# Finding limit using Markov chain.

Here is the question:

I set up the probability matrix as: $$\begin{bmatrix} 0 & 1/3 & 1/3 & 1/3 \\ 0 & 1/2 & 1/2 & 0 \\ 1 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 \\ \end{bmatrix}$$

Then, $$1$$ is an eigenvalue. I calculated the eigenvector for $$1$$ as $$(1,1,1,1)$$. So is the answer $$(1/4,1/4,1/4,1/4)$$ ?

• Instead of looking for an eigenvector of $P$, you should be looking for an eigenvector of $P^T$ (that is, a left-eigenvector of $P$). You should calculate this eigenvector to be $(6,4,5,2)$. – Omnomnomnom Aug 13 at 18:55