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Suppose $A,B$ are conjugate real $n \times n$-matrices with $B = PAP^{-1}$ for some real matrix $P$. Does there exist a real matrix $Q$ with positive determinant such that $B = QAQ^{-1}$? The answer to the question is obviously yes if $n$ is odd since we can just take $Q = -P$. What about even $n$?


I asked this question because I wanted to show that the conjugacy classes of real matrices are path-connected (which is not true, as it turns out) and I had reduced the problem down to the above question.

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No.

A counterexample for $n = 2$ is $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $B = -A$. Then, the only $2\times 2$-matrices $P$ that satisfy $PA = BP$ are those of the form $\begin{pmatrix} a & b \\ b & -a \end{pmatrix}$. Obviously, if such a matrix $P$ has real entries, then its determinant is $\leq 0$.

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    $\begingroup$ This answer makes me sad. Thank you anyways. You saved me a lot of hours. $\endgroup$ – Levi Nov 3 '19 at 23:41

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