# A confuse about Cyclic subspace

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?

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 $$A^{r'}(\alpha)=\sum_{i It doesn't follow that $$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$$.
• 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") – user158796 Feb 23 at 22:58
• I couldn't follow your proof. Which set is linearly dependent / independent and why? Consider specific examples, say $Ae_1=e_2,\ Ae_2=e_1+e_2$. – Berci Feb 23 at 23:14
• 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. – Berci Feb 24 at 6:35