$A \underset{\mathbb{C}}{\sim} B \overset{\text{def}}{\iff} A=C^{-1}BC, \space C\in M_{n}(\mathbb{C})$ and similarly for $\underset{\mathbb{R}}{\sim}$.

I want to prove that $ A \underset{\mathbb{C}}{\sim} B$ for $A,B \in M_{n}(\mathbb{R})$ therefore $A \underset{\mathbb{R}}{\sim} B$.

My idea is that elementary divisors of $A,B$ over $\mathbb{C}$ are the same, and if $(x-z)^k$ is elementary divisor than $(x-\overline{z})^k$ is also elementary divisor $\implies$ $A,B$ have same elementary divisors over $\mathbb{R}$. But i think it's not clear.

  • $\begingroup$ $A,B$ are real matrices and there complex matrix $C$: $A = C^{-1}BC$. I want to prove that exists real matrix $H$: $A = H^{-1}BH$. Im sorry my english is very bad;( $\endgroup$ – qwenty Feb 1 '15 at 23:42
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    $\begingroup$ This is a special case (and hence an abstract duplicate) of the question Similar matrices and field extensions. Since the OP here is perhaps looking for a simple elementary proof for the specific fields $\mathbb R$ and $\mathbb C$, I do not vote to close the question. $\endgroup$ – user1551 Feb 2 '15 at 4:22
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    $\begingroup$ Any book where I can find this exercise? $\endgroup$ – Gauss Jan 25 '18 at 16:26

If $ A \underset{\mathbb{C}}{\sim} B$, there is a matrix $C \in GL_n(\mathbb{C})$ such that $A=C^{-1}BC$.

So $CA=BC$.

$C=P+iQ$ with $P,Q \in M_n(\mathbb{R})$.

If $A \in M_n(\mathbb{R})$ and $B \in M_n(\mathbb{R})$, we have $CA=BC \implies (P+iQ)A=B(P+iQ)\implies PA=BP$ and $QA=BQ$.

The polynomial $\det (P+XQ)$ is not null, because $\det(P+iQ)=\det C \neq 0$.

So, there is a value $\lambda \in \mathbb{R}$ such hat $\det(P+\lambda Q) \neq 0$.

Let $D=P+\lambda Q$, $DA=BD$ because $PA=BP$ and $QA=BQ$.

So, $A=D^{-1}BD$ and $A \underset{\mathbb{R}}{\sim} B$


protected by Community Feb 1 '15 at 23:42

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