I need help with the following question : Let us consider $X$ to be a $\mathbb C $ Banach space . Let $T \in B(X)$ ie. $T$ is a continuous linear map from $X$ to $X$ . define $$P(T)=\sum_{k=o}^{n}a_kT^k$$ with $T^0=id_X$ (identity) .
The claim is that the spectrum of $T : \sigma(T) \subset\{\lambda \in \mathbb C : P(\lambda)= 0\}$ Here $P$ is a polynomial such that $P(T) =0$ .
Thanks.
Let me expand on the setting. Let $\text{Hol}(T)$ be the set of all the functions holomorphic in a neighborhood of $\sigma(T)$. The Riesz functional calculus now says that indeed if I have the polynomials $p$, that $p(T)$ is as expected - of course, for polynomials this is trivial.
Additionally, we have the Banach algebra of bounded operators on $X$, and: Spectral mapping theorem: If $T$ is an element of a Banach algebra $\mathcal A$ and $f$ is holomorphic in a neighborhood of $\sigma(T)$, then $$\sigma(f(T)) = f(\sigma(T)).$$
If we thus select such a polynomial $p$ such that $p(T) = 0$, we get $p(\sigma(T)) = 0$.
It is enough to show the contrapositive, namely, that if $P(\lambda) \neq 0$, then $T - \lambda \operatorname{id}_X$ is invertible. Now, since $P(\lambda) \neq 0$, $x-\lambda$ does not divide the polynomial $P(x)$, so that we may write $P(x) = Q(x)(x-\lambda) + R$ for some polynomial $Q(x)$ and a non-zero remainder $R \in \mathbb{C}$ — since $x-\lambda$ is of degree one, the remainder must be of strictly lower degree, and thus constant. But then, this means that $0 = P(T) = Q(T)(T-\lambda \operatorname{id}_X) + R\operatorname{id}_X$, so that since $T-\lambda \operatorname{id}_X$ and $Q(T)$ commute, we find that $T-\lambda \operatorname{id}_X$ is invertible with bounded inverse $(T-\lambda \operatorname{id}_X)^{-1} = -\tfrac{1}{R}Q(T)$, as required.
