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.



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 $\leq 0$.

| cite | improve this answer | |
  • 2
    $\begingroup$ This answer makes me sad. Thank you anyways. You saved me a lot of hours. $\endgroup$ – Levi Nov 3 '19 at 23:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.