Decomposition of resolvent in projections I am reading the book Perturbation theory for linear operators from Kato. 
He defines (§5 Section 3) for an operator $T : X\to X$ on a finite Banach Space the resolvent as
$$ R(x) = (T- x)^{-1}.$$
He then consideres the Laurent Series of the resolven on a singularity $\lambda$  of $R$ and finds a decomposition of $R$ as
$$ R(x) = -(x-\lambda)^{-1}P - \sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n + \sum_{n=0}^\infty (x-\lambda)^n S^{n+1}$$
on some set $$\Gamma = \{ x\in \mathbb{C} \mid r < |x-\lambda | <r'  \} $$where $P^2 = P$ and $PD=DP=D, \, PS=SP=0$. Since $P$ is a projection one can write
$$ X= U\oplus V$$ with $U=PX$ and $V=(1-P)X$.
The author makes then the following two conclusions:


*

*He claims that $Range(D) = U$. I see that $Range(D) \subset U$ but how can one explain the other inclusion?

*He states:

As the principal part of a Laurent Series at an isolated singularity
  $$  -(x-\lambda)^{-1}P -\sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n$$
  is convergent for $x \neq \lambda$, so that the part of $R(x)$ in $U$
  has only the one singularity $x=\lambda$, and the spectral radius of $D$ must be zero.

I do not understand how he concludes that this part of the Laurent Series has no further singularites (in $\mathbb{C}$ ?)  
 A: To 1)
The operator
$$  [R(x) + (x-\lambda)^{-1} P ] : U \to U$$
is bijective. Therefore, for any $u \in U$ we find a $v\in U$ with $$ [R(x) + (x-\lambda)^{-1} P ]v = u$$ and the following sequence converges to $u$ $$Range(D) \supset \sum_{i=1}^n (x-\lambda)^{-i-1}D^n v \to u.$$
Since $Range(D)$ is a closed subspace we have $u\in Range(D)$ and therefore $Range(D)=U$.
To 2) As explained here Laurent Series of operator-value function the singularites of $R$ are isolated. This means that 
$$ R(x) = -(x-\lambda)^{-1}P - \sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n + \sum_{n=0}^\infty (x-\lambda)^n S^{n+1}$$
converges in a ball $$ B = \{ x \in \mathbb{C} \mid 0 < |x - \lambda |< R \}.$$
For any $x\in\mathbb{C}\backslash\{\lambda\}$ we find a  $x_0 \in \mathbb{C}\backslash\{\lambda\}$ with
$$ | x - \lambda| > | x_0 - \lambda|.$$
If you think of $D$ as a matrix then the convergence of
$$\sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n$$ is
equivalent to the convergence of
$$\sum_{n=1}^\infty (x- \lambda)^{-n-1}[D^n]_{ij}$$ which converges for  $x_0$ and therefore also for $x$ (see comment from T.A.E.).
