A finite order restriction of a Fredholm operator is also a Fredholm operator. 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.
 A: As $D(A)/D(B)$ is finite dimensional choose a finite dimensional complement $C$ of $D(B)$ in $D(A)$. For convenience we take $\lambda =0$, this is of course no restriction. 
Now
$$R(A)=\{ Ax\mid x\in D(B)\oplus C\} = \{A x\mid x\in D(B)\} + \{Ax\mid  x\in C\} = R(B) + A(C),$$
where $A(C)$ is finite dimensional since $C$ is finite-dimensional. Thus $R(B)$ has finite co-dimension in $R(A)$, which has finite co-dimension in $H$ and thus $R(B)$ has finite co-dimension in $H$. You have already seen that $\ker(B)$ is finite dimensional, so the only thing left to check is that $R(B)$ is closed.
Since $B$ is a closed operator you have that its graph $G_B = \{ (y, By) \mid y\in D(B)\}$ is a Banach space wrt the graph norm $\|(y, By)\| = \|y\|+\|By\|$. Since $A$ has closed image you have that $R(A)$ is closed. You may interpret the map $\pi_2:G_B\to H$ as a map $G_B\to R(A)$, which is now a continuous linear map (actually a contraction) between two Banach spaces with image $R(B)\subset R(A)$. We have just seen in the previous paragraph that this map has finite dimensional co-kernel, by standard results the image is then closed (see eg here or here).
