(mathematics) A linear operator where composing it with its adjoint operator in either order gives the same result. Also known as normal transformation.
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In mathematics, especially functional analysis, a normal operator on a complex Hilbert space
is a continuous linear operator

that commutes with its hermitian adjoint N*:

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

; (also, anti-selfadjoint operators:
)
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Normal operators are characterized by the spectral theorem. A compact normal operator (in particular, a normal operator on a finite-dimensional linear space) is unitarily diagonalizable.
Let T be a bounded operator. The following are equivalent.
).
with
rsp.
commute [1] .If N is a normal operator, then N and N* have the same kernel and range. Consequently, the range of N is dense if and only if N is injective. Put in another way, the kernel of a normal operator is the orthogonal complement of its range; thus, 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 N if and only if its complex conjugate
is an eigenvalue of
Eigenvectors of a normal operator corresponding to different eigenvalues are orthogonal, and it stabilizes orthogonal complements to its eigenspaces .[2] 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 generalization in terms of projection-valued measures. Residual spectrum of a normal operator is empty.[2]
The product of normal operators that commute is again normal; this is nontrivial and follows from Fuglede's theorem, which states (in a form generalized by Putnam):
and
are normal operators and if A is a bounded linear operator such that
, then
.The operator norm of a normal operator equals its numerical radius and spectral radius.
A normal operator coincides with its Aluthge transform.
If a normal operator
on a finite-dimensional real or complex Hilbert space (inner product space)
stabilizes a subspace
, then it also stabilizes its orthogonal complement
. (This statement is trivial in the case where
is self-adjoint )
Proof. Denote by
the orthogonal projection onto
. Then the orthogonal projection onto
is
. The fact that
stabilizes
can be expressed as
, or
. The goal is to show that
. Since
is an inner product on the space of endomorphisms of
, it is enough to show that
. This follows from a direct computation, using properties of the trace and of orthogonal projections:
,
.
The same argument goes through for compact normal operators in infinite dimensional Hilbert spaces, where one make use of the Hilbert-Schmidt inner product.[3] However, for bounded normal operators orthogonal complement to a stable subspace may not be stable.[4] It follows that such subspaces cannot be spanned by eigenvectors. Consider, for example, the bilateral shift, which has no eigenvalues. The invariant subspaces of the bilateral shift is characterized by Beurling's theorem.
The notion of normal operators generalizes to an involutive algebra; namely, an element x of an involutive algebra is said to be normal if
. The most important case is when such an algebra is a C*-algebra. A positive element is an example of a normal element.
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
implies that the domain of
is dense, and the equality implies that the domain of
equals that of
, which is not necessarily the case in general.
The spectral theorem still holds for unbounded normal operators, but usually requires a different proof.
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)
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