Next we need to make sure that there is only one left eigenvector with eigenvalue 1. To get an eigenvalue 1, the left eigenvector must be constructed from only ones and zeroes. It is straightforward to see that a vector made up of ones and zeroes can only be an eigenvector with eigenvalue 1 if the matrix element \( M_{ij} = 0 \) for all cases where \( \psi_i \neq \psi_j \). That is we can choose an index \( i \) and take \( \psi_i = 1 \). We require all elements \( \psi_j \) where \( M_{ij} \neq 0 \) to also have the value \( 1 \). Continuing we then require all elements \( \psi_\ell \) $M_{j\ell}$ to have value \( 1 \). Only if the matrix \( M \) can be put into block diagonal form can there be more than one choice for the left eigenvector with eigenvalue 1. We therefore require that the transition matrix not be in block diagonal form. This simply means that we must choose the transition probability so that we can get from any allowed state to any other in a series of transitions.