Classify up to similarity all \(n \times n\) complex matrices \(A\) such that \(A^n = I\).
Solution
Let \[ \zeta=e^{2\pi i/n}. \] The polynomial \[ x^n-1=\prod_{j=0}^{n-1}(x-\zeta^j) \] has distinct roots over \(\mathbb C\), since its derivative \(nx^{n-1}\) has no common root with it. If \(A^n=I\), then the minimal polynomial of \(A\) divides \(x^n-1\), hence is squarefree. Therefore \(A\) is diagonalizable and every eigenvalue is an \(n\)th root of unity [@HK71].
Thus the similarity classes are precisely \[ \boxed{ \operatorname{diag} (I_{m_0},\zeta I_{m_1},\ldots,\zeta^{n-1}I_{m_{n-1}}), \qquad m_j\ge0,\quad \sum_{j=0}^{n-1}m_j=n.} \] The multiplicity vector \((m_0,\ldots,m_{n-1})\) is determined by the characteristic polynomial and hence by the similarity class, so the list is irredundant. Conversely every displayed diagonal matrix has \(n\)th power equal to \(I\). In particular, the number of classes is \[ \binom{2n-1}{n}. \]