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Normal operator

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(on a complex Hilbert space) continuous linear operator
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Inmathematics, especiallyfunctional analysis, anormal operator on acomplexHilbert spaceH{\displaystyle H} is acontinuouslinear operatorN:HH{\displaystyle N\colon H\rightarrow H} thatcommutes with itsHermitian adjointN{\displaystyle N^{\ast }}, that is:NN=NN{\displaystyle N^{\ast }N=NN^{\ast }}.[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 spaceCn{\displaystyle \mathbb {C} ^{n}}.

Properties

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Normal operators are characterized by thespectral theorem. Acompact normal operator (in particular, a normal operator on afinite-dimensionalinner product space) isunitarily diagonalizable.[2]

LetT{\displaystyle T} be a bounded operator. The following are equivalent.

IfN{\displaystyle N} is a bounded normal operator, thenN{\displaystyle N} andN{\displaystyle N^{*}} have the same kernel and the same range. Consequently, the range ofN{\displaystyle N} is dense if and only ifN{\displaystyle 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 operatorNk{\displaystyle N^{k}} coincides with that ofN{\displaystyle N} for anyk.{\displaystyle k.} Every generalized eigenvalue of a normal operator is thus genuine.λ{\displaystyle \lambda } is an eigenvalue of a normal operatorN{\displaystyle N} if and only if its complex conjugateλ¯{\displaystyle {\overline {\lambda }}} is an eigenvalue ofN.{\displaystyle 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.[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):

IfN1{\displaystyle N_{1}} andN2{\displaystyle N_{2}} are normal operators and ifA{\displaystyle A} is a bounded linear operator such thatN1A=AN2,{\displaystyle N_{1}A=AN_{2},} thenN1A=AN2{\displaystyle N_{1}^{*}A=AN_{2}^{*}}.

The operator norm of a normal operator equals itsnumerical radius[clarification needed] andspectral radius.

A normal operator coincides with itsAluthge transform.

Properties in finite-dimensional case

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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 is1HPV. The fact thatT stabilizesV can be expressed as (1HPV)TPV = 0, orTPV =PVTPV. The goal is to show thatPVT(1HPV) = 0.

LetX =PVT(1HPV). 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

XX=PVT(1HPV)2TPV=PVT(1HPV)TPV=PVTTPVPVTPVTPV.{\displaystyle {\begin{aligned}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}.\end{aligned}}}

Now using properties of thetrace and of orthogonal projections we have:

tr(XX)=tr(PVTTPVPVTPVTPV)=tr(PVTTPV)tr(PVTPVTPV)=tr(PV2TT)tr(PV2TPVT)=tr(PVTT)tr(PVTPVT)=tr(PVTT)tr(TPVT)using the hypothesis that T stabilizes V=tr(PVTT)tr(PVTT)=tr(PV(TTTT))=0.{\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^{*})&&{\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 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 on2(Z){\displaystyle \ell ^{2}(\mathbb {Z} )}, which is normal, but has no eigenvalues.

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

Normal elements of algebras

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The notion of normal operators generalizes to an involutive algebra:

An elementx{\displaystyle x} of an involutive algebra is said to be normal ifxx=xx{\displaystyle x^{\ast }x=xx^{\ast }}.

Self-adjoint and unitary elements are normal.

The most important case is when such an algebra is aC*-algebra.

Unbounded normal operators

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

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

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]

Nx=Nx{\displaystyle \|Nx\|=\|N^{*}x\|\qquad }

with

D(N)=D(N).{\displaystyle {\mathcal {D}}(N)={\mathcal {D}}(N^{*}).}

The spectral theorem still holds for unbounded (normal) operators. The proofs work by reduction to bounded (normal) operators.[7][8]

Generalization

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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)

See also

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Notes

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  1. ^In contrast, for the important class ofCreation and annihilation operators of, e.g.,quantum field theory, they don't commute

References

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  1. ^Hoffman, Kenneth;Kunze, Ray (1971),Linear algebra (2nd ed.), Englewood Cliffs, N.J.: Prentice-Hall, Inc., p. 312,MR 0276251
  2. ^Hoffman & Kunze (1971), p. 317.
  3. ^abNaylor, Arch W.; Sell George R. (1982).Linear Operator Theory in Engineering and Sciences. New York: Springer.ISBN 978-0-387-95001-3.Archived from the original on 2021-06-26. Retrieved2021-06-26.
  4. ^Andô, Tsuyoshi (1963). "Note on invariant subspaces of a compact normal operator".Archiv der Mathematik.14:337–340.doi:10.1007/BF01234964.S2CID 124945750.
  5. ^Garrett, Paul (2005)."Operators on Hilbert spaces"(PDF).Archived(PDF) from the original on 2011-09-18. Retrieved2011-07-01.
  6. ^Weidmann, Lineare Operatoren in Hilberträumen, Chapter 4, Section 3
  7. ^Alexander Frei, Spectral Measures, Mathematics Stack Exchange,ExistenceArchived 2021-06-26 at theWayback Machine,UniquenessArchived 2021-06-26 at theWayback Machine
  8. ^John B. Conway, A Course in Functional Analysis, Second Edition, Chapter X, Section §4
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