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As in the title:

If two real (square) matrices are conjugated over $\mathbb{C}$, are they then also conjugated over $\mathbb{R}$?


marked as duplicate by Eric Wofsey, Jyrki Lahtonen Jul 11 '16 at 21:32

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  • $\begingroup$ What does it mean to be conjugate over $\mathbb{R}$? $\endgroup$ – Morgan Rodgers Jul 11 '16 at 21:22
  • $\begingroup$ @MorganRodgers: I think what's meant by "conjugated over a field $k$" here is "conjugate(d) by conjugation with a matrix with entries in $k$". $\endgroup$ – joriki Jul 11 '16 at 21:23
  • $\begingroup$ @joriki Ahh, ok so he's most likely asking if the matrices are similar? $\endgroup$ – Morgan Rodgers Jul 11 '16 at 21:27
  • $\begingroup$ @MorganRodgers: I believe the two terms are essential synonymous; from en.wikipedia.org/wiki/Matrix_similarity, it seems that as far as there's any difference between them, "conjugate" is the more appropriate one. $\endgroup$ – joriki Jul 11 '16 at 21:32
  • 2
    $\begingroup$ See also here for an argument specific to $\Bbb{C}/\Bbb{R}$. $\endgroup$ – Jyrki Lahtonen Jul 11 '16 at 21:34

The answer to this question is yes. For one, it suffices to note that any real matrix is similar to some matrix in real Jordan form, and that any two similar real matrices share such a real Jordan form.

There may be a way to show this directly that doesn't use such powerful techniques, but this certainly works.

  • $\begingroup$ Reference: see for example Horn and Johnson's Matrix Analysis, Theorem 3.4.5. $\endgroup$ – Omnomnomnom Jul 11 '16 at 21:31
  • $\begingroup$ Did you mean Theorem $\endgroup$ – Anonymous Jul 11 '16 at 21:45
  • $\begingroup$ BTW In the proof of, they refer to 1.3.28 and 1.3.29 - these two results make a shorter proof that completely answers my question, without Jordan normal forms. Thank you very much for that reference! $\endgroup$ – Anonymous Jul 11 '16 at 21:45
  • $\begingroup$ Oh, well you're welcome; I guess I missed that. $\endgroup$ – Omnomnomnom Jul 11 '16 at 21:46

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