Spectral theorem

In mathematics, particularly linear algebra and functional analysis, the spectral theorem is any of a number of results about linear operators or matrices. In broad terms, the spectral theorem provides conditions under which an operator or a matrix can be diagonalized (that is, represented as a diagonal matrix in some basis). 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, eigenvalue decomposition, or eigendecomposition, of the underlying vector space on which the operator acts.

Augustin Louis Cauchy proved the spectral theorem for self-adjoint 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.

Finite-dimensional case

Hermitian maps and Hermitian matrices

We begin by considering a Hermitian matrix on Cn or Rn.[clarification needed] More generally we consider a Hermitian map A on a finite-dimensional real or complex inner product space V endowed with a positive definite Hermitian inner product. The Hermitian condition means that for all x, yV,

$\langle A x ,\, y \rangle = \langle x ,\, A y \rangle .$

An equivalent condition is that A = A where A is the hermitian conjugate of A. In the case that A is identified with an Hermitian matrix, the matrix of A can be identified with its conjugate transpose. If A is a real matrix, this is equivalent to AT = A (that is, A is a symmetric matrix).

This condition easily 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 solutions to the characteristic polynomial.)

Theorem. 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 λ1 and eigenvector e1. Then since

$\lambda_1 \langle e_1, e_1 \rangle = \langle A (e_1), e_1 \rangle = \langle e_1, A(e_1) \rangle = \bar\lambda_1 \langle e_1, e_1 \rangle$

we find that λ1 is real. Now consider the space K = span{e1}, the orthogonal complement of e1. By Hermiticity, K is an invariant subspace of A. Applying the same argument to K shows that A has an eigenvector e2K. Finite induction then finishes the proof.

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. The easiest way to prove it is probably to consider A as a Hermitian matrix and use the fact that all eigenvalues of a Hermitian matrix are real.

If one chooses the eigenvectors of A as an orthonormal basis, the matrix representation of A in this basis is diagonal. Equivalently, A can be written as a linear combination of pairwise orthogonal projections, called its spectral decomposition. Let

$V_\lambda = \{\,v \in V: A v = \lambda v\,\}$

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 Vλ where the index ranges over eigenvalues. Let Pλ be the orthogonal projection onto Vλ and λ1, ..., λm the eigenvalues of A, one can write its spectral decomposition thus:

$A =\lambda_1 P_{\lambda_1} +\cdots+\lambda_m P_{\lambda_m}.$

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

Normal matrices

Main article: Normal matrix

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 AA = 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, 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

$A=U D U^*,$

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 Hilbert spaces in general, 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 Hilbert space V. 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. To prove this, we cannot rely on determinants to show existence of eigenvalues, but instead one can use a maximization argument analogous to the variational characterization of eigenvalues. The above spectral theorem holds for real or complex Hilbert spaces.

If the compactness assumption is removed, it is not true that every self adjoint operator has eigenvectors.

The next generalization we consider is that of bounded self-adjoint operators on a Hilbert space. Such operators may have no eigenvalues: for instance let A be the operator of multiplication by t on L2[0, 1], that is,

$[A \varphi](t) = t \varphi(t). \;$
Theorem
[3]
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:HL2μ(X) such that
$U^* T U = A,$
where T is the multiplication operator:
$[T \varphi](x) = f(x) \varphi(x).$
and $\|T\| = \|f\|_\infty$

This 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.

An alternative formulation of the spectral theorem expresses the operator A as an integral of the coordinate function over the operator's spectrum with respect to a projection-valued measure.

$A = \int_{\sigma(A)} \lambda \, d E_{\lambda} .$

When the normal 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.