Let $H$ be a separable Hilbert space and $T:=I-A$, where $A:H\to H$ is a compact operator. If $T^\ast$ is the adjoint operator of $T$ it can be proved that $\ker T$ and $\text{im } T^\ast:=T^\ast (H)$ are closed thanks to the compactness of $A$ (pp. 469-70 here; I can translate for anybody interested).
It is clear, from the very definition of adjoint operator, that $\ker T\perp \text{im } T^\ast$.
Therefore, as Kolmgorov-Fomin's says, in order to prove that $H=\ker T\oplus \text{im } T^\ast$, it is enough to show that no non-null vector can be orthogonal to both $\ker T$ and $\text{im } T^\ast$. Why?
I suppose it is a trivial thing, but the book does not explain such a fact in the chapter on linear operators and I know nothing analogous from finite dimension linear algebra cases. Thank you so much!
Best Answer
Since $\ker T$ and $\operatorname{im} T^\ast$ are closed and orthogonal to each other, the subspace
$$E := \ker T \oplus \operatorname{im} T^\ast$$
of $H$ is closed. Thus we have $E = H$ if an only if $E$ is dense. But a subspace $F$ of $H$ is dense, if and only if $F^\perp = \{0\}$. So showing that
$$(\ker T \oplus \operatorname{im} T^\ast)^\perp = (\ker T)^\perp \cap (\operatorname{im} T^\ast)^\perp = \{0\}$$
shows that $E$ is a dense closed subspace of $H$, i.e. $E = H$.