Minimal polynomial of $T$ and is it diagonalizable? Let $V$ be a vector space over complex numbers with a basis { $v_1,v_2, ..., v_n$} and let $T : V\to V$ be a linear endomorphism given by
$T(v_1)$ = $v_2$, $T(v_2)$ = $v_3$,...,$T$($v_{n-1}$) = $v_n$, $T(v_n)$ = $v_1$
What is the minimal polynomial of $T$? Also how do I show $T$ is diagonalizable?
 A: Your operator $T$ acts on the basis $(v_1,\dots,v_n)$ by a cyclic permutation $\sigma = (1,2,\dots,n)$. That is, $Tv_i = v_{\sigma(i)}$ for all $1 \leq i \leq n$. The powers of $T$ act by the powers of $\sigma$ and since $\sigma^{n} = \operatorname{id}$ we have $T^{n} = \operatorname{id}|_{V}$ so the minimal polynomial of $T$ must divide $x^{n} - 1$. In fact, this is precisely the minimal polynomial of $T$. To see why, note that if $p(x) = a_0 + a_1x + \dots + a_k x^k$ is a polynomial of degree $k < n$ then
$$ p(T)v_1 = a_0 v_1 + a_1 v_2 + a_2 v_3 + \dots + a_k v_{k + 1}. $$
If $p(T) = 0$ then $p(T)v_1 = 0$ but then $a_0 = \dots = a_k = 0$ by the linear independence of the $(v_i)_{i=1}^{k+1}$ so $p = 0$.
Since the roots of the minimal polynomial $x^{n} - 1$ are distinct, this implies that $T$ is diagonalizable.
A: Hint:
$T$ acts on the basis, so it can be interpreted as the circular permutation $(v_1 v_2\dots v_n)$. This shows $T^n =I$. Can you deduce the characteristic polynomial? 
For the minimal polynomial, you have to remember it has the same irreducible factors as the characteristic polynomial, possibly with different multiplicities.
