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| In [[operator theory]], '''quasinormal operators''' is a class of [[bounded operator]]s defined by weakening the requirements of a [[normal operator]].
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| Every quasinormal operator is a [[subnormal operator]]. Every quasinormal operator on a finite-dimensional [[Hilbert space]] is normal.
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| == Definition and some properties ==
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| === Definition ===
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| Let ''A'' be a bounded operator on a Hilbert space ''H'', then ''A'' is said to be '''quasinormal''' if ''A'' commutes with ''A*A'', i.e.
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| :<math>A(A^*A) = (A^*A) A.\,</math>
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| === Properties ===
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| A normal operator is necessarily quasinormal.
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| Let ''A'' = ''UP'' be the [[polar decomposition]] of ''A''. If ''A'' is quasinormal, then ''UP = PU''. To see this, notice that
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| the positive factor ''P'' in the polar decomposition is of the form (''A*A'')<sup>½</sup>, the unique positive square root of ''A*A''. Quasinormality means ''A'' commutes with ''A*A''. As a consequence of the [[continuous functional calculus]] for [[self adjoint operator]]s, ''A'' commutes with ''P'' = (''A*A'')<sup>½</sup> also, i.e.
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| :<math>U P P = P U P.\,</math>
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| So ''UP = PU'' on the range of ''P''. On the other hand, if ''h'' ∈ ''H'' lies in kernel of ''P'', clearly ''UP h'' = 0. But ''PU h'' = 0 as well. because ''U'' is a [[partial isometry]] whose initial space is closure of range ''P''. Finally, the self-adjointness of ''P'' implies that ''H'' is the direct sum of its range and kernel. Thus the argument given proves ''UP'' = ''PU'' on all of ''H''.
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| On the other hand, one can readily verify that if ''UP'' = ''PU'', then ''A'' must be quasinormal. Thus the operator ''A'' is quasinormal if and only if ''UP'' = ''PU''.
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| When ''H'' is finite dimensional, every quasinormal operator ''A'' is normal. This is because that in the finite dimensional case, the partial isometry ''U'' in the polar decomposition ''A'' = ''UP'' can be taken to be unitary. This then gives
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| :<math>A^*A = (UP)^* UP = PU (PU)^* = AA^*.\,</math>
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| In general, a partial isometry may not be extendable to a unitary operator and therefore a quasinormal operator need not be normal. For example, consider the [[unilateral shift]] ''T''. ''T'' is quasinormal because ''T*T'' is the identity operator. But ''T'' is clearly not normal.
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| == Quasinormal invariant subspaces ==
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| It is not known that, in general, whether a bounded operator ''A'' on a Hilbert space ''H'' has a nontrivial invariant subspace. However, when ''A'' is normal, an affirmative answer is given by the [[spectral theorem]]. Every normal operator ''A'' is obtained by integrating the identity function with respect to a spectral measure ''E'' = {''E<sub>B</sub>''} on the spectrum of ''A'', ''σ''(''A''):
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| :<math>A = \int_{\sigma(A)} \lambda d E (\lambda).\,</math> | |
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| For any Borel set ''B'' ⊂ ''σ''(''A''), the projection ''E<sub>B</sub>'' commutes with ''A'' and therefore the range of ''E<sub>B</sub>'' is an invariant subpsace of ''A''.
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| The above can be extended directly to quasinormal operators. To say ''A'' commutes with ''A*A'' is to say that ''A'' commutes with (''A*A'')<sup>½</sup>. But this implies that ''A'' commutes with any projection ''E<sub>B</sub>'' in the spectral measure of (''A*A'')<sup>½</sup>, which proves the invariant subspace claim. In fact, one can conclude something stronger. The range of ''E<sub>B</sub>'' is actually a ''reducing subspace'' of ''A'', i.e. its orthogonal complement is also invariant under ''A''.
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| == References ==
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| *P. Halmos, A Hilbert Space Problem Book, Springer, New York 1982.
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| [[Category:Operator theory]]
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| [[Category:Invariant subspaces]]
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Greetings. Allow me begin by telling you the writer's name - Phebe. One of the issues he loves most is ice skating but he is struggling to find time for it. South Dakota is her birth location but she needs to transfer simply because of her family. Bookkeeping is what I do.
Feel free to surf to my page: http://롤에버.com/