Let \(F\) be a subfield of the field of complex numbers, and let \(A\) and \(B\) be \(n \times n\) matrices over \(F\). Prove that if \(A\) and \(B\) are similar over the field of complex numbers, then they are similar over \(F\). (Hint: Prove that the rational form of \(A\) is the same whether \(A\) is viewed as a matrix over \(F\) or a matrix over \(C\); likewise for \(B\).)
Solution
1. Rational form is unchanged by extending scalars from \(F\) to \(\mathbb C\).
Let \(A\in M_n(F)\). Its rational canonical form over \(F\) is a block diagonal matrix \[ R_A=\operatorname{diag}(C(p_1),\ldots,C(p_r)), \] where the monic invariant factors satisfy \[ p_r\mid p_{r-1}\mid\cdots\mid p_1 \] in the convention of Hoffman–Kunze, and \(C(p_i)\) denotes the companion matrix of \(p_i\) [@HK71]. There is \(P\in GL_n(F)\) with \[ P^{-1}AP=R_A. \]
Regard the same equality over \(\mathbb C\). The polynomials \(p_i\) still have the same divisibility relations in \(\mathbb C[x]\), and the same companion blocks therefore satisfy the defining conditions for rational form over \(\mathbb C\). By uniqueness of the rational canonical form, \(R_A\) is also the rational form of \(A\) when \(A\) is viewed as a complex matrix.
2. Complex similarity forces equality of the two \(F\)-rational forms.
Suppose \(A\) and \(B\) are similar over \(\mathbb C\). Similar matrices have the same rational canonical form over the ground field [@HK71]. Hence their complex rational forms coincide. By step 1, those are exactly the rational forms \(R_A\) and \(R_B\) already obtained over \(F\). Thus \[ R_A=R_B. \] Choose \(P,Q\in GL_n(F)\) with \[ P^{-1}AP=R_A, \qquad Q^{-1}BQ=R_B. \] Then \[ B=Q R_AQ^{-1}=QP^{-1}APQ^{-1}. \] Since \(QP^{-1}\in GL_n(F)\), this is a similarity over \(F\). Therefore \[ \boxed{A\sim_{\mathbb C}B\Longrightarrow A\sim_F B.} \]