Classify up to similarity all \(3 \times 3\) complex matrices \(A\) such that \(A^3 = I\).
Solution
Let \[ \omega=e^{2\pi i/3}. \] Then \[ x^3-1=(x-1)(x-\omega)(x-\omega^2) \] has three distinct roots. Since \(A^3=I\), the minimal polynomial of \(A\) divides \(x^3-1\). Hence it has no repeated root, so \(A\) is diagonalizable [@HK71]. Its eigenvalues can only be \(1,\omega,\omega^2\).
Therefore every such matrix is similar to exactly one matrix \[ \boxed{ \operatorname{diag}(I_a,\omega I_b,\omega^2 I_c), \qquad a,b,c\ge0,\quad a+b+c=3.} \] Two such diagonal matrices are similar if and only if the three eigenvalue multiplicities agree. Thus this parameterization is a complete and irredundant classification. There are \[ \binom{3+3-1}{3}=10 \] similarity classes.