Let \(V\) be an \(n\)-dimensional vector space over a field \(F\), with basis \(\{v_0,\dots,v_{n-1}\}\) and \(T(v_i)=v_{i+1\bmod n}\). Show that \(m_T(x)=x^n-1\), and deduce that \(T\) is diagonalizable exactly when \(x^n-1\) splits into distinct linear factors over \(F\) (in particular, over \(\CC\)).
Solution
Let \(V\) have basis \(v_0,\dots,v_{n-1}\) and \(T(v_i)=v_{i+1\bmod n}\).
1. The minimal polynomial of \(T\) is \(x^n-1\) over every base field.
Since \(T^n(v_i)=v_i\) for every basis vector, \(T^n=I\), so the minimal polynomial \(m_T\) divides \(x^n-1\).
On the other hand, \[ v_0,Tv_0,\dots,T^{n-1}v_0 =v_0,v_1,\dots,v_{n-1} \] is a basis. Thus no nonzero polynomial of degree less than \(n\) can annihilate \(T\): if \[ \sum_{j=0}^{m}a_jT^j=0, \qquad m<n, \] then applying it to \(v_0\) gives \(\sum_{j=0}^{m}a_jv_j=0\), hence every \(a_j=0\). Therefore \(\deg m_T\ge n\). Since \(m_T\mid x^n-1\), both monic of degree \(n\), \[ \boxed{m_T(x)=x^n-1}. \]
2. The operator \(T\) is diagonalizable exactly when \(x^n-1\) splits into distinct linear factors over the base field.
A linear operator is diagonalizable iff its minimal polynomial splits into distinct linear factors. By 1, the minimal polynomial is \(x^n-1\), so this criterion is immediate.
In particular, over \(\CC\) (or any field containing all \(n\)-th roots of unity with characteristic not dividing \(n\)), \(x^n-1\) has \(n\) distinct roots, hence \(T\) is diagonalizable.