A linear operator $T:V\rightarrow V$ has a cyclic vector iff $f_T=m_T$ (minimal polynomial=characteristic polynomial) I want to prove the following statement:

Let linear operator $T:V\rightarrow V$ ($V$ is $n$-dimentional). Then there exists a vector $v$ such that $\left\{ v, Tv, ..., T^{n-1}v \right\}$ form a basis of $V$, iff $f_T=m_T$.

(Actually, the $\Rightarrow$ direction is quite easy. I need to prove the other direction.)
I've seen couple of answers to this problem in this site, all using the "Rational Canonial Form", which is something I wasn't tought in the course (but I am familiar with the Jordan Form).
Moreover, the only proof I've found on the internet is this, which seemed to be promising, until I got to this sentence, which unfortunately seems to be a mistake:

See that $V_i$ is a subspace of $V$, for all $1 \leq i \leq m$, and
  $V=\bigcup_{i=1}^m V_i$. Therefore $V=V_k$, for some $1 \leq k \leq m$.

I don't know if that's true for infinite vector spaces, but for finite ones, such as $\mathbb{F}_p^n$ it's certainly isn't.
Any step towards a proof would be appriciated (as well as an explanation for the suspicious claim stated above). Thank you!
 A: Suppose that the degrees of the characteristic polynomial $p(x)$ and minimal polynomial $m(x)$ of $T:V\to V$ coincide. 
One can express $V$ as a direct sum of the generalised eigenspaces of $T$. If $\lambda_1,\ldots,\lambda_k$ are the eigenvalues of $T$, then
$$
V=V(\lambda_1)\oplus\cdots\oplus V(\lambda_k),
$$ 
where $V(\lambda_j)=\{v\in V: (T-\lambda_j)^\ell v=0\,\,\text{for some $\ell\in\mathbb N$}\}$. Each of these generalised eigenspaces is $T-$invariant. Set $T_j$ the restriction of $T$ to $V(\lambda_k)$, and let $p_j(x)$ and $m_j(x)$ the characteristic and minimal polynomial of $T_j$, respectively. In fact we have that $p(x)=p_1(x)\cdots p_k(x)$ and
$$
m(x)=lcm\{m_1(x),\ldots, m_k(x)\}=m_1(x)\cdots m_k(x),
$$
since the $m_j$'s are pairwise prime. And since, $m(x)$ and $p(x)$ have the same degree, then $m_j(x)=(x-\lambda_j)^{\mu_j}$, where $\mu_j$ is the multiplicity of $\lambda_j$ in $p(x)$.
Hence $(T_j-\lambda_j)^{\mu_j-1}\ne 0$ and $(T_j-\lambda_j)^{\mu_j}= 0$. Thus there exists a $v_j\in V(\lambda_j)$, such that
$$
V(\lambda_j)\ni (T_j-\lambda_j)^{\mu_j-1}v_j\ne 0.
$$
In fact $\{v_j,Tv_j,\ldots,T^{\mu_j-1}v_j\}$ is a basis of $V(\lambda_j)$.
Finally set
$$
v=v_1+\cdots+v_k.
$$
Then $\{v,Tv,\ldots,T^{n-1}v\}$ is a basis of $V$.
Indeed, if 
$$
0=c_0v+c_1Tv+\cdots+c_{n-1}T^{n-1}v=p(T)v,
$$
then $p(T)v_j=0$, for all $j=1,\ldots k$. Hence $(x-\lambda_j)^{\mu_j}$ divides $p(T)$, and hence the characteristic polynomial of $T$ divides $p$. Thus $p\equiv 0$, since $\deg p<n$.
