Suppose $V$ is finite-dimensional, $T\in L(V)$, and $v\in V$ with $v\neq 0$. Let $p$ be a nonzero polynomial of smallest degree such that $p(T)v=0$. Prove that every zero of $p$ is an eigenvalue of $T$.
Approach (1): Suppose $c$ is root of $p$. Then $\exists q\in F[x]$ such that $p=(x-c)q$. So $\deg (p)=\deg (x-c)+\deg (q)=1+\deg (q)$, and $\deg (q)\lt \deg(p)$. Since $p$ is smallest degree such that $p(T)v=0$, we have $q(T)(v)\neq 0$. Let $\alpha= q(T)(v)$. Then $p(T)(v)=(T-cI)q(T)(v)=(T-cI)(\alpha)=0$. So $T(\alpha)=c\alpha$. Hence $c$ is eigenvalue of $T$.
Approach (2): Let $S=\{f\in F[x]\mid f(T)(v)=0\}$. Then $S$ is non zero polynomial ideal. Let $g$ be monic generator of $S$. By uniqueness of monic generator, $p=dg$, for some $d\in F\setminus \{0\}$. Since minimal polynomial of $T$ is in $S$, we have $g|m_T$. If $p(c)=0$, then $g(c)=0$. So $m_T(c)=0$. Since minimal and characteristic polynomial of $T$ have same roots, we have $c$ root of characteristic polynomial of $T$. Thus $c$ is eigenvalue of $T$. Is my proof correct?