I want to prove the following statemaent.
Let $V$ be a $n$-demensional vector space on field $K$ and $f:V\rightarrow V$ be a linear operator. There exists $v\in V$ such that {$v,f(v),f^2(v),\ldots,f^{n-1}(v)$} is a basis of $V$ if the minimal polynomial of $f$ is equal to the characteristic polynomial of $f$.
I've checked this page, but I couldn't understand why $v_i, Tv_i, T^2v_i, \ldots, T^{\mu_{j}-1}v_i$ are linearly independent (in the answer by Yiorgos S. Smyrlis).
Any help is appreciated. Thanks.
Note that $K$ is not neseccarily a algebraically closed field.
Regarding your comment: I will remove the subscript $i$ to simplify notation.
At the relevant point in the proof, we are given that $T|_{V}$ has minimal polynomial $P^m$, where $P$ is irreducible, and we must show the existence of a cyclic generator of $V$, i.e., a vector $v\in V$ such that the minimal degree monic polynomial $Q$ such that $Q[T](v)=0$ has $\deg(Q)=\dim(V)$.
Let $d = \dim(V)$. Note in particular that the degree of this polynomial $Q$ is the dimension of the span of $\{v,Tv,\dots,T^{d-1}v\}$.
As the minimal polynomial of $T|_{V}$ is $P^{m}$, this polynomial $Q$ divides $P^{m}$ for any vector$~v\in V$: only powers of $P$ can occur. It therefore suffices to find a vector$~v\in V$ for which $P^{m-1}[T](v)\neq0$: if $P^{m-1}[T](v)\neq0$, then we know that $Q$ divides $P^m$ but does not divide $P^{m-1}$, which means that we must have $Q = P^m$. However, that means that $\deg(Q) = \deg(P^m) = d$, which means that the set $\{v,Tv,\dots,T^{d-1}v\}$ spans $V$.