)}$ denotes the set of all compact operators on X.
   Example: ${\displaystyle \lambda =0\in \sigma _{\mathrm {ess} ,4}(R)}$ where ${\displaystyle R:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$ is the right shift operator, ${\displaystyle R:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle R:\,e_{j}\mapsto e_{j+1}}$ for ${\displaystyle j\in \mathbb {N} }$ (its kernel is zero, its cokernel is one-dimensional). Note that ${\displaystyle \lambda =0\not \in \sigma _{\mathrm {ess} ,3}(R)}$.
5. The essential spectrum ${\displaystyle \sigma _{\mathrm {ess} ,5}(A)}$ is the union of ${\displaystyle \sigma _{\mathrm {ess} ,1}(A)}$ with all components of ${\displaystyle \mathbb {C} \setminus \sigma _{\mathrm {ess} ,1}(A)}$ that do not intersect with the resolvent set ${\displaystyle \mathbb {C} \setminus \sigma (A)}$. It can also be characterized as ${\displaystyle \sigma (A)\setminus \sigma _{\mathrm {d} }(A)}$.
   Example: consider the operator ${\displaystyle T:\,l^{2}(\mathbb {Z} )\to l^{2}(\mathbb {Z} )}$, ${\displaystyle T:\,e_{j}\mapsto e_{j-1}}$ for ${\displaystyle j\neq 0}$, ${\displaystyle T:\,e_{0}\mapsto 0}$. Since ${\displaystyle \Vert T\Vert =1}$, one has ${\displaystyle \sigma (T)\subset {\overline {\mathbb {D} _{1}}}}$. For any ${\displaystyle z\in \mathbb {C} }$ with ${\displaystyle |z|=1}$, the range of ${\displaystyle T-zI}$ is dense but not closed, hence the boundary of the unit disc is in the first type of the essential spectrum: ${\displaystyle \partial \mathbb {D} _{1}\subset \sigma _{\mathrm {ess} ,1}(T)}$. For any ${\displaystyle z\in \mathbb {C} }$ with ${\displaystyle |z|<1}$, ${\displaystyle T-zI}$ has a closed range, one-dimensional kernel, and one-dimensional cokernel, so ${\displaystyle z\in \sigma (T)}$ although ${\displaystyle z\not \in \sigma _{\mathrm {ess} ,k}(T)}$ for ${\displaystyle 1\leq k\leq 4}$; thus, ${\displaystyle \sigma _{\mathrm {ess} ,k}(T)=\partial \mathbb {D} _{1}}$ for ${\displaystyle 1\leq k\leq 4}$. There are two components of ${\displaystyle \mathbb {C} \setminus \sigma _{\mathrm {ess} ,1}(T)}$: ${\displaystyle \{z\in \mathbb {C}$ :\,|z|>1\}} and ${\displaystyle \{z\in \mathbb {C}$ :\,|z|<1\}} . The component ${\displaystyle \{|z|<1\}}$ has no intersection with the resolvent set; by definition, ${\displaystyle \sigma _{\mathrm {ess} ,5}(T)=\sigma _{\mathrm {ess} ,1}(T)\cup \{z\in \mathbb {C}$ :\,|z|<1\}=\{z\in \mathbb {C} :\,|z|\leq 1\}} .

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 accumulation point, and that any nonzero λ in the spectrum is an eigenvalue.

A bounded operator ${\displaystyle A:\,X\to X}$ is quasinilpotent if ${\displaystyle \lVert A^{n}\rVert ^{1/n}\to 0}$ as ${\displaystyle n\to \infty }$ (in other words, if the spectral radius of A equals zero). Such operators could equivalently be characterized by the condition

${\displaystyle \sigma (A)=\{0\}.}$

An example of such an operator is ${\displaystyle A:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} )}$, ${\displaystyle e_{j}\mapsto e_{j+1}/2^{j}}$ for ${\displaystyle j\in \mathbb {N} }$.

If X is a Hilbert space and T is a self-adjoint operator (or, more generally, 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).

For self-adjoint operators, one can use spectral measures to define a decomposition of the spectrum into absolutely continuous, pure point, and singular parts.

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.