let $ T \in K ( H ) $ be $ T \geqslant 0 $.
What is the best way to show the following statement?
There is a compact, positive and unique operator $ A$ so that $ A^{2} = T $.
( K ( H ) defined compact operator)
compact positive operator
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functional-analysis
operator-theory
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0There are different definitions of what $T\geq 0$ should mean. I am assuming that it implies self-adjointness in the definition you use? – 2017-01-29
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0What have you tired? You can get a positive square root from the functional calculus very easily. – 2017-01-29
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0http://math.stackexchange.com/questions/148162/square-root-of-compact-operator http://math.stackexchange.com/questions/311310/compact-and-self-adjoint-square-root-of-an-operator http://math.stackexchange.com/questions/706525/positive-compact-operator-has-unique-square-root http://math.stackexchange.com/questions/2111827/square-root-of-compact-positive-operator – 2017-01-29