I am unsure if the following Proof is correct would you please care to verify?
SOME PRELIMARNY NOTATION
- $T\in\mathcal{L}(V)$ means that $T$ belongs to the set of linear operators on the vector space $V$.
- Given an operator $T\in\mathcal{L}(V)$ and a polynomial $p(z) = c_0+c_1z+c_2z^2+\cdot\cdot\cdot+c_mz^m$. Then $p(T)$ denotes the operator $p(T) = c_0I+c_1T+c_2T^2+\cdot\cdot\cdot+c_mT^m$
Theorem. Given that $V$ is finite dimensional, $T\in\mathcal{L}(V)$ and $v\in V$ such that $v\neq 0$. and $p$ is the polynomial with the smallest degree such that $p(T)v = 0$ then every zero of $p$ is an eigenvalue of $T$.
Proof. Assume that $p(\lambda) = 0$ equivalently $(z-\lambda)$ is a factor of $p$ consequently for some non-zero polynomial $q$ we have $p(z) = (z-\lambda)q(z)$ and thus the corresponding operator becomes $p(T) = (T-\lambda I)q(T)$ and since $p(T)v = 0$ it follows that $(T-\lambda I)q(T)v = 0$ furthermore since $p$ is the polynomial with the smallest degree satisfying $p(T)v = 0$ it follows that $q(T)v\neq 0$ since $\deg q<\deg p$ consequently $(T-\lambda I)v = 0$ and since $v\neq 0$ it follows that $Tv = \lambda v$ implying that $\lambda$ is an eigenvalue of $T$.
$\blacksquare$
I am skeptical because i did not use the hypothesis that $V$ is finite dimensional and the above proof seems to be suggesting that multiple eigenvalues correspond to the same eigenvector $\huge!$