# Thread: Eigenvectors and transition matrix of markov chains.

1. ## Eigenvectors and transition matrix of markov chains.

Consider a Markov chain { $X_n : n = 0,1,2,...$} with states $S=(1,2)$ and transition matrix

$P= \begin{pmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$

Let $\mathbf{x}$ be a vector such that $\mathbf{x}P= \mathbf{x}$

where $x_1 > 0, x_2 > 0$ and $x_1 + x_2 = 1$

Let $\lambda$ and $\mathbf{y}$ satisfy

$\mathbf{y}P=\lambda \mathbf{y}$, where $|\lambda|<1$.

Let $\mathbf{\rho} = (\rho_1,\rho_2)$ be such that $\rho_1 > 0, \rho_2 > 0 , \rho_1+\rho_2=1$. Show that $\rho$ can be written as

$\mathbf{\rho}= \mathbf{x}+K \mathbf{y}$ for some constant $K$

I get:

$\mathbf{x}, \mathbf{y} \ are \ eigenvectors$, therefore
$\mathbf{\rho}=\alpha \mathbf{x}+ \beta \mathbf{y}$

However I cannot see how to make $\alpha = 1$

2. ## Re: Eigenvectors and transition matrix of markov chains.

The eigenvalues of $P$ are $\lambda_1=1,\lambda_2=-1/4$ . Using the definition of eigenvector, you'll find that necessarily $x=(2/5,3/5)$ and $y=(-\gamma,\gamma)\;\;(\gamma \neq 0)$ . Now, using $\rho_1+\rho_2=1$ it is easy to prove that $(\rho_1,\rho_2)=(2/5,3/5)+K(-\gamma,\gamma)$ for some $K$ .

3. ## Re: Eigenvectors and transition matrix of markov chains.

I am confused as to how you got

$y=(-\gamma, \gamma);\;(\gamma \neq 0)$

4. ## Re: Eigenvectors and transition matrix of markov chains.

Originally Posted by FGT12
I am confused as to how you got $y=(-\gamma, \gamma);\;(\gamma \neq 0)$
$yP=\lambda y$ implies $y$ is an eigenvector of $P$ associated to $\lambda$ . If $|\lambda|<1$ necessarily $\lambda=-1/4$ . Solve the system $y(P+(1/4)I)=0$ .