Eigenspace of linear completely continuous field

Question

Let $H_1$ and $H$ be Hilbert spaces, and $H_1 \subset H$ be a dense and compact inclusion. A linear operator $L: H_1 \to H$ is called a completely continuous field if $L = - A + B$, where $A: H_1 \to H$ is a linear homeomorphism and $B: H_1 \to H$ is a linear compact operator.

Theorem: The eigenspace of $L$ is of finite dimension.

The result is quite similar to compact operator cases, but I am not able to use a similar method to prove it. Could anyone give me some hint on this?

Edit: These are from Section 3.1 of the book "Birfucation theory and applications" by Ma and Wang, and the corresponding paragraphs are attached:

Answer 1

I believe the question is to be understood like this:

Let $i: H_1\to H$ be a compact injection with dense image. An eigenvector of a map $L: H_1\to H$ is a vector $z\in H_1$ so that there is a $\lambda$ with $$L(z) = \lambda \ i(z).$$

Now suppose that $L = A+ B$ where $A:H_1\to H$ is a linear isomorphism and $B : H_1 \to H$ a compact operator. $z$ is an eigenvector of $L$ in the above sense iff: $$(A+B)\ (z) = \lambda \ i(z) \iff \frac1\lambda z = A^{-1}(i-\lambda^{-1}B)\ (z) \text{ or $(A+B)(z)=0$}$$ ie iff it is an eigenvector to the eigenvalue $\frac1\lambda$ of the compact operator $A^{-1}(i-\lambda^{-1}B): H_1\to H_1$ or if it is in the kernel of $A+B$ (corresponding to eigenvalue $0$). The eigenspaces of the first case are all finite dimensional, since $A^{-1}(i-B)$ is compact and the eigenspace of $0$ is also finite dimensional since $A+B$ is Fredholm.