I'm reading a question in this thread
Let $\mathfrak{X}$ be a Banach space, and $K\in \mathcal{K}(\mathfrak{X})$ be a compact operator. Then given $\lambda\in\mathbb{C}\backslash\{0\}$, we can show that
- $\operatorname{ker}(\lambda I - K)$ is finite dimensional
- $\operatorname{ran}(\lambda I - K)$ is closed
Now, if $\mathfrak{M}\subseteq\mathfrak{X}$ is a finite-dimensional subspace, then $\mathfrak{M}$ is complemented in $\mathfrak{X}$, so there exists a closed subspace $\mathfrak{N}\subseteq\mathfrak{X}$ such that $\mathfrak{X} = \mathfrak{M}\oplus\mathfrak{N}$. Let $\mathfrak{R}$ be $\operatorname{ker}(\lambda I - K)$'s complement in $\mathfrak{X}$, and define $$ T : \mathfrak{R}\to \mathfrak{X},\qquad y\mapsto(\lambda I - K)y $$ or in other words $T = (\lambda I - K)|_{\mathfrak{R}}$. It's easy to see that $T$ is injective, and surjective onto it's range, which is closed. Thus $T$ induces an isomorphism $\mathfrak{R}\cong\operatorname{ran}(\lambda I - K)$, and so $$ \mathfrak{X}\cong\operatorname{ker}(\lambda I - K)\oplus\operatorname{ran}(\lambda I - K). $$
I think the reasoning
... $T$ is injective, and surjective onto it's range, ... so $$ \mathfrak{X}\cong\operatorname{ker}(\lambda I - K)\oplus\operatorname{ran}(\lambda I - K). $$
is not true. This is because $T$ is not necessarily injective. So the decomposition $\mathfrak{X}\cong\operatorname{ker}(\lambda I - K)\oplus\operatorname{ran}(\lambda I - K)$ is not true.
Could you confirm if my above understanding is fine?
Of course $T$ is injective.
By definition $\mathfrak R$ is a complement of $\ker T$. If $x\in\mathfrak R$ and $Tx=0$, then $$x\in\mathfrak R\cap\ker T=\{0\}.$$