Minimal polynomial in infinite dimension Let $T$ be a operator on a complex Banach space $E$. Show that there exists a polynomial $P$ such that $P(T)=0$ if and only if the spectrum of $T$ consists in a finite number of eigenvalues.
Firt assume that $P(T)=0$. The spectral theorem shows that : $0=\sigma (P(T))=P(\sigma (T) )$. Thus $\sigma (T)$ is included in the set of roots of $P$, in particular $\sigma (T)$ is finite. But I don't see why these elements are eigenvalues.
For the converse, we can assume without loss of generality that $\sigma (T)=0$ and zero is an eigenvalue. It remains to prove that $T$ is nilpotent. 
Thank you in advance for your help.
 A: If $P[T]=0$ it does not follow that all roots of $P$ are eigenvalues. After all, given such a polynomial and any scalar $\alpha$ we cah form the product $Q=P(X-\alpha)$, and we will also have $Q[T]=0$, while we have ensured that $\alpha$ is a root of$~Q$; however, there is no reason that $\alpha$ (which we chose freely) would have to be an eigenvalue.
On the other hand once $P[T]=0$ for some nonzero polynomial$~P$, there must be a minimal polynomial of$~T$ (the lowest degree monic $P$ with $P[T]=0$), and for this choice all roots of$~P$ will be eigenvalues of$~T$. Given such a root$~\lambda$ we can write $P=(X-\lambda)Q$, and because of its too low degree, $Q[T]\neq0$. Now any nonzero vector$~v$ in the image of $Q[T]$, say $v=Q[T](w)$, satisfies $(T-\lambda I)(v)=((X-\lambda)Q)[T](w)=P[T](w)=0$, so $v$ is an eigenvector of$~T$ for$~\lambda$.
For the converse, I think the following example shows that it is not true (but I must admit not being very much at easy with Banach spaces). In $\ell^\infty$ let $T:(a_n)_{n\in\Bbb N}\mapsto(\frac{a_{n+1}}{n+1})_{n\in\Bbb N}$. Then $T$ does not have any eigenvalues $\lambda\neq0$, since an eigenvector would have to be of the form $c(n!\lambda^n)_{n\in\Bbb N}$ for some $c\neq0$, but that is not an element of $\ell^\infty$. Nonetheless $T$ is not nilpotent.
