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| {{Unreferenced|date=November 2006}}
| | Wilber Berryhill is what his spouse enjoys to contact him and he completely enjoys this title. I've usually loved living in Mississippi. I am truly fond of handwriting but I can't make it my profession really. She works as a journey agent but quickly she'll be on her own.<br><br>Also visit my web blog - psychic love readings - [http://Www.skullrocker.com/blogs/post/10991 http://Www.skullrocker.com/] - |
| 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.
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| ==Definition==
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| 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>N = \begin{bmatrix} A & B\\ 0 & C\end{bmatrix}</math>
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| for some bounded operators
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| :<math>B : H^{\perp} \rightarrow H, \quad \mbox{and} \quad C : H^{\perp} \rightarrow H^{\perp}.</math>
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| ==Normality, quasinormality, and subnormality==
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| === Normal operators ===
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| 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 operator]]s. A unitary operator is an isometry with [[dense set|dense]] [[range (mathematics)|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
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| :<math>H \oplus H</math>
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| by
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| :<math> U = \begin{bmatrix} A & I - AA^* \\ 0 & - A^* \end{bmatrix}.</math>
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| 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''.
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| ===Quasinormal operators===
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| 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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| 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:
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| '''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.
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| 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'',
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| :<math> V = \begin{bmatrix} U & I - UU^* \\ 0 & - U^* \end{bmatrix}
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| = \begin{bmatrix} U & D_{U^*} \\ 0 & - U^* \end{bmatrix}
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| .</math>
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| Define
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| :<math> Q = \begin{bmatrix} P & 0 \\ 0 & P \end{bmatrix}.</math>
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| The operator ''N'' = ''VQ'' is clearly an extension of ''A''. We show it is a normal extension via direct calculation. Unitarity of ''V'' means
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| :<math>N^*N = QV^*VQ = Q^2 = \begin{bmatrix} P^2 & 0 \\ 0 & P^2 \end{bmatrix}.</math>
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| On the other hand,
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| :<math>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}.</math>
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| Because ''UP = PU'' and ''P'' is self adjoint, we have ''U*P = PU*'' and ''D<sub>U*</sub>P = D<sub>U*</sub>P''. Comparing entries then shows ''N'' is normal. This proves quasinormality implies subnormality.
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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.
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| ==Minimal normal extension==
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| === Non-uniqueness of normal extensions ===
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| Given a subnormal operator ''A'', its normal extension ''B'' is not unique. For example, let ''A'' be the unilateral shift, on ''l''<sup>2</sup>('''N'''). One normal extension is the bilateral shift ''B'' on ''l''<sup>2</sup>('''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>
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| where ˆ denotes the zero-th position. ''B'' can be expressed in terms of the operator matrix
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| :<math> B = \begin{bmatrix} A & I - AA^* \\ 0 & A^* \end{bmatrix}.</math> | |
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| Another normal extension is given by the unitary dilation ''B' '' of ''A'' defined above:
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| :<math> B' = \begin{bmatrix} A & I - AA^* \\ 0 & - A^* \end{bmatrix}</math>
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| whose action is described by
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| :<math>
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| 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).
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| </math>
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| ===Minimality===
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| 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*''.)
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| One can show that if two operators ''B''<sub>1</sub> and ''B''<sub>2</sub> are minimal extensions on ''K''<sub>1</sub> and ''K''<sub>2</sub>, respectively, then there exists a unitary operator
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| :<math>U: K_1 \rightarrow K_2.</math>
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| Also, the following interwining relationship holds:
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| :<math>U B_1 = B_2 U. \,</math>
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| This can be shown constructively. Consider the set ''S'' consisting of vectors of the following form:
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| :<math>
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| \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.
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| </math>
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| Let ''K' '' ⊂ ''K''<sub>1</sub> be the subspace that is the closure of the linear span of ''S''. By definition, ''K' '' is invariant under ''B''<sub>1</sub>* and contains ''H''. The normality of ''B''<sub>1</sub> and the assumption that ''H'' is invariant under ''B''<sub>1</sub> imply ''K' '' is invariant under ''B''<sub>1</sub>. Therefore ''K' '' = ''K''<sub>1</sub>. The Hilbert space ''K''<sub>2</sub> can be identified in exactly the same way. Now we define the operator ''U'' as follows:
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| :<math>
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| U \sum_{i=0}^n (B_1^*)^i h_i = \sum_{i=0}^n (B_2^*)^i h_i
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| </math>
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| Because
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| :<math>
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| \langle \sum_{i=0}^n (B_1^*)^i h_i, \sum_{j=0}^n (B_1^*)^j h_j\rangle
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| = \sum_{i j} \langle h_i, (B_1)^i (B_1^*)^j h_j\rangle
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| = \sum_{i j} \langle (B_2)^j h_i, (B_2)^i h_j\rangle
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| = \langle \sum_{i=0}^n (B_2^*)^i h_i, \sum_{j=0}^n (B_2^*)^j h_j\rangle ,
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| </math>
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| , the operator ''U'' is unitary. Direct computation also shows (the assumption that both ''B''<sub>1</sub> and ''B''<sub>2</sub> are extensions of ''A'' are needed here)
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| :<math>\mbox{if} \quad g = \sum_{i=0}^n (B_1^*)^i h_i ,</math>
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| :<math>\mbox{then} \quad U B_1 g = B_2 U g = \sum_{i=0}^n (B_2^*)^i A h_i.</math>
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| When ''B''<sub>1</sub> and ''B''<sub>2</sub> 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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| {{DEFAULTSORT:Subnormal Operator}}
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| [[Category:Operator theory]]
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Wilber Berryhill is what his spouse enjoys to contact him and he completely enjoys this title. I've usually loved living in Mississippi. I am truly fond of handwriting but I can't make it my profession really. She works as a journey agent but quickly she'll be on her own.
Also visit my web blog - psychic love readings - http://Www.skullrocker.com/ -