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I have asked such a question before:

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 ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$

And I gave the answer myself:

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${\mathscr A(α),\cdots,\mathscr A^r(α)}$.

we say set $S$ is linear independent if for any finite subset $\{α_1,\cdots,α_n\} $,if $x_1α_1+\cdots+x_nα_n=0$,then we have $x_1=\cdots=x_n=0$.if exist some finite subset $\{α_1,\cdots,α_n\} $ and Numbers that are not all zeros $x_1,\cdots,x_n$ such that $x_1α_1+\cdots+x_nα_n=0$,we say $S$ is linear dependent.

We say $α$ can be linear expression by $α_1,\cdots,α_n$ if extist $x_1,\cdots,x_n$ such that $α=x_1α_1+\cdots+x_nα_n $

Let $V$ linear space with dimension $n$

Definition: (i)let $α∈V$ the cylic subspace $C_α$ of is the smallest stable subpace under $\mathscr A$(is: the intersection of stable subpace contain $α$) (ii) Minimum polynomial $d_α(λ)$ of $α$ relative to to $\mathscr A$ is Polynomial with minimum degree and first term coefficient of 1 such that $d_α(\mathscr A)(α)=0$.

Here is some relatively theorem: $C_α$ is generated by $\{α,\cdots,\mathscr A^{r}(α),\cdots \}$

My confuse comes from the following theorem : theorem:assume $d_α(λ)=λ^r+a_{r-1}λ^{r-1}+\cdots+a_o$.then $dim C_α=r$ and $\mathscr A^r(α)=-a_0α+\cdots-a_{r-1}\mathscr A^{r-1}(α)$. but from my "proof" above:we have $a_0=\cdots=a_{r-1}=0$. It is seem to impossible,so Where did I go wrong?

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Let $r'$ be the smallest number such that $\alpha, A(\alpha),\dots,A^{r'}(\alpha)$ is linearly dependent.
Then $\alpha,\dots,A^{r'-1}(\alpha)$ are independent, so we can apply the lemma to express $$A^{r'}(\alpha)=\sum_{i<r'}a_iA^i(\alpha)\,.$$ It doesn't follow that $a_i$ must be $0$.
But, by induction we can clearly see that $A^k(\alpha)\in{\rm span}(\alpha,\dots,A^{r'-1}(\alpha))$ for every $k\in\Bbb N$, hence $U={\rm span}(\alpha,\dots,A^{r'-1}(\alpha))$, so $r=\dim U=r'$.

Also observe that $$A^r(\alpha)-a_{r-1}A^{r-1}(\alpha)-\dots -a_1A(\alpha)-a_0\alpha=0\,,$$ and that this polynomial $d_\alpha$ is a factor of the minimal polynomial of $A$.

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  • $\begingroup$ Why? From my "proof" above,i proved $\mathscr A^r(α)$ can be expression by ${\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$,so $a_0=0$,( I didn't find anything wrong from the "proof") $\endgroup$ – user158796 Feb 23 at 22:58
  • $\begingroup$ I couldn't follow your proof. Which set is linearly dependent / independent and why? Consider specific examples, say $Ae_1=e_2,\ Ae_2=e_1+e_2$. $\endgroup$ – Berci Feb 23 at 23:14
  • $\begingroup$ Sorry, I explained the definition of linear independent/dependent. $\endgroup$ – user158796 Feb 24 at 0:21
  • $\begingroup$ No, it's not about the definition. You state certain linear independence/dependence in your argument but they are mostly just not valid. Double check each place. You have a sentence starting with 'From the comment: we know...' What comment? The vectors $A\alpha,\dots,A^r\alpha$ may or may not be independent, but that's irrelevant. However, $\alpha,A\alpha,\dots,A^{r-1}\alpha$ is definitely independent while $A^r\alpha$ depends on them, as clearly deduced in my answer. $\endgroup$ – Berci Feb 24 at 6:35

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