Let \(F\) be a subfield of the field of complex numbers, and let \(A\) be an \(n \times n\) matrix over \(F\). Let \(p\) be the minimal polynomial for \(A\). If we regard \(A\) as a matrix over \(C\), then \(A\) has a minimal polynomial \(f\) as an \(n \times n\) matrix over \(C\). Use a theorem on linear equations to prove \(p = f\). Can you also see how this follows from the cyclic decomposition theorem?
Solution
Let \(p\in F[x]\) be the minimal polynomial of \(A\) over \(F\), and let \(f\in\mathbb C[x]\) be its minimal polynomial after scalar extension.
1. A theorem on linear equations forces \(\deg f\geq\deg p\).
Since \(p(A)=0\) remains true over \(\mathbb C\), minimality of \(f\) gives \[ f\mid p \] in \(\mathbb C[x]\). Hence \(\deg f\leq\deg p\).
Suppose, toward a contradiction, that \(m=\deg f<d=\deg p\). Since \(f\) is monic, write \[ f(x)=x^m+c_{m-1}x^{m-1}+\cdots+c_0. \] The equation \(f(A)=0\) says that the entries of the matrices \(I,A,\ldots,A^m\) satisfy a finite system of linear equations over \(F\) for the unknowns \(c_0,\ldots,c_{m-1}\), with right side \(-A^m\). This system has a solution over \(\mathbb C\).
Row reduction of an augmented matrix with entries in \(F\) uses only field operations in \(F\). Therefore a linear system over \(F\) that is consistent over an extension field is already consistent over \(F\) [@HK71]. Hence there exist \(b_0,\ldots,b_{m-1}\in F\) such that \[ A^m+b_{m-1}A^{m-1}+\cdots+b_0I=0. \] This gives a monic annihilating polynomial over \(F\) of degree \(m<d\), contradicting minimality of \(p\). Thus \(\deg f\ge d\).
2. The two monic minimal polynomials are equal.
Step 1 and the divisibility \(f\mid p\) give \[ \deg f=\deg p. \] Both are monic, so divisibility at equal degree forces \[ \boxed{f=p}. \] In particular the complex minimal polynomial actually has coefficients in \(F\).
3. The cyclic decomposition theorem gives the same conclusion.
Apply the cyclic decomposition theorem over \(F\). In a suitable \(F\)-basis, the matrix of \(A\) is the rational form \[ \operatorname{diag}(C(p_1),\ldots,C(p_r)), \qquad p_r\mid\cdots\mid p_1, \] and the largest invariant factor \(p_1\) is the minimal polynomial \(p\) [@HK71]. The same basis and same block matrix remain valid over \(\mathbb C\), and the same divisibility relations hold in \(\mathbb C[x]\). Hence this is also a rational form over \(\mathbb C\). Its largest invariant factor is still \(p_1\), so the complex minimal polynomial is \(p_1=p\).