# Normal operator

{{ safesubst:#invoke:Unsubst||$N=Refimprove |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} In mathematics, especially functional analysis, a normal operator on a complex Hilbert space H is a continuous linear operator N : HH 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

A normal matrix is the matrix expression of a normal operator on the Hilbert space Cn.

## Properties

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.[2]

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 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 Nk coincides with that of N for any k. Every generalized eigenvalue of a normal operator is thus genuine. λ is an eigenvalue of a normal operator N if and only if its complex conjugate ${\displaystyle {\overline {\lambda }}}$ is an eigenvalue of N*. Eigenvectors of a normal operator corresponding to different eigenvalues are orthogonal, and it stabilizes orthogonal complements to its eigenspaces.[4] 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.[4]

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):

If ${\displaystyle N_{1}}$ and ${\displaystyle N_{2}}$ are normal operators and if A is a bounded linear operator such that ${\displaystyle N_{1}A=AN_{2}}$, then ${\displaystyle N_{1}^{*}A=AN_{2}^{*}}$.

The operator norm of a normal operator equals its numerical radius and spectral radius.

A normal operator coincides with its Aluthge transform.

## Properties in finite-dimensional case

If a normal operator T on a finite-dimensional real 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 1HPV. The fact that T stabilizes V can be expressed as (1HPV)TPV = 0, or TPV = PVTPV. The goal is to show that X := PVT(1HPV) = 0. 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. But first we express XX* in terms of orthogonal projections:

${\displaystyle XX^{*}=P_{V}T({\boldsymbol {1}}_{H}-P_{V})^{2}T^{*}P_{V}=P_{V}T({\boldsymbol {1}}_{H}-P_{V})T^{*}P_{V}=P_{V}TT^{*}P_{V}-P_{V}TP_{V}T^{*}P_{V}}$,

Now using properties of the trace and of orthogonal projections we have:

{\displaystyle {\begin{aligned}\operatorname {tr} (XX^{*})&=\operatorname {tr} \left(P_{V}TT^{*}P_{V}-P_{V}TP_{V}T^{*}P_{V}\right)\\&=\operatorname {tr} (P_{V}TT^{*}P_{V})-\operatorname {tr} (P_{V}TP_{V}T^{*}P_{V})\\&=\operatorname {tr} (P_{V}^{2}TT^{*})-\operatorname {tr} (P_{V}^{2}TP_{V}T^{*})\\&=\operatorname {tr} (P_{V}TT^{*})-\operatorname {tr} (P_{V}TP_{V}T^{*})\\&=\operatorname {tr} (P_{V}TT^{*})-\operatorname {tr} (TP_{V}T^{*})\qquad \qquad {\text{using the hypothesis that }}T{\text{ stabilizes }}V\\&=\operatorname {tr} (P_{V}TT^{*})-\operatorname {tr} (P_{V}T^{*}T)\\&=\operatorname {tr} (P_{V}(TT^{*}-T^{*}T))\\&=0.\end{aligned}}}

The same argument goes through for compact normal operators in infinite dimensional Hilbert spaces, where one make use of the Hilbert-SchmidtTemplate:Dn inner product, defined by tr(AB*) suitably interpreted.[5] However, for bounded normal operators orthogonal complement to a stable subspace may not be stable.[6] It follows that the Hilbert space cannot be spanned by eigenvectors of such an operator. Consider, for example, the bilateral shift (or two-sided shift) acting on ${\displaystyle \ell ^{2}}$, which is normal, but has no eigenvalues.

The invariant subspaces of a shift acting on Hardy space are characterized by Beurling's theorem.

## Normal elements of algebras

The notion of normal operators generalizes to an involutive algebra; namely, an element x of an involutive algebra is said to be normal if xx* = x*x. The most important case is when such an algebra is a C*-algebra. A positive element is an example of a normal element.

## Unbounded normal operators

The definition of normal operators naturally generalizes to some class of unbounded operators. Explicitly, a closed operator N is said to be normal if

${\displaystyle N^{*}N=NN^{*}}$

Here, the existence of the adjoint N* implies that the domain of N is dense, and the equality implies 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:[7]
${\displaystyle \|Nx\|=\|N^{*}x\|}$ with ${\displaystyle {\mathcal {D}}(N)={\mathcal {D}}(N^{*})}$

The spectral theorem still holds for unbounded normal operators, but usually requires a different proof.

## Generalization

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)

## Notes

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3. In contrast, for the important class of Creation and annihilation operators of, e.g., quantum field theory, they don't commute
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7. Weidmann, Lineare Operatoren in Hilberträumen, chapter 4.3, p. 176,

## References

• Hoffman, Kenneth and Kunze, Ray. Linear Algebra. Second Edition. 1971. Prentice-Hall, Inc.