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Let $A,B \in GL(n,\mathbb{C})$ be similar, i.e. $\exists C \in GL(n,\mathbb{C}): A = C^{-1}BC.$ Then i have the following question:

I know, that under certain circumstances a complex matrix can be (unitarily) similar to a real matrix. But if we assume that $A \in GL(n,\mathbb{R})$ is similar to $B$ inside $GL(n,\mathbb{C})$ in the above sense, then does this imply $B \in GL(n,\mathbb{R})$?

In particular, if this were the case, then $A$ and $B$ would be similar inside $GL(n,\mathbb{R})$ according to this question considering the field extension $\mathbb{C}/\mathbb{R}$.

user12345
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  • Do you know what « unitary similar » means? – Plop Aug 24 '22 at 18:50
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    Aren't the matrices $\begin{pmatrix}1&z\0&1\end{pmatrix}$ similar for all (real and) complex numbers $z$? – Greg Martin Aug 24 '22 at 18:51
  • @Plop Two square matrices $A$ and $B$ are unitarily similar matrices if there exists a matrix $P$ s.th. $PA=BP$ and $P$ is a unitary matrix. – user12345 Aug 24 '22 at 19:09
  • @GregMartin Sorry, but i still don´t quite get your example. – user12345 Aug 24 '22 at 19:31
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    @Greg All non-zero numbers $z$ – Ben Grossmann Aug 24 '22 at 19:32
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    Diagonalize the matrix $A=\begin{pmatrix}0&1\-1&0\end{pmatrix}$ over $\mathbb C$ to get $B=\begin{pmatrix}i&0\0&-i\end{pmatrix}$. The columns of $C$ will be (complex) eigenvectors of $A$, corresponding to complex eigenvalues $i, -i$, respectively. –  Aug 24 '22 at 19:53

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The answer is no. As a counterexample, consider $$ A = \pmatrix{1&1\\0&2}, \quad B = \pmatrix{1 & i\\0 & 2}, \quad C = \pmatrix{1&0\\0&i}. $$

Ben Grossmann
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