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 $T$.
 A: Hints:


*

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

*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 $k = 0,1,\dots,n-1$.

*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.  
A: Let dim(V) = n. The minimal polynomial is equal to the characteristic polynomial if and only if
\begin{equation}
  n = \min \{ deg(f) | f \in F[x] \text{ s.t.} f(T)= 0 \},
\end{equation}
if and only if
\begin{equation} 
 \forall f\in F[x], \text{ s.t. } deg(f) =n-1 \text{ and } f(T)=0, \,\,\Rightarrow
 f = 0
\end{equation}
if and only if
\begin{equation}
  \sum_{i=0}^{n-1} a_{i}T^{i} = 0, \Rightarrow \,\, a_{i}=0, i=0, 1, \dots, n-1.
\end{equation}
The latter is equivalent to the existence of a $ v \neq 0 $ s.t.
if $ \sum_{i=0}^{n-1} a_{i}T^{i}(v) = 0 $ then $ a_{0}=a_{1}=\dots = a_{n-1} = 0 $. This is because, if for every nonzero $ v $, $ T^{i}(v) = 0 $, for
$ i = 1, 2, \dots , n-1 $, then for $ f(x) = x + x^{2} + \dots + x^{n-1} $,
$ f(T)= 0 $, and again it contradicts $ m_{T}(x) = \rho_{T}(x) $.
We conclude that, $ m_{T}(x) = \rho_{T}(x) $ if and only if
there exists $ v \neq 0 $ s.t. $ \{ T^{i}(v) \}_{i=0}^{n-1} $
is a basis for $ V $, i.e., $ V $ is $ T $-cyclic.
