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I want to prove the following statement:

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 $QAQ^{-1}=B$

I know that there are some similar questions asked here and I've read all of them. But the point is that I don't want to use determinant approach. All the proofs that one can find here or on the internet are similar to the one you can find in the following link: Similarity of real matrices over $\mathbb{C}$

all of them use this reality that because $P=C+Di$ for some $C,D \in M_n(\mathbb{R})$ and $det(C+Di) \neq 0$ so there exists a $t\in \mathbb{R}$ that $det(C+Dt)\neq0$ and so on.

I don't want to use this approach. Instead of using determinant, I want to use this theorem that a matrix $M$ is invertible if and only if the only answer for the system $MX=0$ is $X=0$.

Now, because the hypothesis says that $P$ is invertible we can say that the only answer for the system $PX=0$ or equivalently $(C+Di)X=0$ is $X=0$. We have:

$$(C+Di)X=0 \Rightarrow CX=0 , DX=0$$ Now, if we can prove that the only answer for the system $CX=0$ or $DX=0$ is $X=0$ then it means that $C$ or $D$ are invertible. Then we can write

$$PAP^{-1}=B \Rightarrow PA=BP$$ $$P=C+Di \Rightarrow CA+DAi=BC+BDi$$ $$CA=BC , DA=BD$$ Then we can conduce that $B=CAC^{-1}$ or $B=DAD^{-1}$ and the problem is solved. But $C$ and $D$ are not necessarily invertible. How can we construct a new real matrix, say $Q$ that $QAQ^-{1}=B$ by using the theorem that I've mentioned above(or some similar theorems) and not using determinant approach?

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F.K
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  • I already answered on this question http://math.stackexchange.com/questions/1722126/if-there-exists-a-p-in-m-n-mathbbc-that-pap-1-b-then-we-can-find-a/1722218#1722218 –  Apr 01 '16 at 13:32
  • Yes, but your approach is similar to the determinant approach which I don't want to use. @B.Igor – F.K Apr 01 '16 at 13:36
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    The thread http://math.stackexchange.com/questions/57242/similar-matrices-and-field-extensions?lq=1 already contains some answers that does not rely on the knowledge of determinant. E.g. Marc van Leeuwen has given a clean and general answer using rational canonical form. – user1551 Apr 02 '16 at 19:17
  • Note that $A,B$ may well be similar and thus have equal spectra (eigenvalues, up to multiplicities), but the shared eigenvalues may nonetheless include conjugate pairs of complex values. Moreover the algebraic multiplicity of an eigenvalue may exceed its geometric multiplicity (although $A,B$ will agree as to both by virtue of similarity over $\mathbb{C}$). The rational canonical form is one concise way of addressing these subtleties. – hardmath Apr 25 '16 at 17:59

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