Minimal polynomial matrix I want to show that $ x^n-1$ is the minimal polynomial of the permutation matrix $P:=(e_2,e_3,....,e_n,e_1)$ where $e_i$ is the i-th unit vector written as a column vector. 
And now I have to show that over an arbitrary field $x_n-1$ is the minimal polynomial. This seems to be tough. I have already proved that $P^n-Id=0$. Does anybody of you know how to do this or has a hint for me? 
 A: Since you observed that $P$ is annihilated by $X^n-1$, it only remains to check that no nonzero polynomial of degree $\leq n-1$ annihilates $P$ to conclude that the minimimal polynomial is $X^n-1$. Since the latter has degree $n$ and divides the characteristic polynomial which is also monic of degree $n$, it will also follow that the characteristic polynomial is $X^n-1$.
So assume that a polynomial of degree $\leq n-1$ annihilates $P$, that is
$$
a_0I_n+a_1P+\ldots+a_{n-1}P^{n-1}=0.
$$
Applying the latter to the first vector $e_1$ of the canonical basis, we get
$$
a_oe_1+a_1Pe_1+\ldots+a_{n-1}P^{n-1}e_1=a_0e_1+a_1e_2+\ldots+ a_{n-1}e_n=0.
$$
That is a linear combination of the basis, whence, by linear independence of the latter, $a_0=a_1=\ldots=a_{n-1}=0$. So no nontrivial polynomial of degree $\leq n-1$ annihilates $P$, which concludes the proof.
Note: you might want to note that this is a special case of a companion matrix $C(q)$, for which in general the characteristic and the minimal polynomials are both equal to $q$.
