Let \(A,B\) be real \(n\times n\) matrices. Suppose \(A=UBU^{-1}\) for some complex invertible matrix \(U\). Show that \(A=VBV^{-1}\) for some real invertible matrix \(V\).
Solution
The similarity relation is equivalent to \[ AU=UB. \] Write \[ U=C+iD, \] where \(C,D\) are real \(n\times n\) matrices.
1. Both real and imaginary parts intertwine \(A\) and \(B\).
Because \(A\) and \(B\) are real, \[ A(C+iD)=(C+iD)B \] becomes \[ AC+iAD=CB+iDB. \] Equality of real and imaginary parts gives \[ AC=CB, \qquad AD=DB. \] Therefore for every real scalar \(t\), \[ A(C+tD)=(C+tD)B. \]
2. Some real linear combination \(C+tD\) is invertible.
Consider the real polynomial \[ p(t)=\det(C+tD). \] It is not the zero polynomial, because evaluating it at the complex number \(t=i\) gives \[ p(i)=\det(C+iD)=\det U\ne0. \] A nonzero real polynomial has only finitely many real roots. Hence there is some real \(t_0\) such that \[ p(t_0)\ne0. \] Thus \[ V=C+t_0D \] is a real invertible matrix.
3. The real matrix \(V\) gives the required similarity.
By step 1, \[ AV=VB. \] Since \(V\) is invertible, \[ \boxed{A=VBV^{-1}.} \] Thus real matrices which are similar over \(\mathbb C\) are already similar over \(\mathbb R\).