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?

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    $\begingroup$ 1. 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. $\endgroup$
    – t.b.
    Commented Jun 25, 2011 at 6:28
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    $\begingroup$ Since 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? $\endgroup$ Commented Jun 25, 2011 at 9:14

1 Answer 1


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.


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