Let $A$ be an invertible operator, and let $K$ be a compact operator in a Banach space. Prove that:
a) $\dim(\ker(A+K)) \lt \infty$
b) $\operatorname{codim}(\operatorname{Im}(A+K)) \lt \infty$
I saw this question in a book that I'm studying from, and the solutions contains the following claims:
a) $A + K = A(I+A^{-1}K)$, and $A^{-1}K)$ is compact as a product of bounded and compact operators. Hence $\dim(\ker(I+A^{-1}K)) \lt \infty$ - and the solution continues. I can't figure out - Why is it true that : $\dim(\ker(I+A^{-1}K)) \lt \infty$ ?
b) $A+K = (I+KA^{-1})A$, and the operator $KA^{-1}$ is compact as a product of bounded and compact operators. Hence $\operatorname{codim}(Im(I+KA^{-1})) \lt \infty$ - and the solution continues.
Same here, why is it true that $\operatorname{codim}(Im(I+KA^{-1})) \lt \infty$ ?
I tried looking at the definitions of compact operators, but either way, taking compact operator $T$ and looking at $T+I$ is not a compact operator anymore...
(a) compact perturbation of identity has only finite-dimensional kernel, because a compact operator has only finite-dimensional eigenspaces with eigenvalues exceeeding $\varepsilon$ in absolute value, for all $\varepsilon>0$. In particular, the compact operator cannot have an infinite-dimensional eigenspace with eigenvalue $-1$.
(b) you missed a few "Im", but anyway, the same reason as (a) gives a finite dimensional cokernel.