Discrete Dynamical System - determine what the model predicts will be the long-term distibution

Question

If I have the following matrix: $$X_{n+1}\begin{pmatrix}1&0\\ 0&0.2\end{pmatrix}X_n$$ and if I also have the following initial state vector: $$X_0=\begin{pmatrix}5\\ 7\end{pmatrix}$$

What does the model predict will be the long-term distribution ?

My assumption to the model:

if we use the initial state vector provided above we know that: $$X_0=\begin{pmatrix}5\\ 7\end{pmatrix},\:where\:5>0\:and\:7\:>0$$ and use it to form a long-term distribution: $$X_0=C_1V_1+C_2V_2$$ $$\downarrow$$ $$X_n=5\left(1\right)^n\begin{pmatrix}1\\ 0\end{pmatrix}+7\left(0.2\right)^n\begin{pmatrix}0\\ 1\end{pmatrix}$$ It is clear that: $$As\:n\rightarrow \infty :\:5\left(1\right)^n\begin{pmatrix}1\\ \:0\end{pmatrix}\:\rightarrow \:\:will\:always\:be\:5.$$ and: $$As\:n\rightarrow \infty :\:7\left(0.2\right)^n\begin{pmatrix}0\\ 1\end{pmatrix}\:\rightarrow \:0$$

My question here is should I consider the first component will be survived in the long-term state ? I doubt this because it's not neither going to be a infinity number or to zero...it's just going to stay at 5 whatever n will be.

Answer 1

yes, the answer that the distribution as $n\to\infty$ is $\pmatrix5\\0$ is correct. The reason it seems fishy to you is that it is too easy. But what the course or professor is going to get to next might be non-diagonal matrices with one eigenvalue of $1$ and another with absolute value less than $1$, for example, $$x_{n+1} = \pmatrix{1 &0\\0.5 & 0.2}x_n$$ with a starting state of $\pmatrix{2\\2}$. The answer there is that the state tends to $\pmatrix{2\\1}$ and the way to find that involves diagonalizing the matrix.

Answer 2

Given that $X_{n+1} = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0.2 \end{array} \right) X_n = A X_n$, we see that $X_1=AX_0$, $X_2=AX_1 = A(A X_0) = A^2 X_0$, and we can obtain the general formula $X_n = A^n X_0$. Observe that $A^n = \left( \begin{array}{cc} 1^n & 0 \\ 0 & 0.2^n \end{array} \right)$. Hence, $X_n = A^n X_0 = \left( \begin{array}{c} 5 \\ 7(0.2^n) \end{array} \right)$. Hence, $X_n$ approaches $\left( \begin{array}{c} 5 \\ 0 \end{array} \right)$ as $n$ approaches infinity.