Suppose the matrix $A \in M_n(\mathbb{F})$ (the set of $n \times n$ matrices with coefficients in $\mathbb{F}[x]$), and the vector $v \in \mathbb{F}^n$. Now let $k>0$ be the smallest integer such that the set of vectors $\{v,Av,A^2v,\ldots,A^kv\}$ is linearly dependent.
Let us now define $V= \operatorname{Span}(\{v,Av,\ldots,A^{k-1}v\}).$ How do we show that this set of vectors spanning $V$ is a basis $\mathcal{B}$? Well, to be a basis for some vector space, we require that the set of vectors span that vector space, which in this case, they do and that the set of vectors be linearly independent. This is where one of my difficulties arises: how is it that $\{v,Av,\ldots,A^{k-1}v\}$ is linearly independent, while $\{v,Av,\ldots,A^kv\}$ is not?
Next, suppose we let the transformation $T:V\rightarrow V$ be induced by multiplication by $A$, i.e. $T(w)= Aw$ for $w \in V$. How exactly do we find the coordinate matrix $[T]_{\mathcal{B}}$? I don't quite remember how to do this. I would guess that it would be something like this, but I'm not sure:
$ [w|Aw|\cdots|A^{k-1}w]_{\mathcal{B}}. $ Where do I go from here?
Finally, I know that the annihilator of $v$ w.r.t. $A$ given as the set
$\{g \in \mathbb{F}[x]: g(A)(v)=0\},$ is an ideal in $\mathbb{F}[x]$. But if I define the annihilator of the vector space $V$ w.r.t. $A$ as
$\{g \in \mathbb{F}[x]: g(A)(w)=0 \forall w \in V\},$ wouldn't both annihilators be the same? If so, how? I would really appreciate some help here.