Subnormal operator

In mathematics, especially operator theory, subnormal operators are bounded operators on a Hilbert space defined by weakening the requirements for normal operators. Some examples of subnormal operators are isometries and Toeplitz operators with analytic symbols.

Definition

Let H be a Hilbert space. A bounded operator A on H is said to be subnormal if A has a normal extension. In other words, A is subnormal if there exists a Hilbert space K such that H can be embedded in K and there exists a normal operator N of the form

$N=\left[\begin{array}{cc}A& B\\ 0& C\end{array}\right]$

for some bounded operators

$B:H^{\perp }\to H,\text{and}C:H^{\perp }\to H^{\perp }.$

Normality, quasinormality, and subnormality

Normal operators

Every normal operator is subnormal by definition, but the converse is not true in general. A simple class of examples can be obtained by weakening the properties of unitary operators. A unitary operator is an isometry with dense range. Consider now an isometry A whose range is not necessarily dense. A concrete example of such is the unilateral shift, which is not normal. But A is subnormal and this can be shown explicitly. Define an operator U on

$H\oplus H$

by

$U=\left[\begin{array}{cc}A& I-A{A}^{\ast }\\ 0& -A^{\ast }\end{array}\right].$

Direct calculation shows that U is unitary, therefore a normal extension of A. The operator U is called the unitary dilation of the isometry A.

Quasinormal operators

An operator A is said to be quasinormal if A commutes with A*A. A normal operator is thus quasinormal; the converse is not true. A counter example is given, as above, by the unilateral shift. Therefore, the family of normal operators is a proper subset of both quasinormal and subnormal operators. A natural question is how are the quasinormal and subnormal operators related.

We will show that a quasinormal operator is necessarily subnormal but not vice versa. Thus the normal operators is a proper subfamily of quasinormal operators, which in turn are contained by the subnormal operators. To argue the claim that a quasinormal operator is subnormal, recall the following property of quasinormal operators:

Fact: A bounded operator A is quasinormal if and only if in its polar decomposition A = UP, the partial isometry U and positive operator P commute.

Given a quasinormal A, the idea is to construct dilations for U and P in a sufficiently nice way so everything commutes. Suppose for the moment that U is an isometry. Let V be the unitary dilation of U,

$V=\left[\begin{array}{cc}U& I-U{U}^{\ast }\\ 0& -U^{\ast }\end{array}\right]=\left[\begin{array}{cc}U& D_{U^{\ast }}\\ 0& -U^{\ast }\end{array}\right].$

Define

$Q=\left[\begin{array}{cc}P& 0\\ 0& P\end{array}\right].$

The operator N = VQ is clearly an extension of A. We show it is a normal extension via direct calculation. Unitarity of V means

$N^{\ast }N=Q{V}^{\ast }VQ=Q^2=\left[\begin{array}{cc}{P}^2& 0\\ 0& P^2\end{array}\right].$

On the other hand,

$N{N}^{\ast }=\left[\begin{array}{cc}U{P}^2{U}^{\ast }+D_{U^{\ast }}{P}^2{D}_{U^{\ast }}& -D_{U^{\ast }}{P}^{2}U\\ -U^{\ast }{P}^2{D}_{U^{\ast }}& U^{\ast }{P}^{2}U\end{array}\right].$

Because UP = PU and P is self adjoint, we have U*P = PU* and D_U*P = D_U*P. Comparing entries then shows N is normal. This proves quasinormality implies subnormality.

For a counter example that shows the converse is not true, consider again the unilateral shift A. The operator B = A + s for some scalar s remains subnormal. But if B is quasinormal, a straightforward calculation shows that A*A = AA*, which is a contradiction.

Minimal normal extension

Non-uniqueness of normal extensions

Given a subnormal operator A, its normal extension B is not unique. For example, let A be the unilateral shift, on l²(N). One normal extension is the bilateral shift B on l²(Z) defined by

$B(\dots ,a_{-1},{\hat{a}}_0,a_1,\dots )=(\dots ,{\hat{a}}_{-1},a_0,a_1,\dots ),$

where ˆ denotes the zero-th position. B can be expressed in terms of the operator matrix

$B=\left[\begin{array}{cc}A& I-A{A}^{\ast }\\ 0& A^{\ast }\end{array}\right].$

Another normal extension is given by the unitary dilation B' of A defined above:

$B^{\prime }=\left[\begin{array}{cc}A& I-A{A}^{\ast }\\ 0& -A^{\ast }\end{array}\right]$

whose action is described by

$B^{\prime }(\dots ,a_{-2},a_{-1},{\hat{a}}_0,a_1,a_2,\dots )=(\dots ,-a_{-2},{\hat{a}}_{-1},a_0,a_1,a_2,\dots ).$

Minimality

Thus one is interested in the normal extension that is, in some sense, smallest. More precisely, a normal operator B acting on a Hilbert space K is said to be a minimal extension of a subnormal A if K' ⊂ K is a reducing subspace of B and H ⊂ K' , then K' = K. (A subspace is a reducing subspace of B if it is invariant under both B and B*.)

One can show that if two operators B₁ and B₂ are minimal extensions on K₁ and K₂, respectively, then there exists a unitary operator

$U:K_1\to K_2.$

Also, the following intertwining relationship holds:

$U{B}_1=B_{2}U.$

This can be shown constructively. Consider the set S consisting of vectors of the following form:

$\sum _{i=0}^n(B_1^{\ast }{)}^i{h}_i=h_0+B_1^{\ast }{h}_1+(B_1^{\ast }{)}^2{h}_2+\cdots +(B_1^{\ast }{)}^n{h}_n\text{where}{h}_i\in H.$

Let K' ⊂ K₁ be the subspace that is the closure of the linear span of S. By definition, K' is invariant under B₁* and contains H. The normality of B₁ and the assumption that H is invariant under B₁ imply K' is invariant under B₁. Therefore, K' = K₁. The Hilbert space K₂ can be identified in exactly the same way. Now we define the operator U as follows:

$U\sum _{i=0}^n(B_1^{\ast }{)}^i{h}_i=\sum _{i=0}^n(B_2^{\ast }{)}^i{h}_i$

Because

$⟨\sum _{i=0}^n(B_1^{\ast }{)}^i{h}_i,\sum _{j=0}^n(B_1^{\ast }{)}^j{h}_j⟩=\sum _{ij}⟨h_i,(B_1{)}^i(B_1^{\ast }{)}^j{h}_j⟩=\sum _{ij}⟨(B_2{)}^j{h}_i,(B_2{)}^i{h}_j⟩=⟨\sum _{i=0}^n(B_2^{\ast }{)}^i{h}_i,\sum _{j=0}^n(B_2^{\ast }{)}^j{h}_j⟩,$

, the operator U is unitary. Direct computation also shows (the assumption that both B₁ and B₂ are extensions of A are needed here)

$\text{if }g=\sum _{i=0}^n(B_1^{\ast }{)}^i{h}_i,$

$\text{then }U{B}_{1}g=B_{2}Ug=\sum _{i=0}^n(B_2^{\ast }{)}^{i}A{h}_i.$

When B₁ and B₂ are not assumed to be minimal, the same calculation shows that above claim holds verbatim with U being a partial isometry.