Find all possible rational canonical forms for a matrix \(A\in M_n(\Bbb Q)\) such that
\(A\) is \(6\times 6\) with minimal polynomial \(q(x) = (x-2)^2(x+3)\).
\(A\) is \(7\times 7\) with \(q(x) = (x^2+1)(x-7)\).
Also find all such forms when \(A \in M_n(\Bbb C)\) instead, and find all possible Jordan Canonical Forms over \(\Bbb C\).
Solution
By the structure theorem for finitely generated torsion modules over the PID \(F[x]\), a matrix \(A\in M_n(F)\) is determined up to similarity by its invariant factors, monic polynomials \(f_1 \mid f_2 \mid \cdots \mid f_k\) of positive degree with \(f_k = q\) the minimal polynomial and \(\sum_j \deg f_j = n\). The rational canonical form is \(\bigoplus_{j=1}^k C(f_j)\), where \(C(f)\) is the companion matrix of \(f\). Over \(\CC\), the Jordan form is the direct sum of the Jordan blocks \(J_a(\lambda)\), one for each elementary divisor \((x-\lambda)^a\), and the elementary divisors are the prime-power factors of the invariant factors. We list each rational canonical form by its invariant factors.
1. (1) Over \(\QQ\), the possible invariant factor lists for \(n=6\), \(q=(x-2)^2(x+3)\) are exactly:
- \((x-2)^2(x+3),\ (x-2)^2(x+3)\);
- \(x-2,\ (x-2)^2,\ (x-2)^2(x+3)\);
- \(x-2,\ (x-2)(x+3),\ (x-2)^2(x+3)\);
- \(x+3,\ (x-2)(x+3),\ (x-2)^2(x+3)\);
- \(x-2,\ x-2,\ x-2,\ (x-2)^2(x+3)\);
- \(x+3,\ x+3,\ x+3,\ (x-2)^2(x+3)\).
The factors \(f_1\mid\cdots\mid f_{k-1}\) are monic divisors of \(q\) of positive degree with degrees summing to \(6-3=3\). The monic divisors of \(q\) of positive degree are \(x-2\), \(x+3\), \((x-2)^2\), \((x-2)(x+3)\), and \(q\). A single factor of degree \(3\) is \(q\), giving list 1. A factor of degree \(2\) preceded by one of degree \(1\) that divides it gives lists 2, 3, and 4, since \(x+3\nmid(x-2)^2\). Three factors of degree \(1\) in a divisibility chain are equal, giving lists 5 and 6.
2. (1) Over \(\CC\), the rational canonical forms are the same six, with Jordan forms respectively\[\begin{aligned} &1.\ J_2(2) \oplus J_2(2) \oplus J_1(-3) \oplus J_1(-3),\\ &2.\ J_1(2) \oplus J_2(2) \oplus J_2(2) \oplus J_1(-3),\\ &3.\ J_1(2) \oplus J_1(2) \oplus J_2(2) \oplus J_1(-3) \oplus J_1(-3),\\ &4.\ J_1(2) \oplus J_2(2) \oplus J_1(-3) \oplus J_1(-3) \oplus J_1(-3),\\ &5.\ J_1(2) \oplus J_1(2) \oplus J_1(2) \oplus J_2(2) \oplus J_1(-3),\\ &6.\ J_2(2) \oplus J_1(-3) \oplus J_1(-3) \oplus J_1(-3) \oplus J_1(-3). \end{aligned}\]
The monic divisors of \(q\) in \(\CC[x]\) are the same as in \(\QQ[x]\) because \(q\) splits over \(\QQ\), so the argument of step 1 gives the same six lists. Factoring each invariant factor into powers of \(x-2\) and \(x+3\) gives the displayed Jordan forms.
3. (2) Over \(\QQ\), the possible invariant factor lists for \(n=7\), \(q=(x^2+1)(x-7)\) are exactly:
- \(x^2+1,\ x^2+1,\ (x^2+1)(x-7)\);
- \(x-7,\ x-7,\ x-7,\ x-7,\ (x^2+1)(x-7)\);
- \(x-7,\ (x^2+1)(x-7),\ (x^2+1)(x-7)\).
Since \(x^2+1\) has no rational root, it is irreducible over \(\QQ\), and the monic divisors of \(q\) of positive degree are \(x-7\), \(x^2+1\), and \(q\). The factors \(f_1\mid\cdots\mid f_{k-1}\) have degrees summing to \(4\). Because \(x-7\nmid x^2+1\), a chain containing \(x^2+1\) below \(q\) consists of copies of \(x^2+1\), giving list 1; a chain containing \(q\) below \(q\) has one further factor of degree \(1\), giving list 3; the remaining chain is four copies of \(x-7\), list 2.
4. (2) Over \(\CC\), the possible Jordan forms are\[\operatorname{diag}(\underbrace{i, \ldots, i}_{m_1}, \underbrace{-i, \ldots, -i}_{m_2}, \underbrace{7, \ldots, 7}_{m_3})\]with \(m_1, m_2, m_3 \ge 1\) and \(m_1 + m_2 + m_3 = 7\); there are \(\binom{6}{2} = 15\) of them. For the rational canonical form with multiplicities \(m_1,m_2,m_3\), the \(j\)th invariant factor counted from the last is \(\prod_{\lambda\in\{i,-i,7\},\ m_\lambda\ge j}(x-\lambda)\), for \(1\le j\le\max(m_1,m_2,m_3)\).
Over \(\CC\), \(q = (x-i)(x+i)(x-7)\) has distinct roots, so \(A\) is diagonalizable with eigenvalues among \(i,-i,7\), and each root of the minimal polynomial is an eigenvalue. The number of compositions of \(7\) into three positive parts is \(\binom62=15\). The elementary divisors are \(m_\lambda\) copies of \(x-\lambda\) for each \(\lambda\); the last invariant factor is the product of one copy of each, and removing it and repeating gives the stated formula.