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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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
1 Answers
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If you are interested in closed ideals, then
for separable $H$ the only ideal is $K(H)$
for non-separable $H$ complete characterization may be read here
If you are inrested in all ideals, then they are between $F(H)$ and $K(H)$ and consisist of operators whose singular values belong to some order ideal in $c_0^+$.
For details see section I.8.7 here.