# Is $I - A$ a compact operator, where $A$ is a compact operator and $I$ is the identity operator?

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?

• The zero operator is compact, so $A=0$ is a counterexample. – Mark Jun 14 '19 at 21:46
• If $A$ is compact, then $I - A$ cannot be compact. – Ben Grossmann Jun 14 '19 at 21:47
• Actually, given that the compact operators are a linear subspace $I-A$ can't be compact for any compact operator $A$. – Mark Jun 14 '19 at 21:48
• 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. – Nate Eldredge Jun 14 '19 at 22:25

## 1 Answer

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.