Let \(A\) and \(B\) be square matrices with rational entries. Suppose \[ CAC^{-1}=B \] for some invertible real matrix \(C\). Prove that one can choose such a conjugating matrix \(C\) with rational entries.
Solution
Suppose \(A,B\in M_n(\QQ)\). Define \[ W_{\QQ} = \{X\in M_n(\QQ):XA=BX\} \] and \[ W_{\RR} = \{X\in M_n(\RR):XA=BX\}. \]
1. The space \(W_{\RR}\) has a basis consisting of matrices in\(W_{\QQ}\).
The equation \[ XA=BX \] is a homogeneous linear system in the \(n^2\) entries of \(X\). Since the entries of \(A\) and \(B\) are rational, every coefficient of this system is rational.
Row reduction over \(\QQ\) produces a parametrization of its solutions whose coefficient vectors are rational. The same row-reduced system describes the solutions after allowing the free parameters to range over \(\RR\). Hence there are matrices \[ X_1,\ldots,X_d\in W_{\QQ} \] which form a \(\QQ\)-basis of \(W_{\QQ}\) and an \(\RR\)-basis of \(W_{\RR}\).
2. Let\[ C_0\in GL_n(\RR) \]be a matrix satisfying\[ C_0A=BC_0. \]There are real numbers \(t_1^{(0)},\ldots,t_d^{(0)}\) such that\[ C_0 = \sum_{j=1}^d t_j^{(0)}X_j. \]
The relation \[ C_0AC_0^{-1}=B \] from the hypothesis is equivalent to \[ C_0A=BC_0. \] Thus \(C_0\in W_{\RR}\). Step 1 says that \(X_1,\ldots,X_d\) is a basis of \(W_{\RR}\), so \(C_0\) has the displayed coordinates.
3. Define\[ P(t_1,\ldots,t_d) = \det \left( \sum_{j=1}^d t_jX_j \right). \]Then\[ P\in\QQ[t_1,\ldots,t_d] \]and \(P\) is not the zero polynomial.
Every entry of every \(X_j\) is rational. The determinant is a polynomial with integer coefficients in the matrix entries, so \(P\) has rational coefficients.
By step 2, \[ P(t_1^{(0)},\ldots,t_d^{(0)}) = \det C_0 \neq 0. \] Therefore \(P\) is not identically zero.
4. If\[ Q\in\QQ[x_1,\ldots,x_d] \]is a nonzero polynomial, then there is a point\[ q\in\QQ^d \]such that\[ Q(q)\neq0. \]
Proceed by induction on \(d\).
For \(d=1\), a nonzero polynomial has only finitely many roots, while \(\QQ\) is infinite. Hence some rational number is not a root.
Suppose the statement holds for \(d-1\). Write \[ Q(x_1,\ldots,x_d) = \sum_{k=0}^m Q_k(x_1,\ldots,x_{d-1})x_d^k, \] where at least one coefficient polynomial \(Q_k\) is nonzero. By the induction hypothesis, choose \[ (q_1,\ldots,q_{d-1})\in\QQ^{d-1} \] so that this nonzero coefficient does not vanish. Then \[ x_d \longmapsto Q(q_1,\ldots,q_{d-1},x_d) \] is a nonzero one-variable polynomial. It has only finitely many roots, so some \(q_d\in\QQ\) is not a root.
5. There are rational numbers \(q_1,\ldots,q_d\) such that the matrix\[ C = \sum_{j=1}^d q_jX_j \]is invertible.
3. Choose\[ (q_1,\ldots,q_d)\in\QQ^d \]with\[ P(q_1,\ldots,q_d)\neq0. \]By the definition of \(P\),\[ \det C = P(q_1,\ldots,q_d) \neq 0. \]Thus \(C\) is invertible. Since the \(X_j\) and the coefficients \(q_j\) arerational, every entry of \(C\) is rational.
6. The rational matrix \(C\) from step 5 satisfies\[ \boxed{ CAC^{-1}=B. } \]
Each \(X_j\) lies in \(W_{\QQ}\), so every rational linear combination of the \(X_j\) also lies in \(W_{\QQ}\). Hence \[ CA=BC. \] Step 5 says that \(C\) is invertible. Multiplying the last equality on the right by \(C^{-1}\) gives \[ CAC^{-1}=B. \]
7. Q.E.D.
Step 6 gives an invertible conjugating matrix with rational entries.