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.
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.
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.
Let
T
T
T\star
\|Tx\|=\|T*x\|
x
\|Tx\|2=\langleT*Tx,x\rangle=\langleTT*x,x\rangle=\|T*x\|2
T
T
T=T1+iT2
T1:=
T+T* | |
2 |
iT2:=
T-T* | |
2 |
,
T1T2=T2T1.
If
N
N
N*
N
N
Nk
N
k.
λ
N
\overline{λ}
N*.
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
N1
N2
A
N1A=AN2,
* | |
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.
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 1H−PV. The fact that T stabilizes V can be expressed as (1H−PV)TPV = 0, or TPV = PVTPV. The goal is to show that PVT(1H−PV) = 0.
Let X = PVT(1H−PV). 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. First we note that
\begin{align} XX*&=PVT(\boldsymbol{1}H-
2 | |
P | |
V) |
T*PV\\ &=PVT(\boldsymbol{1}H-PV)T*PV\\ &=PVTT*PV-PVTPVT*PV. \end{align}
Now using properties of the trace and of orthogonal projections we have:
\begin{align} \operatorname{tr}(XX*)&=\operatorname{tr}\left(
*P | |
P | |
V |
-PVTP
*P | |
V |
\right)\\ &=
*P | |
\operatorname{tr}(P | |
V) |
-\operatorname{tr}(PVTP
*P | |
V) |
\\ &=
2TT | |
\operatorname{tr}(P | |
V |
*)-
*) | |
\operatorname{tr}(P | |
VT |
\\ &=
*) | |
\operatorname{tr}(P | |
VTT |
-\operatorname{tr}(PVTP
*) | |
VT |
\\ &=
*) | |
\operatorname{tr}(P | |
VTT |
-
*) | |
\operatorname{tr}(TP | |
VT |
&&usingthehypothesisthatTstabilizesV\\ &=
*) | |
\operatorname{tr}(P | |
VTT |
-
*T) | |
\operatorname{tr}(P | |
VT |
\\ &=
*-T | |
\operatorname{tr}(P | |
V(TT |
*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.[3] However, for bounded normal operators, the orthogonal complement to a stable subspace may not be stable.[4] 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
\ell2
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
N*N=NN*.
Equivalently normal operators are precisely those for which[5]
\|Nx\|=\|N*x\|
l{D}(N)=l{D}(N*).
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)