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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*.^{[1]}

Normal operators are important because the spectral theorem holds for them. The class of normal operators is well understood. Examples of normal operators are

- unitary operators:
*N**=*N*^{−1} - Hermitian operators (i.e., self-adjoint operators):
*N**=*N* - skew-Hermitian operators:
*N**= −*N* - positive operators:
*N*=*MM**for some*M*(so*N*is self-adjoint).

A normal matrix is the matrix expression of a normal operator on the Hilbert space **C**^{n}.

Normal operators are characterized by the spectral theorem. A compact normal operator (in particular, a normal operator on a finite-dimensional inner product space) is unitarily diagonalizable.^{[2]}

Let be a bounded operator. The following are equivalent.

- is normal.
- is normal.
- for all (use ).
- The self-adjoint and anti–self adjoint parts of commute. That is, if is written as with and then
^{[note 1]}

If is a 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 of projection-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 from Fuglede's theorem, which states (in a form generalized by Putnam):

- If and are normal operators and if is a bounded linear operator such that then .

The operator norm of a normal operator equals its numerical radius^{[clarification needed]} and spectral radius.

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 *P _{V}* be the orthogonal projection onto

Let *X* = *P _{V}T*(

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.^{[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, 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

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^{[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)