I'm stuck on the following practice question, and I'm uncertain of how to proceed with the proof. We've only very briefly touched on annihilators in class, and I really don't understand how to combine them with eigenvectors and eigenvalues. Any help would be appreciated.
Q. Consider a vector space $V$ with $\dim(V)=n$ and $T \in \operatorname{End}_K(V)$
(a) Show that $\operatorname{Ann}(V)=\operatorname{Ann}(u)$ for some eigenvector $u$ of $T$ if and only if $T$ is diagonalizable with exactly one eigenvalue.
(b) Show that $\operatorname{Ann}(V)=\operatorname{Ann}(u)$ for some eigenvector $u$ of $T$ if and only if $T=\lambda I$
EDIT For a vector space V over a field K, we denote $\operatorname{Ann}(u)=\{p(x) \in K[x] ; p(T)(u)=0 \}$.
We denote $\operatorname{Ann}(V)=\{p(x) \in K[x] ; p(T)=0 \}$.
Note that $p(T) = 0$ if and only if $p$ is divisible by the minimal polynomial of $T$, which we can call $m_T(x)$. So, we can write $\operatorname{Ann}(V) = \{m_T(x)q(x) : q \in K[x]\}$, the set of all polynomials divisible by $m_T(x)$.
Now, if $u$ is an eigenvector with eigenvalue $\lambda$, then for any $p$, $p(T)u = p(\lambda) u$. It follows that $p(T)u = 0$ iff $p(\lambda) = 0$, which is to say that $x - \lambda$ divides $p$. So, we can write $\operatorname{Ann}(u) = \{(x - \lambda)q(x) : q \in K[x]\}$, the set of all polynomials divisible by $x - \lambda$ (alternatively, we can think of this as the reapplication of the same argument over an invariant subspace).
Now, if these two sets are the same, what can we deduce about $m_T$?