rate of convergence of absorbing markov chain Let $G$ be a biconnected and non-bipartite graph. I can simulate a random walk on this graph with a markov chain. The stochastic matrix is $M = AD^{-1}$, where $A$ is the adjacency matrix of $G$ and $D$ is a diagonal matrix with $D_{ii} = degree(i)$. 
Because $G$ is connected and non-bipartite there is a unique stationary distribution $\pi$ so that  
$$\|\pi - M^i p\|_2 \le c (1-\lambda)^i $$
where $\lambda$ is the difference between the first and second eigenvalue of $A$, called spectral gap. So there is a lower bound to the rate of convergence towards the stationary distribution, depending on the spectral gap.
I can modify the graph, so that there is a single sink node, that once entered can never be left. This graph $G'$ is no longer undirected. I get an absorbing markov chain where one state is absorbing and all other states are transient. There still is a unique stationary distribution $\pi'$, where the probability of being in at sink node is 1.
So my question is, can I get similar bound to the rate of convergence of the absorbing markov process, based on the spectral gap of $G$?
 A: By the way the expression you have with $1-\lambda $ involving the spectral gap, is just the second largest eigenvalue of the transition matrix.
In the case where each node has an edge to the absorbing node the rate of decay will change to $N/(N+1)$ where N is the total number of nodes in the original graph. You can see this by observing that the original eigenvectors of the original transition matrix will still work if we augment them with a zero in the last position. However since the submatrix corresponding to staying within the original grasps nodes must get scaled down in size by the factor involving the degree so will the  eigenvalues, so the old steady state eigenvector which had eigenvalue 1, now has the eigenvalue I wrote above.
In the case where the number of edges to the absorbing state is much smaller than the size of the graph I think you could make a quasi steady state approximation. Assume that within the original graph things are distributed according to the old stationary distribution. Say there is just one node , say labeled $i$, with an edge to the absorbing node. Then on average if you start in the original graph your probability of moving to the absorbing state in a given turn will be $\pi_i/(D_i + 1)$ where $D_i $ is the original degree of node i. So the probability of staying in the subgraph is 1 minus the above, and that should in the limit of the subgraph being large approximate your decay rate.
