Identity plus finite rank has index $0$ I'm supposed to prove the strong Fredholm alternative in the form $$\text{Ind}(1-K)=0$$
for any compact operator $K:H\to H$ where $H$ is a Hilbert space and $$\text{Ind}(T):=\text{dim Ker }T+\text{dim Coker }T=\text{dim Ker }T+\text{dim (Ran }T)^\perp.$$
The hint is that I should first prove the equality $\text{Ind}(1-F)=0$ for a finite rank operator $F$ directly. I tried a few things but nothing promising so far. Can someone give me a hint?
 A: I tried to reduce the question to the finite-dimensional case. Is the following argument valid?
Let $F:H\to H$ be a finite rank operator on a Hilbert space $H$. Then $$G:=\text{Ker }F\cap(\text{Ran }F)^\perp=\text{Ker }F\cap\text{Ker }F^*\subset H$$ is a subspace of finite codimension i.e. $\text{dim }G^\perp=:n<\infty$. Moreover $F(G^\perp)=F(H)\subset G^\perp$ since $F=0$ on $G$ and $\langle f,F e\rangle=\langle F^* f,e\rangle=0$ for all $f\in G$. Thus it is possible to split $1+F$ to $(1+F|_{G^\perp})\oplus1$ on $G^\perp\oplus G$. Then $$\text{dim Ker }(1+F)-\text{dim Ker} (1+F^*)=\text{dim Ker }(1+F|_{G^\perp})-\text{dim Ker} (1+F^*|_{G^\perp})$$ since $\text{Ker }1$ is trivial.
But now we are in the finite dimensional case and can use that $$\text{dim Ker }(1+F|_{G^\perp})+\text{dim Ran} (1+F|_{G^\perp})=n$$ and hence \begin{align}\text{ind}(1+F)&=\text{dim Ker }(1+F|_{G^\perp})-\text{dim (Ran} (1+F|_{G^\perp}))^\perp\\ &=\text{dim Ker }(1+F|_{G^\perp})-(n-\text{dim Ran} (1+F|_{G^\perp}))=0.\end{align}
What do you think?
