Basin of attraction (Eigenvectors)

Question

First of all, the transition matrix is given as: $$ M= \begin{matrix} \frac 45 & 0 & 0 \\ 0 & \frac 65 & 0 \\ 0 & 0 & 1 \\ \end{matrix} $$ The problem (e) in https://i.stack.imgur.com/AXJ48.jpg seems very unclear to me. I don't really understand what they mean by basin of attraction and the term never appears in the book issued by my university. I managed to find the eigenvalues of the matrix and their corresponding eigenvectors. which are as follows:

$ \lambda_1 = 1 $

$ \lambda_2 = \frac 65 $

$ \lambda_3 = \frac 45 $

$ v_1=\begin{matrix} 0\\ 0\\ 1\\ \end{matrix} $

$ v_2=\begin{matrix} 0\\ 1\\ 0\\ \end{matrix} $

$ v_3=\begin{matrix} 1\\ 0\\ 0\\ \end{matrix} $

I'm really not sure about this step, but I think that by using the eigenvector corresponding to $\lambda = 1$ I managed to answer (d) which is essential for solving (e). In other words, to my understanding, the state $(x_s, y_s, z_s)$ to satisfy the equation in (d) must be the eigenvector corresponding to the eigenvalue $\lambda = 1$, which is $(0, 0, 1)$. Please correct me if I'm wrong here. Otherwise, how do I go about solving (e)? I am clueless as of where to even start.

Answer 1

The basin of attraction means all the starting points ${(x,y,z)}^T$ from which repeated multiplication by the matrix $M$ will converge toward the equilibrium state ${x_s,y_x,z_s}$.

So call your starting vector ${(x,y,z)}^T$. The $x$ will get multiplied $4/5$ infinity many times and go to zero. The $y$ will get multiplied by $6/5$ infinitely many times and diverge to infinity. The z will get multiplied by 1 and remain z.

So to reach an equilibrium state, y will have to equal 0. x and z can be anything, and they'll converge to ${(0,0,z)}^T$

So, you equilibrium state is ${0,0,z}^T$, and the basin of attraction of that state is ${x,0,z}$ with any x.