Suppose \(A,B\) are real \(n\times n\) matrices and \(C\) is an invertible complex \(n\times n\) matrix such that \[ CAC^{-1}=B. \] Find an invertible real \(n\times n\) matrix \(D\) such that \[ DAD^{-1}=B. \]
Solution
Write \[ C=X+iY \] with \[ X,Y\in M_n(\RR). \]
1. The real matrices \(X\) and \(Y\) satisfy\[ XA=BX \qquad\text{and}\qquad YA=BY. \]
The hypothesis \[ CAC^{-1}=B \] is equivalent to \[ CA=BC. \] Substitute \(C=X+iY\): \[ (X+iY)A = B(X+iY). \] Since \(A\) and \(B\) are real, equality of real and imaginary parts gives the two displayed relations.
2. For every real number \(t\), the real matrix\[ D_t\coloneqq X+tY \]satisfies\[ D_tA=BD_t. \]
By step 1, \[ \begin{aligned} D_tA &= XA+tYA\\ &= BX+tBY\\ &= B(X+tY)\\ &= BD_t. \end{aligned} \]
3. The polynomial\[ p(t)\coloneqq\det(X+tY)\in\RR[t] \]is not identically zero.
Evaluate the polynomial at the complex number \(i\): \[ p(i) = \det(X+iY) = \det C. \] The matrix \(C\) is invertible, so \(\det C\neq0\). Hence \(p\) cannot be the zero polynomial.
4. There is a real number \(t_0\) such that\[ \det(X+t_0Y)\neq0. \]
By step 3, \(p\) is a nonzero polynomial. It has only finitely many real roots, so choose any real \(t_0\) outside that finite set.
5. The real matrix\[ \boxed{D=X+t_0Y} \]is invertible and satisfies\[ DAD^{-1}=B. \]
Step 4 gives \(\det D\neq0\), so \(D\) is invertible. Step 2 gives \[ DA=BD. \] Multiplying on the right by \(D^{-1}\) yields \[ DAD^{-1}=B. \]
6. Q.E.D.
Step 5 constructs the required invertible real matrix.