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
The rational canonical form of \(A\) is determined by its invariant factors \(a_1\mid a_2\mid\cdots\mid a_k\), which are monic, satisfy \(a_k=q\), and have \(\sum_i\deg a_i=n\).
1. For \(n=6\) and \(q=(x-2)^2(x+3)\), the invariant-factor lists, over \(\QQ\) and over \(\CC\) alike, are\[\begin{aligned} &q,\ q; &&(x-2),\ (x-2)^2,\ q; &&(x-2),\ (x-2)(x+3),\ q;\\ &(x+3),\ (x-2)(x+3),\ q; &&(x-2),\ (x-2),\ (x-2),\ q; &&(x+3),\ (x+3),\ (x+3),\ q. \end{aligned}\]
The factors \(a_1,\dots,a_{k-1}\) divide \(q\) and have degrees summing to \(3\). The proper monic divisors of \(q\) are \(x-2\), \(x+3\), \((x-2)^2\) and \((x-2)(x+3)\). A chain of two factors has degrees \(1,2\), giving \((x-2)\mid(x-2)^2\), \((x-2)\mid(x-2)(x+3)\) or \((x+3)\mid(x-2)(x+3)\); a chain of three factors of degree \(1\) is constant, giving \((x-2)^3\) or \((x+3)^3\); a single factor of degree \(3\) is \(q\). Since \(q\) splits over \(\QQ\), the lists over \(\CC\) are the same.
2. The corresponding Jordan forms over \(\CC\) are, in the same order,\[\begin{aligned} &J_2(2)^{\oplus2}\oplus J_1(-3)^{\oplus2}, &&J_2(2)^{\oplus2}\oplus J_1(2)\oplus J_1(-3), &&J_2(2)\oplus J_1(2)^{\oplus2}\oplus J_1(-3)^{\oplus2},\\ &J_2(2)\oplus J_1(2)\oplus J_1(-3)^{\oplus3}, &&J_2(2)\oplus J_1(2)^{\oplus3}\oplus J_1(-3), &&J_2(2)\oplus J_1(-3)^{\oplus4}. \end{aligned}\]
Each invariant factor \(\prod_\lambda(x-\lambda)^{e_\lambda}\) contributes the Jordan blocks \(J_{e_\lambda}(\lambda)\) with \(e_\lambda>0\).
3. For \(n=7\) and \(q=(x^2+1)(x-7)\), the invariant-factor lists over \(\QQ\) are\[(x^2+1),\ (x^2+1),\ q;\qquad (x-7),\ (x-7),\ (x-7),\ (x-7),\ q;\qquad (x-7),\ q,\ q.\]
The monic divisors of \(q\) in \(\QQ[x]\) are \(1\), \(x-7\), \(x^2+1\) and \(q\). The factors before \(q\) form a divisibility chain with degrees summing to \(4\): four copies of \(x-7\), two copies of \(x^2+1\), or \(x-7\) followed by \(q\).
4. Over \(\CC\), every such \(A\) is diagonalizable with Jordan form \(\operatorname{diag}(i^{(m_1)},(-i)^{(m_2)},7^{(m_3)})\), where \(m_1,m_2,m_3\ge1\) and \(m_1+m_2+m_3=7\); there are \(\binom62=15\) of them.
Over \(\CC\), \(q=(x-i)(x+i)(x-7)\) has distinct roots, so every Jordan block has size \(1\), and each root of \(q\) is an eigenvalue. The multiplicities form a composition of \(7\) into three positive parts.
5. Over \(\CC\), the rational canonical forms for \(n=7\) correspond to the same \(15\) triples: with \(k=\max(m_1,m_2,m_3)\), the invariant factors are\[a_j=(x-i)^{[j>k-m_1]}(x+i)^{[j>k-m_2]}(x-7)^{[j>k-m_3]},\qquad 1\le j\le k,\]where \([P]\) is \(1\) if \(P\) holds and \(0\) otherwise.
The elementary divisors are \(m_1\) copies of \(x-i\), \(m_2\) of \(x+i\) and \(m_3\) of \(x-7\); the \(j\)th invariant factor is the product over the eigenvalues whose number of copies is at least \(k-j+1\).