Describe all conjugacy classes of \(3\times3\) matrices \(A\in M_3(\QQ)\) satisfying \[ A^4-A^3-A+I=0. \] Give a representative of each class.
Solution
Factor the annihilating polynomial: \[ x^4-x^3-x+1=(x-1)(x^3-1)=(x-1)^2(x^2+x+1). \] Over \(\QQ\), the factors \[ x-1,\qquad x^2+x+1 \] are irreducible. Therefore the minimal polynomial of \(A\) divides \[ (x-1)^2(x^2+x+1). \]
View \(\QQ^3\) as a \(\QQ[x]\)-module via \(x\cdot v=Av\). The possible elementary divisors have total degree \(3\), with \((x-1)\)-primary blocks of size at most \(2\) and an \((x^2+x+1)\)-primary block of size at most \(1\).
There are exactly three possibilities:
1. Three size-\(1\) \((x-1)\) blocks:\[ A\sim I_3. \]
2. One size-\(2\) \((x-1)\) block and one size-\(1\) block:\[ A\sim \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}. \]
3. One \((x-1)\) block and one block for \(x^2+x+1\):\[ A\sim \begin{pmatrix} 1&0&0\\ 0&0&-1\\ 0&1&-1 \end{pmatrix}. \]The lower-right block is a companion matrix for \(x^2+x+1\).
These three matrices have distinct minimal polynomials \[ x-1, \qquad (x-1)^2, \qquad (x-1)(x^2+x+1), \] so they are pairwise nonconjugate. Hence these are exactly the conjugacy classes.