Linear operator $T$ on $F^2$ defined by $[T]_B=\begin{bmatrix}0&0\\ 1&0 \\ \end{bmatrix}$ has cyclic vector for $T$ In Linear algebra by Hoffman and Kunze,it is given that "an example of an operator which has a cyclic vector is the linear operator $T$ on $F^2$ which is represented in the standard ordered basis by the matrix $\bigl[ \begin{smallmatrix} 0 & 0 \\ 1 & 0 \end{smallmatrix} \bigr]$"
My question is HOW?
My argument:Since the characteristic polynomial of the matrix is $x^2$,so its minimal polynomial is $x$.So,the characteristic polynomial and the minimal polynomial are different.By making use of Theorem:  $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same.... 
Hence,$T$ cannot have the cyclic vector.
Please help!!
P.S.

Please explain stuff  in yellow highlighted text!!
 A: Let $v=(1,0)^T$. Then $Tv=(0,1)^T$. Thus $\{v,Tv\}$ is a basis of $F^2$.
Conclusion: $v$ is a cyclic vector.
A: Characteristic polynomial of matrix $\begin{bmatrix}0&0\\ 1&0\\ \end{bmatrix}$ is $x^2$. So it’s minimal polynomial is $x$ or $x^2$. Certainly $x$ is not minimal polynomial, because $x$ don’t annihilate that matrix. Thus $x^2$ is minimal polynomial. The theorem you stated is correct but we can’t use result of that theorem, since it’s proof comes later in the book.
Author explicitly showed $e_1$ is cyclic vector of $T$, i.e. $F^2=Z(e_1;T)=\{g(T)(e_1)\mid g\in F[x]\}$. Let $\beta =(a,b)\in F^2$. Note $T(e_2)=(0,0)$. So $\beta =a\cdot e_1+b\cdot e_2=a\cdot e_1+b\cdot T(e_1)$. Take $g=a+bx$. Clearly $g(T)(e_1)=\beta \in Z(e_1;T)$. Thus $F^2\subseteq Z(e_1;T)$. Trivially $F^2\supseteq Z(e_1;T)$. Hence $F^2=Z(e_1;T)$.
Since $T(e_2)=(0,0)=0\cdot e_2$, we have $e_2$ is eigenvector of $T$ corresponding to eigenvalue $0$. By example 1 section 7.1, $\dim Z(e_2;T)=1$. Thus $Z(e_2;T)\neq F^2$. Hence $e_2$ is not cyclic vector of $T$.
Edit: Well, I recently proved the theorem OP used. It’s proof is not trivial.
