# equilibrium of the matrix markov

Suppose to have the following 2 x 2 Markov Matrix.

$$\begin{pmatrix} .8 & .1\\ .2 & .9 \end{pmatrix}$$

Considering the system

$$\begin{pmatrix} .8 \\ .2 \end{pmatrix} \cdot 150 + \begin{pmatrix} .1\\ .9 \end{pmatrix} \cdot 150 = \begin{pmatrix} a\\ b \end{pmatrix}$$

when $$a= 100$$ and $$b = 200$$ the entire system is in equilibrium.

Is there a theorem about reaching (always?) an equilibrium point for a Markov matrix?

• Note $\begin{bmatrix}1\\2\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $\lambda=1$. That can help you find the equilibrium vector. Note that the other eigenvalue will be $0.7$ so as $n \to \infty$ $(0.7)^n \to 0$. Jun 25, 2020 at 7:19
• Thanks Anurag. I know the eigenvector (1 2)^T is the equilibrium vector. I can not find a theorem about this. Jun 25, 2020 at 7:21

This (stochastic) matrix $$A$$ has two eigen values $$\lambda_1=1$$ and $$\lambda_2=0.7$$ so it is diagonalizable, i.e. $$A=P\begin{bmatrix}1&0\\0&0.7\end{bmatrix}P^{-1}=PDP^{-1}$$, where $$P=\begin{bmatrix}1&-1\\2&1\end{bmatrix}$$
To get the equilibrium vector. We want to compute $$v$$ such that $$\lim_{n \to \infty}A^nv=v$$.
Now $$v=c_1v_1+c_2v_2$$, where $$v_1,v_2$$ are eigenvectors corresponding to the eigenvalues $$1$$ and $$0.7$$.
Thus $$A^nv=c_1A^nv_1+c_2A^nv_2 =c_1(1)^nv_1+c_2(0.7)^nv_2$$ $$\lim_{n \to \infty}A^nv=c_1v_1$$ So the multiple of first eigenvector (for eigenvalue $$1$$) is the equilibrium vector.