In mathematics, especially functional analysis, a normal operator on a complex Hilbert space H is a continuous linear operator N : H -> H that commutes with its hermitian adjoint N*, that is: NN* = N*N.
Normal operators are important because the spectral theorem holds for them. The class of normal operators is well understood. Examples of normal operators are
A normal matrix is the matrix expression of a normal operator on the Hilbert space Cn.
Let T be a bounded operator. The following are equivalent.
If N is a normal operator, then N and N* have the same kernel and the same range. Consequently, the range of N is dense if and only if N 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 Nk coincides with that of N for any k. Every generalized eigenvalue of a normal operator is thus genuine. ? is an eigenvalue of a normal operator N if and only if its complex conjugate is an eigenvalue of N*. Eigenvectors of a normal operator corresponding to different eigenvalues are orthogonal, and a normal operator stabilizes the orthogonal complement of each of its eigenspaces. 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 of projection-valued measures. The residual spectrum of a normal operator is empty.
The product of normal operators that commute is again normal; this is nontrivial, but follows directly from Fuglede's theorem, which states (in a form generalized by Putnam):
A normal operator coincides with its Aluthge transform.
If a normal operator T on a finite-dimensional real[clarification needed] or complex Hilbert space (inner product space) H stabilizes a subspace V, then it also stabilizes its orthogonal complement V?. (This statement is trivial in the case where T is self-adjoint.)
Proof. Let PV be the orthogonal projection onto V. Then the orthogonal projection onto V? is 1H-PV. The fact that T stabilizes V can be expressed as (1H-PV)TPV = 0, or TPV = PVTPV. The goal is to show that PVT(1H-PV) = 0.
Let X = PVT(1H-PV). Since (A, B) ? tr(AB*) is an inner product on the space of endomorphisms of H, it is enough to show that tr(XX*) = 0. First we note that
Now using properties of the trace 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 the Hilbert-Schmidt inner product, defined by tr(AB*) suitably interpreted. However, for bounded normal operators, the orthogonal complement to a stable subspace may not be stable. It follows that the Hilbert space cannot in general be spanned by eigenvectors of a normal operator. Consider, for example, the bilateral 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 by Beurling's theorem.
The notion of normal operators generalizes to an involutive algebra:
An element x of an involutive algebra is said to be normal if xx* = x*x.
Self-adjoint and unitary elements are normal.
The most important case is when such an algebra is a C*-algebra.
The definition of normal operators naturally generalizes to some class of unbounded operators. Explicitly, a closed operator N is said to be normal if we can write
Here, the existence of the adjoint N* requires that the domain of N be dense, and the equality includes the assertion that the domain of N*N equals that of NN*, which is not necessarily the case in general.
Equivalently normal operators are precisely those for which
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)