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Assume that $A,B \in M_n(\mathbb{R})$ and there is an invertible matrix $P \in M_n(\mathbb{C})$ that $PAP^{-1}=B$. Prove that there is an invertible matrix $Q\in M_n(\mathbb{R})$ that $QAQ^{-1}=B$

(By the symbol $M_n(F)$, we mean all $n\times n $ matrices over the field $F$.)

Please note the change in field from $\mathbb{C}$ to $\mathbb{R}$

copper.hat
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F.K
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  • I presume you are missing something, just take $Q=P$. You need to fix the question, it is truly trivial as is. – copper.hat Mar 31 '16 at 17:44
  • So, it is needed to prove that the imaginary part of each element of $P$ is $0$. How to prove it? @copper.hat – F.K Mar 31 '16 at 17:47
  • My apologies, I missed the change of field. – copper.hat Mar 31 '16 at 17:48
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    Are you the same person as the user Petar? He asked the same question a moment ago: http://math.stackexchange.com/questions/1722091/real-and-complex-invertible-matrices – user251257 Mar 31 '16 at 17:57
  • This can be done using real Jordan forms. I wonder if there's a solution that avoids that result. – Ben Grossmann Mar 31 '16 at 18:00
  • @user251257 No I'm not that one! just a funny coincidence! – F.K Mar 31 '16 at 18:00
  • This is not so easy to prove in my opinion. See http://math.stackexchange.com/questions/57242/similar-matrices-and-field-extensions?lq=1 for a more genral question. –  Mar 31 '16 at 18:02
  • We might also note that $Q\mapsto QA - BQ$ is a linear map, and that our $Q$s lie in the kernel of this map. – Ben Grossmann Mar 31 '16 at 18:02
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    @Omnomnomnom there's an elementary proof in the question linked by Phoemuex – Gabriel Romon Mar 31 '16 at 18:04

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I'm not sure but I think my solution is correct. Generalized statement: $L/K$ - fields extension, $A,B\in M_{n}(K)$, $A\sim B$ over $L$. We need to prove, that $A\sim B$ over K. $$A\sim B \;\text over \;L \:\Longleftrightarrow\:\left(A-\lambda E\right)\approx\left(B-\lambda E\right) (\approx\text - matrix\; equivalence)\; \text over\; L[\lambda]$$ $$\left(A-\lambda E\right)\approx diag(A-\lambda E)\;\; \text and \;\left(B-\lambda E\right)\approx diag(B-\lambda E)$$ over $K[\lambda]$, $$diag(A-\lambda E)\approx diag(B-\lambda E)$$ over $L[\lambda]$ and $$diag(A-\lambda E)\sim diag(B-\lambda E)$$ over $K$ (permutation of diagonal elements), that means $$diag(A-\lambda E)\approx diag(B-\lambda E)$$ over $K[\lambda]$, that means $$\left(A-\lambda E\right)\approx\left(B-\lambda E\right)$$ over $K[\lambda]$ and $A\sim B$ over $K$.