I have a nilpotent matrix, $A$, which is a bounded linear operator on a Banach space. Let $(I-\lambda A)$ be invertible, then for which range of $\lambda$ is $(I-\lambda A)^{-1}$ well-defined?
Answer:
Since $A$ is a bounded linear operator on a Banach space, then the spectral radius $r(A) = \sup\{|\lambda|\ |\lambda\in\sigma(A)\}$ satisfies $$ r(A) = \lim_{n\to\infty} \|A^n\|^{1/n}.$$ So the spectrum of the nilpotent operator $A$ is countable and finite, and contains only zeros.
So $\sigma(A)=\{0,0,0,0,0,0,...0,\ldots\}$
But according to the definition of well-defined, some value of $\lambda$ should assign a unique value for the operator. Here we have that, but infinitely many times (for infinitely many zeros).
Does this mean the zero-spectrum makes $(I-\lambda A)^{-1}$ well-defined?
My asumption is that since $\sigma(A)=\{,0,0,0,\ldots,0,\ldots\}$ then we simply have $(I-\lambda A)^{-1}=I$, which is "well-defined".
Since $ \sigma(A)= \{0\}$, the operator $ \lambda I -A$ is invertible for all $ \lambda \ne 0.$
If $ \lambda \ne 0,$ then $I- \lambda A=\lambda ( \frac{1}{\lambda}I-A) $ is invertible. If $ \lambda =0$, then $I- \lambda A=I$
Consquence: $I- \lambda A$ is invertible for all $ \lambda.$
Note that, in general, $(I-\lambda A)^{-1}=I$ is not (!) valid.