Inlinear algebra andfunctional analysis, aspectral theorem is a result about when alinear operator ormatrix can bediagonalized (that is, represented as adiagonal matrix in some basis). This is extremely useful because computations involving a diagonalizable matrix can often be reduced to much simpler computations involving the corresponding diagonal matrix. The concept of diagonalization is relatively straightforward for operators onfinite-dimensional vector spaces but requires some modification for operators on infinite-dimensional spaces. In general, the spectral theorem identifies a class oflinear operators that can be modeled bymultiplication operators, which are as simple as one can hope to find. In more abstract language, the spectral theorem is a statement about commutativeC*-algebras. See alsospectral theory for a historical perspective.

Examples of operators to which the spectral theorem applies areself-adjoint operators or more generallynormal operators onHilbert spaces.

The spectral theorem also provides acanonical decomposition, called thespectral decomposition, of the underlying vector space on which the operator acts.

Augustin-Louis Cauchy proved the spectral theorem forsymmetric matrices, i.e., that every real, symmetric matrix is diagonalizable. In addition, Cauchy was the first to be systematic aboutdeterminants.^[1][2] The spectral theorem as generalized byJohn von Neumann is today perhaps the most important result ofoperator theory.

This article mainly focuses on the simplest kind of spectral theorem, that for aself-adjoint operator on a Hilbert space. However, as noted above, the spectral theorem also holds for normal operators on a Hilbert space.

We begin by considering aHermitian matrix on${\displaystyle {\mathbb{C}}^n}$ (but the following discussion will be adaptable to the more restrictive case ofsymmetric matrices on${\displaystyle {\mathbb{R}}^n}$). We consider aHermitian mapA on a finite-dimensionalcomplexinner product spaceV endowed with apositive definitesesquilinearinner product${\displaystyle ⟨\cdot ,\cdot ⟩}$. The Hermitian condition on${\displaystyle A}$ means that for allx,y ∈V,$${\displaystyle ⟨Ax,y⟩=⟨x,Ay⟩.}$$

An equivalent condition is thatA^* =A, whereA^* is theHermitian conjugate ofA. In the case thatA is identified with a Hermitian matrix, the matrix ofA^* is equal to itsconjugate transpose. (IfA is areal matrix, then this is equivalent toA^T =A, that is,A is asymmetric matrix.)

This condition implies that all eigenvalues of a Hermitian map are real: To see this, it is enough to apply it to the case whenx =y is an eigenvector. (Recall that aneigenvector of a linear mapA is a non-zero vectorv such thatAv =λv for some scalarλ. The valueλ is the correspondingeigenvalue. Moreover, theeigenvalues are roots of thecharacteristic polynomial.)

We provide a sketch of a proof for the case where the underlying field of scalars is thecomplex numbers.

By thefundamental theorem of algebra, applied to thecharacteristic polynomial ofA, there is at least one complex eigenvalueλ₁ and corresponding eigenvectorv₁, which must by definition be non-zero. Then since$${\displaystyle {\lambda }_1⟨v_1,v_1⟩=⟨A(v_1),v_1⟩=⟨v_1,A(v_1)⟩={\overline{\lambda }}_1⟨v_1,v_1⟩,}$$ we find thatλ₁ is real. Now consider the space${\displaystyle {\mathcal{K}}^{n-1}=\text{span}(v_1{)}^{\perp }}$, theorthogonal complement ofv₁. By Hermiticity,${\displaystyle {\mathcal{K}}^{n-1}}$ is aninvariant subspace ofA. To see that, consider any${\displaystyle k\in {\mathcal{K}}^{n-1}}$ so that${\displaystyle ⟨k,v_1⟩=0}$ by definition of${\displaystyle {\mathcal{K}}^{n-1}}$. To satisfy invariance, we need to check if${\displaystyle A(k)\in {\mathcal{K}}^{n-1}}$. This is true because,${\displaystyle ⟨A(k),v_1⟩=⟨k,A(v_1)⟩=⟨k,{\lambda }_1{v}_1⟩=0}$. Applying the same argument to${\displaystyle {\mathcal{K}}^{n-1}}$ shows thatA has at least one real eigenvalue${\displaystyle {\lambda }_2}$ and corresponding eigenvector${\displaystyle v_2\in {\mathcal{K}}^{n-1}\perp v_1}$. This can be used to build another invariant subspace${\displaystyle {\mathcal{K}}^{n-2}=\text{span}(\{v_1,v_2\}{)}^{\perp }}$. Finite induction then finishes the proof.

