Given the characteristic polynomial, find the minimal polynomial. The Question:
Let $f(x) = x^n+\alpha_{n-1}x^{n-1}+\alpha_{n-2}x^{n-2} + \cdots + \alpha_0$, and let $A$ be the matrix
\begin{pmatrix}
0 & 1 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & 1 & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 0 & 1 \\
-\alpha_0 & -\alpha_1 & -\alpha_2 & -\alpha_3 & \cdots & -\alpha_{n-2} & -\alpha_{n-1}
\end{pmatrix}
(i) Find the characteristic polynomial
(ii) Find the minimal polynomial

My Attempt:
(i) No problem here, I found that $\chi _A (x) = (-1)^n f(x)$.
(ii) No idea here, I only know that $m_A(x) \, | \, \chi _A (x)$.
Assuming that $f \in \bar {\Bbb F} [x]$, we can factorize $f$ into linear factors $$f(x) = (\lambda_1 - x)(\lambda_2 - x) \cdots (\lambda_n - x)$$
where the $\lambda_i$ are the eigenvalues of $A$, and then I am still stuck.
Any hints?

EDIT:
OK, after some experiments with smaller matrices, it seems that the minimal polynomial is always the full polynomial, i.e. $m_A(x) = f(x)$.
I have tried induction on the size of the matrix, but it does not seem to work either.
 A: Look at what happens to the matrix as you take powers.  Those ones on the superdiagonal move up a diagonal, there are still zeros above them, and more or less anything goes below them.  Thus, in order to get a non-trivial linear combination of powers of this matrix to be zero, you need to push the diagonal of ones out of the matrix entirely...otherwise the highest power appearing will have a super-super-....-super diagonal (possibly just the upper right corner) of ones that no other power can subtract out.  It's fairly easy to see that you need to take the $n^{th}$ power to clear all the ones.  Thus, the minimal polynomial must have degree at least $n$.  Since it divides the characteristic polynomial, which also has degree $n$, they must be equal.
A: I prefer to see this by considering the space $V$ of row vectors of length $n$, with the standard basis
$$
v_{0} = (1, 0, \dots, 0), \quad
v_{1} = (0, 1, \dots, 0),\quad
\quad\dots,\quad
v_{n-1} = (0, 0, \dots, 1).
$$
The matrix $A$ acts on $V$ by multiplication on the right. 
Clearly $v_{i} A = v_{i+1}$ for $0 \le i < n-1$, so that $v_{0} A^{i} = v_{i}$ for $0 < i \le n-1$.
Take any polynomial
$$
c = c_{0} + c_{1} x + \dots + c_{n-1} x^{n-1}
$$ 
of degree less than $n$, and suppose
$$
0 = c(A)
=c_{0} I + c_{1} A + \dots + c_{n-1} A^{n-1},
$$ 
where $I$ is the $n \times n$ identity matrix.
Then
$$
0 = v_{0} c(A)
=
v_{0} (c_{0} I + c_{1} A + \dots + c_{n-1} A^{n-1})
=
c_{0} v_{0} + c_{1} v_{1} + \dots + c_{n-1} v_{n-1}.
$$
Since the $v_{i}$ are linearly independent, all $c_{i}$ are zero, so that the minimal polynomial has degree at least $n$. Since the minimal polynomial divides the characteristic polynomial, and the latter has degree $n$, the two must coincide.
