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Inmathematics, especiallyfunctional analysis, anormal operator on acomplexHilbert space is acontinuouslinear operator thatcommutes with itsHermitian adjoint, that is:.[1]
Normal operators are important because thespectral theorem holds for them. The class of normal operators is well understood. Examples of normal operators are
Anormal matrix is the matrix expression of a normal operator on the Hilbert space.
Normal operators are characterized by thespectral theorem. Acompact normal operator (in particular, a normal operator on afinite-dimensionalinner product space) isunitarily diagonalizable.[2]
Let be a bounded operator. The following are equivalent.
If is a bounded normal operator, then and have the same kernel and the same range. Consequently, the range of is dense if and only if is injective.[clarification needed] Put in another way, the kernel of a normal operator is the orthogonal complement of its range. It follows that the kernel of the operator coincides with that of for any Every generalized eigenvalue of a normal operator is thus genuine. is an eigenvalue of a normal operator if and only if its complex conjugate is an eigenvalue of Eigenvectors of a normal operator corresponding to different eigenvalues are orthogonal, and a normal operator stabilizes the orthogonal complement of each of its eigenspaces.[3] This implies the usual spectral theorem: every normal operator on a finite-dimensional space is diagonalizable by a unitary operator. There is also an infinite-dimensional version of the spectral theorem expressed in terms ofprojection-valued measures. The residual spectrum of a normal operator is empty.[3]
The product of normal operators that commute is again normal; this is nontrivial, but follows directly fromFuglede's theorem, which states (in a form generalized by Putnam):
The operator norm of a normal operator equals itsnumerical radius[clarification needed] andspectral radius.
A normal operator coincides with itsAluthge transform.
If a normal operatorT on afinite-dimensional real[clarification needed] or complex Hilbert space (inner product space)H stabilizes a subspaceV, then it also stabilizes its orthogonal complementV⊥. (This statement is trivial in the case whereT is self-adjoint.)
Proof. LetPV be the orthogonal projection ontoV. Then the orthogonal projection ontoV⊥ is1H−PV. The fact thatT stabilizesV can be expressed as (1H−PV)TPV = 0, orTPV =PVTPV. The goal is to show thatPVT(1H−PV) = 0.
LetX =PVT(1H−PV). Since (A,B) ↦ tr(AB*) is aninner product on the space of endomorphisms ofH, it is enough to show that tr(XX*) = 0. First it is noted that
Now using properties of thetrace and of orthogonal projections we have:
The same argument goes through for compact normal operators in infinite dimensional Hilbert spaces, where one make use of theHilbert-Schmidt inner product, defined by tr(AB*) suitably interpreted.[4] However, for bounded normal operators, the orthogonal complement to a stable subspace may not be stable.[5] It follows that the Hilbert space cannot in general be spanned by eigenvectors of a normal operator. Consider, for example, thebilateral shift (or two-sided shift) acting on, which is normal, but has no eigenvalues.
The invariant subspaces of a shift acting on Hardy space are characterized byBeurling's theorem.
The notion of normal operators generalizes to an involutive algebra:
An element of an involutive algebra is said to be normal if.
Self-adjoint and unitary elements are normal.
The most important case is when such an algebra is aC*-algebra.
The definition of normal operators naturally generalizes to some class of unbounded operators. Explicitly, a closed operatorN is said to be normal if
Here, the existence of the adjointN* requires that the domain ofN be dense, and the equality includes the assertion that the domain ofN*N equals that ofNN*, which is not necessarily the case in general.
Equivalently normal operators are precisely those for which[6]
with
The spectral theorem still holds for unbounded (normal) operators. The proofs work by reduction to bounded (normal) operators.[7][8]
The success of the theory of normal operators led to several attempts for generalization by weakening the commutativity requirement. Classes of operators that include normal operators are (in order of inclusion)