# How can I get the symbolic steady state vector of a Markov Chain?

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Shih-Wen Liu on 17 Jun 2022
Commented: Bruno Luong on 8 Aug 2022
Hello, does anyone know how to obtain the symbolic steady state vector (i.e. the long-term probability of each state) of this Markov Chain example in MATLAB? At the end of this demonstration, it does not show how can I further get a steady state vector?
It will be very appreciated if you can help me with this problem.

John D'Errico on 7 Aug 2022
Edited: John D'Errico on 7 Aug 2022
Easy, peasy. For example, given a simple Markov process, described by the 3x3 transition matrix T.
T = [.5 .2 .3;.1 .4 .5;.1 .1 .8]
T = 3×3
0.5000 0.2000 0.3000 0.1000 0.4000 0.5000 0.1000 0.1000 0.8000
There are no absorbing states. We can see this is indeed the transition matrix of a Markov chain. One good test is the rows all sum to 1, and none of the elements are greater than 1, or less than zero.
sum(T,2)
ans = 3×1
1 1 1
What are the steady-state probabilities?
[V,D] = eig(T')
V = 3×3
0.2357 0.4082 0.0000 0.2357 0.4082 -0.7071 0.9428 -0.8165 0.7071
D = 3×3
1.0000 0 0 0 0.4000 0 0 0 0.3000
Take eigenvector that corresponds to the unit eigenvalue. In this case, it is the first eigenvector.
P = V(:,1)';
Normalize so the elements sum to 1.
format long g
P = P/sum(P)
P = 1×3
0.166666666666666 0.166666666666667 0.666666666666667
Those are the steady state probabilites for this system. We can see that this does not change P.
P*T
ans = 1×3
0.166666666666666 0.166666666666667 0.666666666666667
I won't do your homework for you, but you can easily enough see how to proceed from here.
Bruno Luong on 8 Aug 2022
You don't need to compute eigen value, you can compute this, possibly easier in symbolic way:
ss = null(T.'-eye(size(T))).';
ss = ss/sum(ss)

R2021b

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