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: $e^{Q} = S$?

3$\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Steve Huntsman Jul 24 '10 at 22:31

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 Speyer Jul 25 '10 at 2:13

1$\begingroup$ More generic than "Analogy question" is "Question". Or also "This question" if one wants to be somehow more precise. $\endgroup$ – Pietro Majer Jul 25 '10 at 4:36

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 Protsak Jul 25 '10 at 6:14

3$\begingroup$ 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 Huntsman Jul 25 '10 at 13:17
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 nondiagonal elements must be nonnegative. 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.)

$\begingroup$ They are frequently called Qmatrices. $\endgroup$ – Steve Huntsman Jul 25 '10 at 13:57

$\begingroup$ Thanks! When Googling "infinitesimally stochastic", the first hit with a definition seemed to be this one: arxiv.org/abs/1002.4773 $\endgroup$ – Mike Stay Jul 25 '10 at 16:01
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 unitarystochastic 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 finitedimensional 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. “Unitarystochastic matrix ensembles and spectral statistics”. J. Phys. A 34, 8485 (2001).

$\begingroup$ 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? $\endgroup$ – Michael Hoffman Jul 24 '10 at 23:23

1$\begingroup$ I don't think this answer is trivial. I think the answer in my comment above is trivial. $\endgroup$ – Steve Huntsman Jul 24 '10 at 23:49
