Subnormal operator: Difference between revisions

Revision as of 17:34, 24 October 2009

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{{Unreferenced|date=November 2006}}

In [[mathematics]], especially [[operator theory]], '''subnormal operators''' are [[bounded operator]]s on a [[Hilbert space]] defined by weakening the requirements for [[normal operator]]s. Some examples of subnormal operators are [[isometry|isometries]] and [[Toeplitz operator]]s 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

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:<math>B : H^{\perp} \rightarrow H, \quad \mbox{and} \quad C : H^{\perp} \rightarrow H^{\perp}.</math>

== Normality, quasinormality, and subnormality ==

=== Normal operators ===

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:<math> U = \begin{bmatrix} A & I - AA^* \\ 0 & - A^* \end{bmatrix}.</math>

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 operator|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.

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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 ===

=== 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 ''l''²('''N'''). One normal extension is the bilateral shift ''B'' on ''l''²('''Z''') defined by

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:<math>B (\cdots, a_{-1}, {\hat a_0}, a_1, \cdots) = (\cdots, {\hat a_{-1}}, a_0, a_1, \cdots),</math>

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

:<math> B = \begin{bmatrix} A & I - AA^* \\ 0 & A^* \end{bmatrix}.</math>

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</math>

=== 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' '' &sub;⊂ ''K'' is a reducing subspace of ''B'' and ''H'' &sub;⊂ '' K' '', then ''K' '' = ''K''. (A subspace is a reducing subspace of ''B'' if it is invariant under both ''B'' and ''B*''.)

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' '' &sub; ''K'' is a reducing subspace of ''B'' and ''H'' &sub; '' 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 ''B''₁ and ''B''₂ are minimal extensions on ''K''₁ and ''K''₂, respectively, then there exists an unitary operator

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</math>

Let ''K' '' &sub;⊂ ''K''₁ be the subspace that is the closure of the linear span of ''S''. By definition, ''K' '' is invariant under ''B''₁* and contains ''H''. The normality of ''B''₁ and the assumption that ''H'' is invariant under ''B''₁ imply ''K' '' is invariant under ''B''₁. Therefore ''K' '' = ''K''₁. The Hilbert space ''K''₂ can be identified in exactly the same way. Now we define the operator ''U'' as follows:

:<math>

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When ''B''₁ and ''B''₂ are not assumed to be minimal, the same calculation shows that above claim holds verbatim with ''U'' being a [[partial isometry]].

{{DEFAULTSORT:Subnormal Operator}}

[[Category:Operator theory]]

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