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$A,B$ are real matrices, similiar in complex field, then they are similar in the real field. This is well-known. My problem is that if we know there exists an invertible complex matrix $C$ such that $C^{-1}AC=B$, can we use $A,B,C$ to find a real matrix $D$ such that $D^{-1}AD=B$.

Let $C=C_1+C_2i$, where $C_1$ and $C_2$ are real matrices, then $AC_1=C_1B$ and $AC_2=C_2B$. What next?

mathlander
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xldd
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  • And it is also true $A(C_1+\alpha C_2)=(C_1+\alpha C_2)B$ for any real $\alpha$. – kabenyuk Dec 16 '22 at 11:47
  • @kabenyuk then why for some $\alpha$, $C_1+\alpha C_2$ is invertable? – xldd Dec 16 '22 at 11:59
  • @kabenyuk Can we use $A,B,C$ to compute $D$? Is there a solution form? – xldd Dec 16 '22 at 12:50
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    This is true not just for real matrices viewed as complex matrices, but for matrices with entries in a field $F$ that are viewed in a larger field $E$: conjugacy over $E$ implies conjugacy over $F$. See Theorem 4 in https://kconrad.math.uconn.edu/blurbs/linmultialg/potdiagonalizable.pdf. – KCd Dec 25 '22 at 03:47

1 Answers1

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Consider the polynomial $f(x)=\operatorname{det}(C_1+C_2x)$. We have $f(i)\neq0$. Then $f$ is a nonzero polynomial. Hence there exists $x_0\in\mathbb{R}$ that $f(x_0)\neq0$.

To find a particular matrix $D$ you will need to compute $f(k)$ for $k=0,1,\ldots,n$. Since the degree $f$ is not greater than $n$, there exists $k$ that $f(k)\neq0$.

Note two hours later.

A much more general situation is discussed in this post Similar matrices and field extensions.

kabenyuk
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  • Can we use $A,B,C$ to compute $D$? Is there a solution form? – xldd Dec 16 '22 at 12:50
  • As $D$ we can thus take one of the matrices: $C_1,C_1+C_2,\ldots C_1+nC_2$. Which one? There seems to be no general formula. At least I don't know it. – kabenyuk Dec 16 '22 at 14:12