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What are the necessary and sufficient conditions for a real $n \times n$ matrix to have $n$ distinct real eigenvalues?

Ideally I'm looking for a test that does not require (and is hopefully more efficient than) computing the eigenvalues.

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  • $\begingroup$ There are many ways to answer this question. One possible answer is that all the blocks in Jordan Canonical Form are $1\times 1$, and are distinct and real. $\endgroup$ – vadim123 Jun 30 '14 at 14:56
  • $\begingroup$ ... It's characteristic polynomial has $n$ distinct real roots? ... (This is equivalent to vadim123's and Peter Franek's condition, but differently computed.) (Thanks to Peter Franek for correcting also me for dropping "real" the first time.) $\endgroup$ – Eric Towers Jun 30 '14 at 14:59
  • $\begingroup$ @vadim123 Yes, but finding the Jordan Canonical Form is equivalent to just computing the eigenvalues. Is there some test I can use to know without calculating what the eigenvalues actually are? $\endgroup$ – user160914 Jun 30 '14 at 15:00
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    $\begingroup$ @user 160914 If the matrix is real symmetric, then it has $n$ real eigenvalues (not necessarily different). I don't think there is an equivalent reformulation that is simpler than "$n$ distinct real eigenvalues". $\endgroup$ – Peter Franek Jun 30 '14 at 15:01
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    $\begingroup$ It is easy to use standard methods to find out if a polynomial has no repeated roots. $p(x)$ has no repeated roots if and only if $p(x)$ and $p'(x)$ are not relatively prime. $\endgroup$ – Thomas Andrews Jun 30 '14 at 15:08

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