Let $A:D(A) \subseteq H \to H$ and $B:D(B) \subseteq H \to H$ be closed linear operators on a Hilbert space $H$ such that $A$ is a finite order extension of $B$, that is, $B \subseteq A$ and $\mbox{dim } D(A)/D(B) < \infty$. I need to show that if $\lambda \notin \sigma(A)$ and $\lambda I-A$ is a Fredholm operator, then $\lambda I-B$ is also a Fredholm operator.
There is a hint: Since $A$ is a finite order extension of $B$, the difference of the resolvents is of finite order.
But, I don't know how can I use the hint. Can I say that $\lambda \notin \sigma(B)$?
My attempt
Since $\lambda I-A$ is a Fredholm operator, by definition of a Fredholm operator we know that $\mbox{dim} (\ker (\lambda I-A))<\infty$, $\mbox{dim} (H/R(\lambda I-A))<\infty$ and the range $R(\lambda I-A)$ of $\lambda I-A$ is closed in $H$. From here we get that $\mbox{dim} (\ker (\lambda I-B))<\infty$.
Thank you for any help you can privide me.
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