Compact operator problem: $I-T$ is onto, then $I-T$ is invertible?

Question

I want to show the following:

Let $T$ be a compact operator in a Hilbert space $H$. If $I-T$ is onto, then $I-T$ is invertible.

Would you show me how to prove this argument? Or please tell me some references concerning this theorem.

Thanks, would you give me some more direct proof for this specific problem?

Answer 1

The following comes from Rudin's Functional Analysis 2ed. Theorem 4.25(a)

Let $X$ be a Banach space, with $T\in\mathcal{B}(X)$ compact. If $\lambda\neq0$ then $\dim\mathcal{N}(T-\lambda I)=\dim X/\mathcal{R}(T-\lambda I),$ and both are finite.

In fact, Rudin says a lot more, but this is all we need.

By hypothesis, $I-T$ is onto, and therefore $\dim X/\mathcal{R}(I-T)=0$. But then $$ \dim\mathcal{N}(I-T)=\dim X/\mathcal{R}(I-T)=0, $$ so $I-T$ is one-to-one, and therefore $I-T$ is invertible.