The matrix representation ofA in a basis of eigenvectors is diagonal, and by the construction the proof gives a basis of mutually orthogonal eigenvectors; by choosing them to be unit vectors one obtains an orthonormal basis of eigenvectors.A can be written as a linear combination of pairwise orthogonal projections, called itsspectral decomposition. Let$${\displaystyle V_{\lambda }=\{v\in V:Av=\lambda v\}}$$be the eigenspace corresponding to an eigenvalue${\displaystyle \lambda }$. Note that the definition does not depend on any choice of specific eigenvectors. In general,V is the orthogonal direct sum of the spaces${\displaystyle V_{\lambda }}$ where the${\displaystyle \lambda }$ ranges over thespectrum of${\displaystyle A}$.

When the matrix being decomposed is Hermitian, the spectral decomposition is a special case of theSchur decomposition (see the proof in case ofnormal matrices below).

The spectral decomposition is a special case of thesingular value decomposition, which states that any matrix${\displaystyle A\in {\mathbb{C}}^{m\times n}}$ can be expressed as${\displaystyle A=U\mathrm{\Sigma }{V}^{\ast }}$, where${\displaystyle U\in {\mathbb{C}}^{m\times m}}$ and${\displaystyle V\in {\mathbb{C}}^{n\times n}}$ areunitary matrices and${\displaystyle \mathrm{\Sigma }\in {\mathbb{R}}^{m\times n}}$ is a diagonal matrix. The diagonal entries of${\displaystyle \mathrm{\Sigma }}$ are uniquely determined by${\displaystyle A}$ and are known as thesingular values of${\displaystyle A}$. If${\displaystyle A}$ is Hermitian, then${\displaystyle A^{\ast }=A}$ and${\displaystyle V\mathrm{\Sigma }{U}^{\ast }=U\mathrm{\Sigma }{V}^{\ast }}$ which implies${\displaystyle U=V}$.

The spectral theorem extends to a more general class of matrices. LetA be an operator on a finite-dimensional inner product space.A is said to benormal ifA^*A =AA^*.

One can show thatA is normal if and only if it is unitarily diagonalizable using theSchur decomposition. That is, any matrix can be written asA =UTU^*, whereU is unitary andT isupper triangular.IfA is normal, then one sees thatTT^* =T^*T. Therefore,T must be diagonal since a normal upper triangular matrix is diagonal (seenormal matrix). The converse is obvious.

In other words,A is normal if and only if there exists aunitary matrixU such that$${\displaystyle A=UD{U}^{\ast },}$$whereD is adiagonal matrix. Then, the entries of the diagonal ofD are theeigenvalues ofA. The column vectors ofU are the eigenvectors ofA and they are orthonormal. Unlike the Hermitian case, the entries ofD need not be real.

In the more general setting of Hilbert spaces, which may have an infinite dimension, the statement of the spectral theorem forcompactself-adjoint operators is virtually the same as in the finite-dimensional case.

As for Hermitian matrices, the key point is to prove the existence of at least one nonzero eigenvector. One cannot rely on determinants to show existence of eigenvalues, but one can use a maximization argument analogous to the variational characterization of eigenvalues.

If the compactness assumption is removed, then it isnot true that every self-adjoint operator has eigenvectors. For example, the multiplication operator${\displaystyle M_x}$ on${\displaystyle L^2([0,1])}$ which takes each${\displaystyle \psi (x)\in L^2([0,1])}$ to${\displaystyle x\psi (x)}$ is bounded and self-adjoint, but has no eigenvectors. However, its spectrum, suitably defined, is still equal to${\displaystyle [0,1]}$, see spectrum of bounded operator.

