# Spectral theorem

In mathematics, particularly linear algebra and functional analysis, a **spectral theorem** is a result about when a linear operator or matrix can be diagonalized (that is, represented as a diagonal 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 on finite-dimensional vector spaces but requires some modification for operators on infinite-dimensional spaces. In general, the spectral theorem identifies a class of linear operators that can be modeled by multiplication operators, which are as simple as one can hope to find. In more abstract language, the spectral theorem is a statement about commutative C*-algebras. See also spectral theory for a historical perspective.

Examples of operators to which the spectral theorem applies are self-adjoint operators or more generally normal operators on Hilbert spaces.

The spectral theorem also provides a canonical decomposition, called the **spectral decomposition**, of the underlying vector space on which the operator acts.

Augustin-Louis Cauchy proved the spectral theorem for symmetric matrices, i.e., that every real, symmetric matrix is diagonalizable. In addition, Cauchy was the first to be systematic about determinants.^{[1]}^{[2]} The spectral theorem as generalized by John von Neumann is today perhaps the most important result of operator theory.

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

An equivalent condition is that *A*^{*} = *A*, where *A*^{*} is the Hermitian conjugate of *A*. In the case that *A* is identified with a Hermitian matrix, the matrix of *A*^{*} can be identified with its conjugate transpose. (If *A* is a real matrix, then this is equivalent to *A*^{T} = *A*, that is, *A* is a symmetric matrix.)

This condition implies that all eigenvalues of a Hermitian map are real: it is enough to apply it to the case when *x* = *y* is an eigenvector. (Recall that an eigenvector of a linear map *A* is a (non-zero) vector *x* such that *Ax* = *λx* for some scalar *λ*. The value *λ* is the corresponding eigenvalue. Moreover, the eigenvalues are roots of the characteristic polynomial.)

**Theorem**. If *A* is Hermitian, then there exists an orthonormal basis of *V* consisting of eigenvectors of *A*. Each eigenvalue is real.

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

By the fundamental theorem of algebra, applied to the characteristic polynomial of *A*, there is at least one eigenvalue *λ*_{1} and eigenvector *e*_{1}. Then since

The spectral theorem holds also for symmetric maps on finite-dimensional real inner product spaces, but the existence of an eigenvector does not follow immediately from the fundamental theorem of algebra. To prove this, consider *A* as a Hermitian matrix and use the fact that all eigenvalues of a Hermitian matrix are real.

The matrix representation of *A* 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 its **spectral decomposition**. Let

be the eigenspace corresponding to an eigenvalue *λ*. Note that the definition does not depend on any choice of specific eigenvectors. *V* is the orthogonal direct sum of the spaces *V*_{λ} where the index ranges over eigenvalues.

In other words, if *P*_{λ} denotes the orthogonal projection onto *V*_{λ}, and *λ*_{1}, ..., *λ*_{m} are the eigenvalues of *A*, then the spectral decomposition may be written as

The spectral decomposition is a special case of both the Schur decomposition and the singular value decomposition.

The spectral theorem extends to a more general class of matrices. Let *A* be an operator on a finite-dimensional inner product space. *A* is said to be normal if *A*^{*}*A* = *AA*^{*}. One can show that *A* is normal if and only if it is unitarily diagonalizable. Proof: By the Schur decomposition, we can write any matrix as *A* = *UTU*^{*}, where *U* is unitary and *T* is upper-triangular.
If *A* is normal, then one sees that *TT*^{*} = *T*^{*}*T*. Therefore, *T* must be diagonal since a normal upper triangular matrix is diagonal (see normal matrix). The converse is obvious.

In other words, *A* is normal if and only if there exists a unitary matrix *U* such that

where *D* is a diagonal matrix. Then, the entries of the diagonal of *D* are the eigenvalues of *A*. The column vectors of *U* are the eigenvectors of *A* and they are orthonormal. Unlike the Hermitian case, the entries of *D* need not be real.

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

**Theorem**. Suppose *A* is a compact self-adjoint operator on a (real or complex) Hilbert space *V*. Then there is an orthonormal basis of *V* consisting of eigenvectors of *A*. Each eigenvalue is real.

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 is *not* true that every self-adjoint operator has eigenvectors.

When the self-adjoint operator in question is compact, 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. The significance of this result is that multiplication operators are in many ways easy to understand.

**Theorem**.^{[6]} Let *A* be a bounded self-adjoint operator on a Hilbert space *H*. Then there is a measure space (*X*, Σ, *μ*) and a real-valued essentially bounded measurable function *f* on *X* and a unitary operator *U*:*H* → *L*^{2}_{μ}(*X*) such that

The spectral theorem is the beginning of the vast research area of functional analysis called operator theory; see also the spectral measure.

There is also an analogous spectral theorem for bounded normal operators on Hilbert spaces. The only difference in the conclusion is that now *f* may be complex-valued.

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

Many important linear operators which occur in analysis, such as differential operators, are unbounded. There is also a spectral theorem for self-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 the Fourier transform; the multiplication operator is a type of Fourier 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.