Prove a Vector set is linear independent? 
problem:Let $V$ be a linear space,$\mathscr A $ is a linear translation on $V$.choose $0≠α∈V$.let $S=\{\mathscr A^k(α):k≥0\}$ and $U=span(S)$, prove that:
(i) $U$ is stable under $\mathscr A$.
(ii) assume $dim(U)=r$,then ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is a basis of $U$ and find out the metric of  $\mathscr A|_U$ under basis ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$.

My attempt
For(i) i know we only check $\mathscr A$ map the element of $S$ to $U$.
For(ii) i don't know how to do it.I try yes the definition of linear independent to prove it,but i failed. Then i want to use a proposition : if there is k such
that  $\mathscr A^{k-1}(α)≠0,\mathscr A^{k}(α)=0$  then ${α,\mathscr A(α),\cdots,\mathscr A^{k-1}(α)}$ is linear independent,i want to use $dim(U)=r$ to prove that k is exist, but i also failed.
 A: 
lemma: if ${α_1,α_2,\cdots,α_n}$ is linear independent,${α_1,α_2,\cdots,α_n,β}$ is linear dependent,then $β$ can be linear expression by ${α_1,α_2,\cdots,α_n}$,but $α_i$ can't be linear expression by the other.

From the comment:we know${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is linear dependent and ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α),\mathscr A^r(α)}$ is linear dependent.
We write $\mathscr A^r(α)=x_0α+\cdots+x_{r-1}\mathscr A^{r-1}(α)$.
We have$(\mathscr A(α),\cdots,\mathscr A^r(α))⊆ (α,\cdots,\mathscr A^{r-1}(α))$. If ${\mathscr A(α),\cdots,\mathscr A^r(α)}$ is independent,then $α$ can be linear expression by ${\mathscr A(α),\cdots,\mathscr A^r(α)}$,by lemma,it is impossible, so ${\mathscr A(α),\cdots,\mathscr A^r(α)}$ is linear dependent,thus $\mathscr A^r(α)$ can be expression by ${\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$,so $x_0=0$.
we know${\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is linear dependent and ${\mathscr A(α),\cdots,\mathscr A^{r-1}(α),\mathscr A^r(α)}$ is linear dependent. Use similarly way with above,we can get $x_1=0$,this  process can continues. So $x_0=\cdots=x_{r-1}=0$.
So $\mathscr A^r(α)=0$.
Remark:$(\mathscr A(α),\cdots,\mathscr A^r(α))$ is subspace generated by${\mathscr A(α),\cdots,\mathscr A^r(α)}$.
