Let $H$ be a Hilbert Space and $K: H \to H$ be an compact operator. By Fredholm's Alternative, i know that $T:= I – K$ is injective iff is onto. Moreover, $T$ is a Fredholm operator of zero index, that is, $\dim \mathrm{Ker} (T) = \mathrm{codim} \ T(H)$.
My question is:
It's possible to write $H = \mathrm{Ker} (T) \ \oplus T(H)$? I think that this separation is natural, since $T$ is a compact pertubation of identity. But i don't know how i can prove this.
Any help is welcome.
Best Answer
No. Here is a counterexample for $H=\mathbb R^2$: Choose $K$ such that $$ T = \begin{pmatrix}0&1\\0&0\end{pmatrix}. $$ I think the claim is true for self-adjoint operators: range of $T$ is closed, then we get this decomposition by the closed range theorem adjusted to Hilbert spaces, so that the decomposition is even orthogonal.
On the way to the spectral theorem for compact operators on Banach spaces, one proves that $$ H = \ker(T^p) \oplus \operatorname{ran}(T^p) $$ holds for some finite $p$, where ``ran'' means image/range of $T^p$.