Subnormal operator

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In 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.

1 Definition
2 Normality, quasinormality, and subnormality
- 2.1 Normal operators
- 2.2 Quasinormal operators
3 Minimal normal extension
- 3.1 Non-uniqueness of normal extensions
- 3.2 Minimality

[edit] 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

$N = \begin{bmatrix} A & B\\ 0 & C\end{bmatrix}$

for some bounded operators

$B : H^{\perp} \rightarrow H, \quad \mbox{and} \quad C : H^{\perp} \rightarrow H^{\perp}.$

[edit] Normality, quasinormality, and subnormality

[edit] Normal operators

Every normal operator is subnormal by definition, but the converse is not true in general. An 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

$H \oplus H$

$U = \begin{bmatrix} A & I - AA^* \\ 0 & - A^* \end{bmatrix}.$

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.

[edit] 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,

$V = \begin{bmatrix} U & I - UU^* \\ 0 & - U^* \end{bmatrix} = \begin{bmatrix} U & D_{U^*} \\ 0 & - U^* \end{bmatrix} .$

Define

$Q = \begin{bmatrix} P & 0 \\ 0 & P \end{bmatrix}.$

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

$N^*N = QV^*VQ = Q^2 = \begin{bmatrix} P^2 & 0 \\ 0 & P^2 \end{bmatrix}.$

On the other hand,

$N N^* = \begin{bmatrix} UP^2U^* + D_{U^*} P^2 D_{U^*} & -D_{U^*}P^2 U \\ -U^* P^2 D_{U^*} & U^* P^2 U \end{bmatrix}.$

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 normality.

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.

[edit] Minimal normal extension

[edit] 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

$B (\cdots, a_{-1}, {\hat a_0}, a_1, \cdots) = (\cdots, {\hat a_{-1}}, a_0, a_1, \cdots),$

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

$B = \begin{bmatrix} A & I - AA^* \\ 0 & A^* \end{bmatrix}.$

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

$B' = \begin{bmatrix} A & I - AA^* \\ 0 & - A^* \end{bmatrix}$

whose action is described by

$B' (\cdots, a_{-2}, a_{-1}, {\hat a_0}, a_1, a_2, \cdots) = (\cdots, - a_{-2}, {\hat a_{-1}}, a_0, a_1, a_2, \cdots).$

[edit] 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 an unitary operator

$U: K_1 \rightarrow K_2.$

Also, the following interwining relationship holds:

$U B_1 = B_2 U. \,$

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

$\sum_{i=0}^n (B_1^*)^i h_i = h_0+ B_1 ^* h_1 + (B_1^*)^2 h_2 + \cdots + (B_1^*)^n h_n \quad \mbox{where} \quad h_i \in H.$

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:

$U \sum_{i=0}^n (B_1^*)^i h_i = \sum_{i=0}^n (B_2^*)^i h_i$

Because

$\langle \sum_{i=0}^n (B_1^*)^i h_i, \sum_{j=0}^n (B_1^*)^j h_j\rangle = \sum_{i j} \langle h_i, (B_1)^i (B_1^*)^j h_j\rangle = \sum_{i j} \langle (B_2)^j h_i, (B_2)^i h_j\rangle = \langle \sum_{i=0}^n (B_2^*)^i h_i, \sum_{j=0}^n (B_2^*)^j h_j\rangle ,$