Let $X$ be a vector space and let $T\colon X\to X$ be a Fredholm operator.
Fix $V$ a finite dimensional subspace of $X$ such that $T(X)+V=X$.
Define $S\colon X\oplus V\to X$ by the formula $S(x,v) = Tx+v$.
It is clear that $S$ is surjective and that $\ker S = \{(x,v) : Tx=-v\}$.
I was told that $S$ is a Fredholm operator and that $\mbox{ind}(S)=\mbox{ind}(T)$, so of course we must have $$\dim\ker S = \mbox{ind}(T) \ \dot{=} \ \dim\ker T – \dim X/T(X)$$ but I am not able to prove it by calculating the dimension of $\ker S$. Is there a way to do this?
Thanks in advance!
PS. You can assume that $X$ is Banach, or even Hilbert, and that $T$ is a bounded Fredholm operator (therefore $T(X)$ is closed).
Best Answer
The result is false. Just take $T$ to have negative index ($\dim\ker S$ is always nonnegative).
For instance, let $X=\ell_2$ and $T$ be the right-shift operator, i.e. $T(e_k)=e_{k+1}$. We have $\ker T=\{0\}$ and $T(\ell_2)=\mbox{span}\{e_k : k\geq2\}$. Thus $\mbox{ind}(T)=-1$.
On the other hand, if we let $V=\mbox{span}\{e_1\}$ (so that $T(\ell_2)+V=\ell_2$), the operator $S\colon\ell_2\oplus V\to \ell_2$ is just concatenation $$S((a_1,a_2,a_3,\cdots),(b,0,0,\cdots)) = (b,a_1,a_2,a_3,\cdots)$$ We then have that $S$ is also injective, in such a way that $\mbox{ind}(S)=0$.