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?

  • 1
    $\begingroup$ 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 $n \to \infty$ $(0.7)^n \to 0$. $\endgroup$
    – Anurag A
    Jun 25, 2020 at 7:19
  • $\begingroup$ Thanks Anurag. I know the eigenvector (1 2)^T is the equilibrium vector. I can not find a theorem about this. $\endgroup$
    – Gianni
    Jun 25, 2020 at 7:21

1 Answer 1


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.


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