# Dimension of kernel of Fredholm operator

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?

PS. You can assume that $$X$$ is Banach, or even Hilbert, and that $$T$$ is a bounded Fredholm operator (therefore $$T(X)$$ is closed).
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 $$\mbox{ind}(S)=0$$.