Prove that $T$ has a cyclic vector iff its minimal and characteristic polynomials are the same

Let $k$ be an algebraically closed field and $V$ be a finite-dimensional $k$-vector space of dimension $n$.

Let $T:V \rightarrow V$ be a $k$-linear endomorphism of $V$. A vector $v \in V$ is called a cyclic vector for $T$ if the set of vectors $\{T^nv: n \in \mathbb{Z}, n \geqslant 0\}$ span $V$.

1 Show that if $v \in V$ is a cyclic vector, then $\{v, Tv,\cdots, T^{n-1}v\}$ form a basis for $V$.

2 If $T$ admits a cyclic vector, and $A:V\rightarrow V$ is a linear map commuting with $T$, show that there exists a polynomial $P(x) \in k[x]$ such that $A=P(T)$.

3 Show that a cyclic vector for $T$ exists if and only if the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$.

Hints:

1. By the Cayley-Hamilton theorem, $T^kv$ is in the span of $\{v,Tv,\dots,T^{n-1}v\}$ for any $k \geq n$

2. By 1, there exists a polynomial $T$ of degree $n-1$ such that $$ Av = p(T)v $$ Then, note that $AT^kv = T^k Av = T^kp(T)v = p(T)T^k v$ for $k = 0,1,\dots,n-1$.

3. If the minimal polynomial is of lower degree, then $T^{n-1}$ is a linear combination of $\{I,T,\dots,T^{n-2}\}$, so there can be no cyclic vector. The opposite implication is tricky; it suffices to consider the Jordan form or rational canonical form of $T$, but that may be excessive.