The next generalization we consider is that ofbounded self-adjoint operators on a Hilbert space. Such operators may have no eigenvectors: for instance letA be the operator of multiplication byt on${\displaystyle L^2([0,1])}$, that is,^[3]$${\displaystyle [Af](t)=tf(t).}$$

This operator does not have any eigenvectorsin${\displaystyle L^2([0,1])}$, though it does have eigenvectors in a larger space. Namely thedistribution${\displaystyle f(t)=\delta (t-t_0)}$, where${\displaystyle \delta }$ is theDirac delta function, is an eigenvector when construed in an appropriate sense. The Dirac delta function is however not a function in the classical sense and does not lie in the Hilbert spaceL²[0, 1]. Thus, the delta-functions are "generalized eigenvectors" of${\displaystyle A}$ but not eigenvectors in the usual sense.

In the absence of (true) eigenvectors, one can look for a "spectral subspace" consisting of analmost eigenvector, i.e, a closed subspace${\displaystyle V_E}$ of${\displaystyle V}$ associated with aBorel set${\displaystyle E\subset \sigma (A)}$ in thespectrum of${\displaystyle A}$. This subspace can be thought of as the closed span of generalized eigenvectors for${\displaystyle A}$ with eigenvalues in${\displaystyle E}$.^[4] In the above example, where${\displaystyle [Af](t)=tf(t),}$ we might consider the subspace of functions supported on a small interval${\displaystyle [a,a+\epsilon ]}$ inside${\displaystyle [0,1]}$. This space is invariant under${\displaystyle A}$ and for any${\displaystyle f}$ in this subspace,${\displaystyle Af}$ is very close to${\displaystyle af}$. Each subspace, in turn, is encoded by the associated projection operator, and the collection of all the subspaces is then represented by aprojection-valued measure.

One formulation of the spectral theorem expresses the operatorA as an integral of the coordinate function over the operator's spectrum${\displaystyle \sigma (A)}$ with respect to a projection-valued measure.^[5]$${\displaystyle A={\int }_{\sigma (A)}\lambda d\pi (\lambda ).}$$When the self-adjoint operator in question iscompact, this version of the spectral theorem reduces to something similar to the finite-dimensional spectral theorem above, except that the operator is expressed as a finite or countably infinite linear combination of projections, that is, the measure consists only of atoms.

An alternative formulation of the spectral theorem says that every bounded self-adjoint operator is unitarily equivalent to a multiplication operator, a relatively simple type of operator.

Multiplication operators are a direct generalization of diagonal matrices. A finite-dimensional Hermitian vector space${\displaystyle V}$ may be coordinatized as the space of functions${\displaystyle f:B\to \mathbb{C}}$ from a basis${\displaystyle B}$ to the complex numbers, so that the${\displaystyle B}$-coordinates of a vector are the values of the corresponding function${\displaystyle f}$. The finite-dimensional spectral theorem for a self-adjoint operator${\displaystyle A:V\to V}$ states that there exists an orthonormal basis of eigenvectors${\displaystyle B}$, so that the inner product becomes thedot product with respect to the${\displaystyle B}$-coordinates: thus${\displaystyle V}$ is isomorphic to${\displaystyle L^2(B,\mu )}$ for the discrete unit measure${\displaystyle \mu }$ on${\displaystyle B}$. Also${\displaystyle A}$ is unitarily equivalent to the multiplication operator${\displaystyle [Tf](v)=\lambda (v)f(v)}$, where${\displaystyle \lambda (v)}$ is the eigenvalue of${\displaystyle v\in B}$: that is,${\displaystyle A}$ multiplies each${\displaystyle B}$-coordinate by the corresponding eigenvalue${\displaystyle \lambda (v)}$, the action of a diagonal matrix. Finally, theoperator norm${\displaystyle |A|=|T|}$ is equal to the magnitude of the largest eigenvector${\displaystyle |\lambda {|}_{\mathrm{\infty }}}$.

The spectral theorem is the beginning of the vast research area of functional analysis calledoperator theory; see alsospectral measure.

