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Let $A,B \in M_{n}( R )$. (They're both in real numbers.) There is an invertible matrix $P \in M_{n}(C)$ (P is in complex number), where $PAP^{-1} = B$. Show that there is an invertible matrix $Q \in M_{n}(R)$ where $QAQ^{-1} = B$.

I've tried this way, but it doesn't work: $$P = C + iD$$ $$P^{-1} = E + iF$$ where $C$ and $E$ are real parts and $iD$ and $iF$ are imaginary parts. $$PAP^{-1} = (C+iD)A(E+iF) = B$$ $$\Rightarrow CAE - DAF + i(DAE + CAF) = B$$ B, A, C, F, E are real, so $DAE + CAF = 0$ $$\Rightarrow B = CAE - DAF$$

also $$PP^{-1} = I$$ $$\Rightarrow (C+iD)(E+iF) = I$$ $$\Rightarrow CE - DF + i(DE + CF) = I$$ again $DE+CF = 0$

I'm trying to change $CAE - DAF$ into $QAQ^{1}$, but I can't.

Shadi Taghian
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  • Do you know about the rational canonical form? – Ethan Alwaise Apr 03 '16 at 04:25
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    The answer is here: http://math.stackexchange.com/questions/1129628/similarity-of-real-matrices-over-mathbbc. If you resort to a more advanced treatment of similarity, then it is easy. $A$ is similar to $B$ over some field if and only if they have the same elementary divisors. So if $A$ and $B$ are over the reals, their elementary divisors over $\mathbb{C}$ can just be multiplied to give elementary divisors over $\mathbb{R}$. – Christiaan Hattingh Apr 04 '16 at 18:56

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