subnormal operator

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

:

for some bounded operators

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

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

:$H\; \backslash oplus\; H$

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.

An operator A is said to be quasinormal if A commutes with A*A.{{citation|author=John B. Conway|title=The Theory of Subnormal Operators|url=https://books.google.com/books?id=Ho7yBwAAQBAJ|accessdate=15 June 2017|year=1991|publisher=American Mathematical Soc.|isbn=978-0-8218-1536-6|page=29|chapter=11}} 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.{{citation|author1=John B. Conway|author2=Robert F. Olin|title=A Functional Calculus for Subnormal Operators II|url=https://books.google.com/books?id=yQXUCQAAQBAJ|accessdate=15 June 2017|year=1977|publisher=American Mathematical Soc.|isbn=978-0-8218-2184-8|page=51}}

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,

:

= \begin{bmatrix} U & D_{U^*} \\ 0 & - U^* \end{bmatrix}

.

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 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.

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

:$B\; (\backslash ldots,\; a\_\{-1\},\; \{\backslash hat\; a\_0\},\; a\_1,\; \backslash ldots)\; =\; (\backslash ldots,\; \{\backslash hat\; a\_\{-1\}\},\; a\_0,\; a\_1,\; \backslash ldots),$

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

:$$

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

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*.){{citation|author=John B. Conway|title=The Theory of Subnormal Operators|url=https://books.google.com/books?id=Ho7yBwAAQBAJ&pg=PA38|accessdate=15 June 2017|year=1991|publisher=American Mathematical Soc.|isbn=978-0-8218-1536-6|pages=38–}}

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

:$U:\; K\_1\; \backslash rightarrow\; K\_2.$

Also, the following intertwining relationship holds:

:$U\; B\_1\; =\; B\_2\; U.\; \backslash ,$

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 \text{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

:$$

\left\langle \sum_{i=0}^n (B_1^*)^i h_i, \sum_{j=0}^n (B_1^*)^j h_j\right\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

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

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

:$\backslash text\{if\; \}\; g\; =\; \backslash sum\_\{i=0\}^n\; (B\_1^*)^i\; h\_i\; ,$

:$\backslash text\{then\; \}\; U\; B\_1\; g\; =\; B\_2\; U\; g\; =\; \backslash sum\_\{i=0\}^n\; (B\_2^*)^i\; A\; h\_i.$

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.

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Category:Operator theory

Category:Linear operators