# Solving for equilibrium distribution (symbolic) by matrix multiplication

I have a huge transition matrix (81x81). The matrix is too huge to paste here, so I store it in this notebook. (There are constraints on the symbols: $0<p_b<1$ and $0<p_g<1$. If further constraints are needed, $0<p_b<0.5<p_g<1$) The terms are either in symbols or rational numbers, or the mixture of both. I tried to solve for its equilibrium distribution using LinearSolve, but the execution lasted for one day and ended up using up all work PC's physical RAM and swap space (16G in total) but no results could be obtained. Now I'm thinking about getting the equilibrium distribution by matrix multiplications, since for a transition matrix $\mathbb{M}$:

$$\lim_{n\rightarrow\infty}\mathbb{M}^n=\mathbb{A}.$$

The rows of $\mathbb{A}$ is the equilibrium distribution, provided the equilibrium exists. Operationally, I will repeatedly multiply the transition matrix and do Simplify. I need the symbolic expressions rather than the numerical results for particular $p_b$ and $p_g$. I would like to know if this method will work and if anyone has done the same before.

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Apparently the limit doesn't exist for all values of pband pg. If you take the element {1,1} of the successive powers of mat you get an increasing order (although non-monotonous) poly on pb and pg –  belisarius Dec 18 '13 at 6:29
@belisarius,sorry I forgot to mention the constraints on $p_b$ and $p_g$. Please see the updated question. –  wdg Dec 18 '13 at 6:47
It is a stochastic matrix only if pb = pq. –  Daniel Lichtblau Dec 18 '13 at 16:46
@DanielLichtblau, I use the convention $\pi^{t}\mathbb{M}=\pi^{t+1}$. Every row sums up to one, and every entry is non-negative, so it is a stochastic matrix even if $p_b\neq p_g$. –  wdg Dec 19 '13 at 4:46
I did not follow that note re convention. What is π^t? I guess M is your matrix, but I don't know what the formula means. –  Daniel Lichtblau Dec 19 '13 at 16:53