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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3$\begingroup$ The zero operator is compact, so $A=0$ is a counterexample. $\endgroup$ – Mark Jun 14 '19 at 21:46
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2$\begingroup$ If $A$ is compact, then $I - A$ cannot be compact. $\endgroup$ – Ben Grossmann Jun 14 '19 at 21:47
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3$\begingroup$ 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 '19 at 21:48
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2$\begingroup$ 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. $\endgroup$ – Nate Eldredge Jun 14 '19 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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1$\begingroup$ @Kamal Here is discussion on the sum of two compact operators math.stackexchange.com/questions/1011095/… $\endgroup$ – AnyAD Jun 15 '19 at 6:00
