Let H be a Hilbert space. A bounded operatorA 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 operatorN of the form for some bounded 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 denserange. 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 by 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 decompositionA = UP, the partial isometryU and positive operatorP 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, Define The operator N = VQ is clearly an extension of A. We show it is a normal extension via direct calculation. Unitarity of V means On the other hand, Because UP = PU and P is self adjoint, we have U*P = PU* and DU*P = DU*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, letA be the unilateral shift, on l2. One normal extension is the bilateral shiftB on l2 defined by where ˆ denotes the zero-th position. B can be expressed in terms of the operator matrix Another normal extension is given by the unitary dilation B' of A defined above: whose action is described by
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. One can show that if two operators B1 and B2 are minimal extensions on K1 and K2, respectively, then there exists a unitary operator Also, the following intertwining relationship holds: This can be shown constructively. Consider the set Sconsisting of vectors of the following form: 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: Because , the operator U is unitary. Direct computation also shows 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.
