The complex matrix exponential of a Hermitian matrix is unitary: $e^{-iH} = U$. Is there a name or a characterization for matrices Q whose real exponential is stochastic: $e^{-Q} = S$?
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3$\begingroup$ en.wikipedia.org/wiki/… $\endgroup$– Steve HuntsmanCommented Jul 24, 2010 at 22:31
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6$\begingroup$ I disagree with the suggestion that this question be closed. It's a basic question, with a simple answer, but if you don't know the field it's not obvious where to look this up and the subject is hardly one which most mathematicians cover in grad school. As per this discussion, tea.mathoverflow.net/discussion/506/… , consider me a vote against closing. $\endgroup$– David E SpeyerCommented Jul 25, 2010 at 2:13
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1$\begingroup$ More generic than "Analogy question" is "Question". Or also "This question" if one wants to be somehow more precise. $\endgroup$– Pietro MajerCommented Jul 25, 2010 at 4:36
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1$\begingroup$ I join David Speyer in voting against closing. Note that Pietro's answer addresses a stronger condition, namely, that the matrix exponent of any positive multiple of $A$ is stochastic. $\endgroup$– Victor ProtsakCommented Jul 25, 2010 at 6:14
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4$\begingroup$ The grumpy old man in me can't resist: an order of magnitude more people surely learn about Markov processes than coherent sheaves or motivic homotopy theory or whatever "most" mathematicians supposedly study at any level. $\endgroup$– Steve HuntsmanCommented Jul 25, 2010 at 13:17
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2 Answers
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A matrix $A$ such that $\exp(tA)$ is (right) stochastic for all $t > 0$ should be called a "generator of a semigroup of stochastic matrices" or an "infinitesimally stochastic matrix". Clearly, since $A=\lim_{t\to0} (\exp(tA)-I)/t$, (i) the sum of the elements in each row of $A$ has to be 0, and (ii) all non-diagonal elements must be non-negative. Conversely, a matrix $A$ satisfing (i) and (ii), for large enough $n$ produces a stochastic matrix $I+A/n$, hence $(I+A/n)^n$ and $\exp(A)=\lim_{n\to\infty}(I+A/n)^n$ are also stochastic (and so is $\exp(tA)$). That said, I would have a look at the results of a Google search with "infinitesimally stochastic" (I can't do it now).
(Edit: as observed, the above is a stronger condition than the one you asked for; though it's a more close analog to your example.)
answered Jul 25, 2010 at 3:50
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$\begingroup$ They are frequently called Q-matrices. $\endgroup$ Commented Jul 25, 2010 at 13:57
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$\begingroup$ Thanks! When Googling "infinitesimally stochastic", the first hit with a definition seemed to be this one: arxiv.org/abs/1002.4773 $\endgroup$ Commented Jul 25, 2010 at 16:01
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Although I've voted to close because there is a trivial answer, based on your interest in analogies between quantum and statistical physics I think the following may be of interest to you. Since a comment isn't appropriate here I've CW'd this.
Let $u$ be a generic unitary matrix, so that $\sum_j u_{ij}\bar u_{kj} = \delta_{ik}$. If we set $v_{ij} := |u_{ij}|^2$, then it is easy to show that $v$ is a doubly stochastic matrix (though not all doubly stochastic matrices are of this form [1]). Indeed such a matrix is called a unitary-stochastic transition [2] or unistochastic [1] matrix. When one starts with a unitary matrix that is the propagator representing a time evolution operator associated to some Hamiltonian acting on a finite-dimensional Hilbert space, then taking the squared norms yields the associated transition matrix.
[1] See appendix A of Pakonski, P. et al. “Classical 1D maps, quantum graphs and ensembles of unitary matrices”. J. Phys. A 34, 9303 (2001).
[2] Marshall, A. W., and Olkin, I. Inequalities: Theory of Majorization and Its Applications. Academic Press (1979). Cited in [1] and in Tanner, G. “Unitary-stochastic matrix ensembles and spectral statistics”. J. Phys. A 34, 8485 (2001).
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$\begingroup$ while this is a "trivial" answer, I'm not sure it's "vote to close" status. Badly titled, but not vote to close. I mean how is one supposed to look up that answer effectively? $\endgroup$ Commented Jul 24, 2010 at 23:23
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1$\begingroup$ I don't think this answer is trivial. I think the answer in my comment above is trivial. $\endgroup$ Commented Jul 24, 2010 at 23:49
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$\begingroup$ I was referring to that one $\endgroup$ Commented Jul 25, 2010 at 1:26