$\sigma(T)$ is finite $\iff \exists p\in P: p(T) = 0$ The following is an old exam problem and I can not figure out how to solve it. I have been gathering some facts related to the problem, see below.
Problem
Let $H$ be a separable Hilbert space and let $T: H \to H$ be a compact symmetric linear map. Let $\sigma(T)$ denote the spectrum of $T$ and $P$ the set of polynomials.

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*Prove:  $\sigma(T)$ is finite $\iff \exists p\in P: p(T) = 0$

*Find a separable Hilbert space $H$ and a compact linear map $C:H\to H$ with finite spectrum such that there is no $p\in P$ with $p(C) = 0$
What I know

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*There exists an orthonormal base $\{z_1,z_2,..\}$ for $H$, where $z_i$ are eigenvectors of $T$ corresponding to real eigenvalues.


*$\sigma(T) \subset \{ \lambda : |\lambda| \le |||T||\}$


*For any $p \in P, \sigma(p(T)) = p(\sigma(T)) $


*The eigenspace corresponding to any $\lambda \neq 0$ is finite dimensional.


*For $\lambda \neq 0$, the null spaces of $T_\lambda, T_\lambda^2, T_\lambda^3, ..$ are finite dimensional. ($T_\lambda = T - \lambda I$)
 A: *

*Suppose $\sigma(T)$ is finite. Say $\sigma(T)=\{\lambda_j\}_{j=1}^n$. Put $$p(\lambda)=\Pi_j(\lambda-\lambda_j).$$ Then $\sigma(p(T))=p(\sigma(T))=\{0\}$. Note that $T$ is self-adjoint, all $\lambda_j$ are real. This ensure $p(T)$ is self-adjoint. There is an important fact that: for self-adjoint(may not compact) elements $A$, $\|A\|=r(A)$, here $r(A)$ denotes the Spectral radius of $A$. So $\|p(T)\|=r(p(T))=0$, i.e. $p(T)=0$.

*Conversly, assume there exists some nonzero polynomial $p$, such that $p(T)=0$. Then $p(\sigma(T))=\sigma(p(T))=\{0\}$, which shows $$\sigma(T)\subseteq Z(p).$$ Here $Z(p)$ denotes the zero set of $p$, which is finite by Fundamental theorem of Algebra.

*Now we present a compact operator with finite spectrum such that there is no nonzero $p$ with $p(T)=0$. Consider Volterra operator $$V: L^2[0,1]\to L^2[0,1]: f(x) \mapsto \int_0^xf(t)dt.$$ It is indeed Compact, and $\sigma(V)=\{0\}$, of course finite. Note that: $$V^n 1=\frac{1}{n!}x^n.$$
Polynomials $\{1,x,x^2,...\}$ are linear independent. This shows $p(V)1\neq 0$, hence $p(V)\neq 0$
