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I know that the identity is not a compact operator in an infinite dimensional space, is the difference composition of the identity operator with a compact operator a compact one?

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    $\begingroup$ The zero operator is compact, so $A=0$ is a counterexample. $\endgroup$
    – Mark
    Jun 14, 2019 at 21:46
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    $\begingroup$ If $A$ is compact, then $I - A$ cannot be compact. $\endgroup$ Jun 14, 2019 at 21:47
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    $\begingroup$ Actually, given that the compact operators are a linear subspace $I-A$ can't be compact for any compact operator $A$. $\endgroup$
    – Mark
    Jun 14, 2019 at 21:48
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    $\begingroup$ The intuition is that a compact operator is "small" and the identity (or any operator with infinite-dimensional closed range) is "big". If you perturb something big by something small, the result is still big. $\endgroup$ Jun 14, 2019 at 22:25

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Identity on a finite dimensional space is compact, so it is best to specify what sort of space you are talking about.

If $1-A$ was a compact operator, then $1-A+A$ would be compact.

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