Let $A$ be a stochastic matrix, that is, the entries are non-negative and each row adds to $1$. Assume that it is primitive, that is, $A^n$ has only positive entries for sufficiently large $n$. We can use $A$ as the transition matrix of a random walk on a directed graph, with the entry $A_{ij}$ giving the probability of moving directly to node $j$ from node $i$. The vector with all entries 1 is a right eigenvector of $A$ with eigenvalue 1, while the left eigenvector gives the fraction of time spent at each node. Given a row vector $\bf v$ containing the probabilities of occupying each node at some time, the corresponding vector for the subsequent time is $\bf v A.$ The primitivity condition ensures that at late times the distribution relaxes to this left eigenvector, at a rate determined (for almost all initial states) by the second largest (in magnitude) eigenvalue. I am not assuming that $A$ is symmetric, so apart from the leading eigenvalue at $1$, the eigenvalues may be complex.

Now consider $B$, a first minor of $A$, that is, obtained by removing a row (say the $i$th) and corresponding ($i$th) column from $A$. This describes a Markov chain or walk on a directed graph with a single absorbing state $i$. The Perron-Frobenius theorem says $B$ has a (possibly multiple) eigenvalue $0\leq\lambda<1$ of maximum magnitude, which corresponds to the rate of absorption. The question is whether this absorption rate can equal a relaxation rate of the original Markov chain, that is:

Can $\lambda$ also be an eigenvalue of $A$?

I cannot find any such example and so conjecture that the answer is false. One approach would be to use interlacing results, but these appear to be valid only for symmetric matrices (hard to imagine interlacing complex eigenvalues!) and I would like as general conditions on $A$ as possible, perhaps even a countable Markov chain?