Linear span proof 
Let $\mathbb{F}$ be a field, $A \in \cal{M}_{n\times n}(\mathbb{F})$
  and $W=\left\{B \in {\cal{M}}_{n\times n}(\mathbb{F})|AB=BA \right\}$.
  Suppose there exists a column vector $v \in  \mathbb{F}^n$ such that
  $\left\{v,Av,A^2v,..., A^{n-1}v\right\}$ is a basis for
  $\mathbb{F}^n$. Prove that $\left\{I,A,A^2,...,A^{n-1}\right\}$ is a
  basis for $W$.
  $(\cal{M}_{n\times n}(\mathbb{F})$ is the vector space comprising all $n\times n$ matrices over $\mathbb{F}.)$

I'm done with the linear independence part, but to prove $\left\{I,A,A^2,...,A^{n-1}\right\}$ is indeed a basis I would need to prove $B=c_0I+c_1A+...+c_{n-1}A^{n-1}$ for some scalar $c_i$. However, since $Bv \in \mathbb{F}^n$ and $\left\{v,Av,A^2v,..., A^{n-1}v\right\}$ is a basis for $\mathbb{F}^n$, I only have $Bv=(c_0I+c_1A+...+c_{n-1}A^{n-1})v$. How do I proceed from here to "eliminate" $v$ from the above equation? Would be very much appreciated if someone could shed me some light on this, thanks!

Hint
$$Bx = \sum_{k=0}^{n-1} \lambda_k A^k v$$
For any $x$ (since we have a basis). Now check $ABx$ and $BAx$ and compare coefficients.
