5
$\begingroup$

This question already has an answer here:

Suppose I have matrices $A,B\in\mathrm{Mat}_n(\mathbf{R})$ which are conjugate in $\mathrm{Mat}_n(\mathbf{C})$ in the sense that there is $S\in\mathrm{GL}_n(\mathbf{C})$ with $A=SBS^{-1}$. Then is it possible to choose $S$ with this property and $S\in\mathrm{GL}_n(\mathbf{R})$?

this fact is claimed in the answers to An element of $SL(2,\mathbb{R})$

$\endgroup$

marked as duplicate by user26857, Watson, Oscar Cunningham, C. Falcon, Thomas Jul 7 '16 at 15:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

Write $S=R+iI$, with $R,I\in M_n(\mathbb R)$, for the real and imaginary parts of $S$. By hypothesis, $AS=SB$ and so $AR=BR$ and $AI=IB$. Thus, $$\forall t\in\mathbb R, A(R+tI)=(R+tI)B$$ We will be done once we show that there are $t\in\mathbb R$ such that $R+tI$ is invertible. This follows from the observation that $$z\mapsto\det(R+zI)$$ is polynomial in $z$, nonzero since it is nonzero for $z=i$, and thus only has finitely many roots, and in particular, $(R+tI)$ is invertible for all but a finite number of $t\in\mathbb R$.


The same proof technique implies that if two matrices with coefficients in an infinite field $k(\subset K)$ are conjugate in some larger field $K$, then they are already conjugate in $k$.

$\endgroup$
  • $\begingroup$ thank you, this is very kind. nice proof by the way $\endgroup$ – Rudi Oct 14 '14 at 10:26
  • $\begingroup$ The proof seems to rely on the the particular case of field extension $\mathbb R\subset\mathbb C$, and I don't see how one can extend it to the general case (which I agree that it is also true). $\endgroup$ – user26857 Dec 27 '15 at 11:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.