Normal operator

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.

• T is normal.
• T* is normal.
• ||Tx|| = ||T*x|| for all x (use ${\displaystyle \|Tx\|^{2}=\langle T^{*}Tx,x\rangle =\langle TT^{*}x,x\rangle =\|T^{*}x\|^{2}}$).
• The self-adjoint and anti–self adjoint parts of T commute. That is, if we write ${\displaystyle T\equiv T_{1}+iT_{2}}$ with ${\displaystyle T_{1}:={\frac {T+T^{*}}{2}}}$ and ${\displaystyle i\,T_{2}:={\frac {T-T^{*}}{2}}}$, then ${\displaystyle T_{1}T_{2}=T_{2}T_{1}}$.[3]

If N is a normal operator, then N and N* have the same kernel and the same 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 N^k 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 a normal operator stabilizes the orthogonal complement of each of 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 version of the spectral theorem expressed in terms of projection-valued measures. The residual spectrum of a normal operator is empty.[4]

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 ${\displaystyle N_{1}}$ 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.

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 P_V be the orthogonal projection onto V. Then the orthogonal projection onto V^⊥ is 1_H−P_V. The fact that T stabilizes V can be expressed as (1_H−P_V)TP_V = 0, or TP_V = P_VTP_V. The goal is to show that P_VT(1_H−P_V) = 0.

Let X = P_VT(1_H−P_V). 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

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

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

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

The same argument goes through for compact normal operators in infinite dimensional Hilbert spaces, where one make use of the Hilbert-Schmidt inner product, defined by tr(AB*) suitably interpreted.[5] However, for bounded normal operators, the orthogonal complement to a stable subspace may not be stable.[6] 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 ${\displaystyle \ell ^{2}}$, which is normal, but has no eigenvalues.

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

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 we can write

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

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[7]

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

with

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

The spectral theorem still holds for unbounded (normal) operators. The proofs work by reduction to bounded (normal) operators.[8][9]

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)

• Quasinormal operators
• Subnormal operators
• Hyponormal operators
• Paranormal operators
• Normaloids