Let \(A,B\) be real \(n\times n\) matrices. Suppose there is a nonsingular complex matrix \(C\) such that \[ CAC^{-1}=B. \] Show that there exists a nonsingular real matrix \(D\) satisfying \[ DAD^{-1}=B. \]
Solution
1. Write\[ C=X+iY \]with \(X,Y\) real \(n\times n\) matrices. Then\[ XA=BX \qquad\text{and}\qquad YA=BY. \]
The relation \[ CAC^{-1}=B \] is equivalent to \[ CA=BC. \] Substituting \(C=X+iY\) gives \[ XA+iYA=BX+iBY. \] Because all four matrices \(XA,YA,BX,BY\) are real, equality of real and imaginary parts gives the two displayed identities.
2. For every real number \(t\), the real matrix\[ D_t=X+tY \]satisfies\[ D_tA=BD_t. \]
Using step 1, \[ \begin{aligned} D_tA &= XA+tYA\\ &= BX+tBY\\ &= B(X+tY)\\ &= BD_t. \end{aligned} \]
3. The polynomial\[ p(t)=\det(X+tY)\in\RR[t] \]is not identically zero.
Viewed as a polynomial over \(\CC\), its value at \(t=i\) is \[ p(i) = \det(X+iY) = \det C \neq0, \] because \(C\) is nonsingular. Therefore \(p\) is not the zero polynomial.
4. There exists a real number \(t_0\) such that\[ D=X+t_0Y \]is nonsingular.
By step 3, \(p\) is a nonzero real polynomial. Hence it has only finitely many real roots. Choose a real number \(t_0\) that is not a root. Then \[ \det D = p(t_0) \neq0, \] so \(D\) is nonsingular.
5. The real nonsingular matrix \(D\) from step 4 satisfies\[ \boxed{DAD^{-1}=B}. \]
Step 2 gives \[ DA=BD. \] Since \(D\) is invertible by step 4, right multiplication by \(D^{-1}\) yields \[ DAD^{-1}=B. \]
6. Q.E.D.
Step 5 gives the required real similarity matrix.