$\mathscr{B}$-matrix of T and Characteristic Polynomial I'm having a difficult time trying to figure out this proof problem.  Any advice on first steps?

Let A be an $n\times n$ matrix satisfying the matrix equation

$A^{n} + c_{1}A^{n-1}+c_{2}A^{n-2} + . . . + c_{n-1}A + c_{n}I_{n} = 0 \cdot I_{n}$

and suppose v is a vector in $\mathbb{R}^n$ such that

$\mathscr{B}$ = {v, Av, . . . , $A^{n-1}$ v}

is a basis for $\mathbb{R}^n$.  Let T be the linear transformation whose standard matrix is A.
(a) Determine, with proof, the $\mathscr{B}$-matrix of T.
(b) Find the characteristic polynomial of A.

 A: Let $v_k = A^k v$, $k=0,...,n-1$.
Then note that $Av_k = v_{k+1}$, $k=0,...,n-1$, and
\begin{eqnarray}
A v_{n-1} &=&  A^n v \\
&=& -(c_{1}A^{n-1}+c_{2}A^{n-2} + . . . + c_{n-1}A + c_{n}I)v \\
&=& -(c_1v_{n-1}+c_2 v_{n-2}+...+c_{n-1}v_1+ c_n v_0)
\end{eqnarray}
Hence in the basis ${\cal B}$, that is, $v_0,...,v_{n-1}$, $T$ has the form:
$$C=\begin{bmatrix}
0 & 0 & 0& \cdots & -c_n\\
1 & 0 & & \cdots & -c_{n-1}\\
0 & 1 & 0 & \cdots & -c_{n-2}\\
\vdots & & \ddots& & \vdots \\
0 & \cdots & 0 & 1 & -c_1 \\
\end{bmatrix}$$
Computing $s \mapsto \det(s I -C)$ is straightforward and gives
$\det(s I -C) = c_n+c_{n-1}s + ... + c_1 s^{n-1} + s^n$. (One way is to start in the bottom right corner of the matrix and use induction on $\det(s I -C)$.)
Addendum: Let $V= \begin{bmatrix} v_0 & \cdots & v_{n-1} \end{bmatrix}$. Then, just by using the relationships above, we have $A V = V C$. Since $v_0,...,v_{n-1}$ is a basis, $V$ is invertible and we can write $C = V^{-1} A V$, and hence $A$ is similar to the ${\cal B}$ matrix of $T$.
Then we have $\det (sI -A) = \frac{1}{\det V} \det (sI-A) \det V = \det (V^{-1}(sI-A) V) = \det(sI -C)$. In general, the characteristic polynomial depends on $T$, not the particular matrix representation $A, $ or $C$.
