I want to show that the set of all compact operators $K(H)$ is the unique ideal in $B(H)$. Is there any relation between invertibility and compactness of an operator?
Set of all compact operators $K(H)$ is the unique ideal in $B(H)$?
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functional-analysis
operator-theory
hilbert-spaces
ideals
compact-operators
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51. Assuming $H$ is separable, it's the unique proper *closed* ideal. (the operators of finite rank are also an ideal, as well as the Schatten $p$-ideals, for instance) 2. An invertible operator is compact if and only if $H$ is finite-dimensional. – 2011-06-25
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1Since this is a standard result explained in textbooks on operator algebras, is it homework? What have you tried, where are you stuck, what hints did you get, what auxilliary results are you allowed to use? – 2011-06-25