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# Normal operator

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 Title: Normal operator Author: World Heritage Encyclopedia Language: English Subject: Collection: Operator Theory Publisher: World Heritage Encyclopedia Publication Date:

### Normal operator

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. 

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.

## Contents

• Properties 1
• Properties in finite-dimensional case 2
• Normal elements of algebras 3
• Unbounded normal operators 4
• Generalization 5
• References 6

## 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.

Let T be a bounded operator. The following are equivalent.

• T is normal.
• T* is normal.
• ||Tx|| = ||T*x|| for all x (use \|Tx\|^2 = \langle T^*Tx, x \rangle = \langle TT^*x, x \rangle = \|T^*x\|^2).
• The selfadjoint and anti-selfadjoint parts of T commute. That is, if we write T\equiv T_1+i T_2 with T_1:=\frac{T+T^*}{2} and i\,T_2:=\frac{T-T^*}{2},, then T_1T_2=T_2T_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. 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. λ is an eigenvalue of a normal operator N if and only if its complex conjugate \overline{\lambda} 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):

If N_1 and N_2 are normal operators and if A is a bounded linear operator such that N_1 A = A N_2, then N_1^* A = A N_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:

XX^* = P_VT(\boldsymbol{1}_H-P_V)^2T^*P_V= P_VT(\boldsymbol{1}_H-P_V)T^*P_V = P_VTT^*P_V - P_VTP_VT^*P_V,

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

\begin{align} \operatorname{tr}(XX^*) &= \operatorname{tr} \left ( P_VTT^*P_V - P_VTP_VT^*P_V \right ) \\ &= \operatorname{tr}(P_VTT^*P_V) - \operatorname{tr}(P_VTP_VT^*P_V) \\ &= \operatorname{tr}(P_V^2TT^*) - \operatorname{tr}(P_V^2TP_VT^*) \\ &= \operatorname{tr}(P_VTT^*) - \operatorname{tr}(P_VTP_VT^*) \\ &= \operatorname{tr}(P_VTT^*) - \operatorname{tr}(TP_VT^*) \qquad\qquad\text{using the hypothesis that } T \text{ stabilizes } V\\ &= \operatorname{tr}(P_VTT^*) - \operatorname{tr}(P_VT^*T) \\ &= \operatorname{tr}(P_V(TT^*-T^*T)) \\ &= 0. \end{align}

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 orthogonal complement to a stable subspace may not be stable. 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 \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:

An element x of an involutive algebra is said to be normal if xx* = x*x.

Selfadjoint and unitary elements are normal.

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

## 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 we can write

N^*N = NN^*.

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

with