That the maximum eigenvalue is 1 follows immediately from the property that \( \sum_i M_{ik} = 1 \). Similarly the minimum eigenvalue can be -1, but only if \( M_{kk} = 0 \) and the magnitude of all the other elements \( \psi_i^{\rm min} \) of the eigenvector that multiply nonzero elements \( M_{ik} \) are -1.
Let's first see what the properties of \( M \) must be to eliminate any -1 eigenvalues. To have a -1 eigenvalue, the left eigenvector must contain only \( \pm 1 \) and \( 0 \) values. Taking in turn each \( \pm 1 \) value as the maximum, so that it corresponds to the index \( k \), the nonzero \( M_{ik} \) values must correspond to \( i \) index values of the eigenvector which have opposite sign elements. That is, the \( M \) matrix must break up into sets of states that always make transitions from set A to set B ... back to set A. In particular, there can be no rejections of these moves in the cycle since the -1 eigenvalue requires \( M_{kk}=0 \). To guarantee no eigenvalues with eigenvalue -1, we simply have to make sure that there are no cycles among states. Notice that this is generally trivial since such cycles cannot have any rejections at any stage. An example of such a cycle is sampling a noninteracting Ising spin. If the transition is taken to flip the spin, and the energy difference is zero, the Boltzmann factor will not change and the move will always be accepted. The system will simply flip from up to down to up to down ad infinitum. Including a rejection probability or using a heat bath algorithm immediately fixes the problem.