There is also an analogous spectral theorem for boundednormal operators on Hilbert spaces. The only difference in the conclusion is that now${\displaystyle \lambda }$ may be complex-valued.

There is also a formulation of the spectral theorem in terms ofdirect integrals. It is similar to the multiplication-operator formulation, but more canonical.

Let${\displaystyle A}$ be a bounded self-adjoint operator and let${\displaystyle \sigma (A)}$ be the spectrum of${\displaystyle A}$. The direct-integral formulation of the spectral theorem associates two quantities to${\displaystyle A}$. First, a measure${\displaystyle \mu }$ on${\displaystyle \sigma (A)}$, and second, a family of Hilbert spaces${\displaystyle \{H_{\lambda }\},\lambda \in \sigma (A).}$ We then form the direct integral Hilbert space$${\displaystyle {\int }_{\mathbf{R}}^{\oplus }{H}_{\lambda }d\mu (\lambda ).}$$The elements of this space are functions (or "sections")${\displaystyle s(\lambda ),\lambda \in \sigma (A),}$ such that${\displaystyle s(\lambda )\in H_{\lambda }}$ for all${\displaystyle \lambda }$. The direct-integral version of the spectral theorem may be expressed as follows:^[7]

The spaces${\displaystyle H_{\lambda }}$ can be thought of as something like "eigenspaces" for${\displaystyle A}$. Note, however, that unless the one-element set${\displaystyle \lambda }$ has positive measure, the space${\displaystyle H_{\lambda }}$ is not actually a subspace of the direct integral. Thus, the${\displaystyle H_{\lambda }}$'s should be thought of as "generalized eigenspace"—that is, the elements of${\displaystyle H_{\lambda }}$ are "eigenvectors" that do not actually belong to the Hilbert space.

Although both the multiplication-operator and direct integral formulations of the spectral theorem express a self-adjoint operator as unitarily equivalent to a multiplication operator, the direct integral approach is more canonical. First, the set over which the direct integral takes place (the spectrum of the operator) is canonical. Second, the function we are multiplying by is canonical in the direct-integral approach: Simply the function${\displaystyle \lambda \mapsto \lambda }$.

A vector${\displaystyle \phi }$ is called acyclic vector for${\displaystyle A}$ if the vectors${\displaystyle \phi ,A\phi ,A^2\phi ,\dots }$ span a dense subspace of the Hilbert space. Suppose${\displaystyle A}$ is a bounded self-adjoint operator for which a cyclic vector exists. In that case, there is no distinction between the direct-integral and multiplication-operator formulations of the spectral theorem. Indeed, in that case, there is a measure${\displaystyle \mu }$ on the spectrum${\displaystyle \sigma (A)}$ of${\displaystyle A}$ such that${\displaystyle A}$ is unitarily equivalent to the "multiplication by${\displaystyle \lambda }$" operator on${\displaystyle L^2(\sigma (A),\mu )}$.^[8] This result represents${\displaystyle A}$ simultaneously as a multiplication operatorand as a direct integral, since${\displaystyle L^2(\sigma (A),\mu )}$ is just a direct integral in which each Hilbert space${\displaystyle H_{\lambda }}$ is just${\displaystyle \mathbb{C}}$.

Not every bounded self-adjoint operator admits a cyclic vector; indeed, by the uniqueness in the direct integral decomposition, this can occur only when all the${\displaystyle H_{\lambda }}$'s have dimension one. When this happens, we say that${\displaystyle A}$ has "simple spectrum" in the sense ofspectral multiplicity theory. That is, a bounded self-adjoint operator that admits a cyclic vector should be thought of as the infinite-dimensional generalization of a self-adjoint matrix with distinct eigenvalues (i.e., each eigenvalue has multiplicity one).

Although not every${\displaystyle A}$ admits a cyclic vector, it is easy to see that we can decompose the Hilbert space as a direct sum of invariant subspaces on which${\displaystyle A}$ has a cyclic vector. This observation is the key to the proofs of the multiplication-operator and direct-integral forms of the spectral theorem.

