Surjectivity of the sum of two bounded operators

Question

Let $H$ be a Hilbert space.

Let $A\in B\left( H\right) $ be surjective, and let $Q\in B\left( H\right) $ be quasi-nilpotent such that $AQ=QA$

Prove that $A+Q$ is surjective.

Thank you.

Answer 1

I can do this if $Q$ is nilpotent, which is not what is asked, but will maybe give you an idea.

Anyway, if $Q$ is nilpotent, let $n$ be such that $Q^n = 0$.

Now, choose any $y \in H$, and let $z$ be such that $A^n z = y$ (if $A$ is surjective, so is $A^n$).

Then, I claim that $x = \sum_{j=0}^{n-1}(-1)^jA^{n-1-j}Q^jz$ satisfies $(A+Q)x=y$: $$ (A+Q)x = \sum_{j=0}^{n-1}(-1)^jA^{n-j}Q^jz + \sum_{j=0}^{n-1}(-1)^jQA^{n-1-j}Q^{j}z\\ = \sum_{j=0}^{n-1}(-1)^jA^{n-j}Q^jz + \sum_{j=0}^{n-1}(-1)^jA^{n-1-j}Q^{j+1}z \quad (AQ=QA)\\ = A^nz + \sum_{j=1}^{n-1}(-1)^jA^{n-j}Q^jz + \sum_{j=0}^{n-2}(-1)^jA^{n-1-j}Q^{j+1}z + (-1)^{n-1}Q^nz.\\ $$

The first term is $y$, the final term is $0$ due to nilpotency, and the middle terms will cancel each other (obvious after reindexing).