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The "largest" eigenvalue $1$ of a stochastic matrix is well-characterized by the classical Perron-Frobenius theorem. In particular, it gives sufficient conditions for the eigenvalue $1$ to be simple.

Are there any sufficient conditions known when all (or at least all real) eigenvalues of such a matrix are simple?

Thank you very much.

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    $\begingroup$ Similar problems in continuous variable (that is, for stochastic kernels rather than stochastic matrices) are considered to be difficult, so I doubt there are general methods to show simplicity of eigenvalues. The only thing that comes to my mind is a result for strictly totally positive matrices; see Section 6 here. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 29, 2018 at 9:21
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    $\begingroup$ If the matrix is irreducible, then not only the eigenvalue $1$, but all eigenvalus of modulus $1$ are simple. For the eigenvalues of smaller modulus I second @MateuszKwaśnicki's comment that the question is likely to be very difficult to answer. $\endgroup$
    – Jochen Glueck
    Commented Jul 29, 2018 at 10:05
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    $\begingroup$ Also: why do you need to prove that all eigenvalues are simple? Probably it's worth asking yourself if the result that you need holds also for matrices with non-simple eigenvalues... $\endgroup$
    – Federico Poloni
    Commented Jul 29, 2018 at 12:48
  • $\begingroup$ @Randomguy: Just as for matrices: one requires that the arguments are arranged in an increasing order; in your example, $x_1 < x_2$. I do not think total positivity has been applied to study simplicity of eigenvalues in continuous case. Actually, when I wrote my first comment, I figured out that one should perhaps try this approach. Unfortunately, I believe that most interesting kernels are not totally positive. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 29, 2018 at 18:41

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