One important application of the spectral theorem (in whatever form) is the idea of defining afunctional calculus. That is, given a function${\displaystyle f}$ defined on the spectrum of${\displaystyle A}$, we wish to define an operator${\displaystyle f(A)}$. If${\displaystyle f}$ is simply a positive power,${\displaystyle f(x)=x^n}$, then${\displaystyle f(A)}$ is just the${\displaystyle n}$-th power of${\displaystyle A}$,${\displaystyle A^n}$. The interesting cases are where${\displaystyle f}$ is a nonpolynomial function such as a square root or an exponential. Either of the versions of the spectral theorem provides such a functional calculus.^[9] In the direct-integral version, for example,${\displaystyle f(A)}$ acts as the "multiplication by${\displaystyle f}$" operator in the direct integral:$${\displaystyle [f(A)s](\lambda )=f(\lambda )s(\lambda ).}$$That is to say, each space${\displaystyle H_{\lambda }}$ in the direct integral is a (generalized) eigenspace for${\displaystyle f(A)}$ with eigenvalue${\displaystyle f(\lambda )}$.

Many important linear operators which occur inanalysis, such asdifferential operators, areunbounded. There is also a spectral theorem forself-adjoint operators that applies in these cases. To give an example, every constant-coefficient differential operator is unitarily equivalent to a multiplication operator. Indeed, the unitary operator that implements this equivalence is theFourier transform; the multiplication operator is a type ofFourier multiplier.

In general, spectral theorem for self-adjoint operators may take several equivalent forms.^[10] Notably, all of the formulations given in the previous section for bounded self-adjoint operators—the projection-valued measure version, the multiplication-operator version, and the direct-integral version—continue to hold for unbounded self-adjoint operators, with small technical modifications to deal with domain issues. Specifically, the only reason the multiplication operator${\displaystyle A}$ on${\displaystyle L^2([0,1])}$ is bounded, is due to the choice of domain${\displaystyle [0,1]}$. The same operator on, e.g.,${\displaystyle L^2(\mathbb{R})}$ would be unbounded.

The notion of "generalized eigenvectors" naturally extends to unbounded self-adjoint operators, as they are characterized asnon-normalizable eigenvectors. Contrary to the case ofalmost eigenvectors, however, the eigenvalues can be real or complex and, even if they are real, do not necessarily belong to the spectrum. Though, for self-adjoint operators there always exist a real subset of "generalized eigenvalues" such that the corresponding set of eigenvectors iscomplete.^[11]

• Hahn-Hellinger theorem – Linear operator equal to its own adjointPages displaying short descriptions of redirect targets
• Spectral theory of compact operators
• Spectral theory of normal C*-algebras
• Borel functional calculus
• Spectral theory
• Matrix decomposition
• Canonical form
• Jordan decomposition, of which the spectral decomposition is a special case.
• Singular value decomposition, a generalisation of spectral theorem to arbitrary matrices.
• Eigendecomposition of a matrix
• Wiener–Khinchin theorem

• Sheldon Axler,Linear Algebra Done Right, Springer Verlag, 1997
• Hall, B.C. (2013),Quantum Theory for Mathematicians, Graduate Texts in Mathematics, vol. 267, Springer,Bibcode:2013qtm..book.....H,ISBN 978-1461471158
• Paul Halmos,"What Does the Spectral Theorem Say?",American Mathematical Monthly, volume 70, number 3 (1963), pages 241–247Other link
• de la Madrid Modino, R. (2001).Quantum mechanics in rigged Hilbert space language (PhD thesis). Universidad de Valladolid.
• M. Reed andB. Simon,Methods of Mathematical Physics, vols I–IV, Academic Press 1972.
• G. Teschl,Mathematical Methods in Quantum Mechanics with Applications to Schrödinger Operators,https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/, American Mathematical Society, 2009.
• Valter Moretti (2017).Spectral Theory and Quantum Mechanics; Mathematical Foundations of Quantum Theories, Symmetries and Introduction to the Algebraic Formulation 2nd Edition. Springer.ISBN 978-3-319-70705-1.

• Spectral theory
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