The stochastic matrix $$ A=\left( \begin{array} [c]{cc}% 0.9 & 0.2\\ 0.1 & 0.8 \end{array} \right) $$
has a dominant eigenvectors of $u_{1}=\left( \begin{array} [c]{c}% 2\,\\ 1\, \end{array} \right) $. The second eigenvector is $u_{2}=\left( \begin{array} [c]{c}% 1\,\\ -1 \end{array} \right) $. The corresponding eigenvalues are $1$ and $0.7$, respectively. I am trying to visualize what happens when we repeatedly multiply a vector by $A$. $$ Ax,~A^{2}x,~... $$
Let $x$ be any vector that is not an eigenvector. In the matrix product $Ax$, the matrix $A$ transforms $x$ and changes its direction. Loosely speaking, the direction of $Ax$ is closer to the direction of $u_{1}$, the dominant eigenvector, than the original $x$. We can repeat this transformation and in the limit, we have that the two vectors coincide. $$ \lim_{k\rightarrow\infty}A^{k}x=u_{1}% $$ In the figure below, the directions of the two eigenvectors are indicated by black dashed lines. The orange line is perpendicular to the dominant eigenvector. I have two questions.
Consider the two vectors a and b (both in blue, lower right). Repeated multiplication of $a$ by $A$ will take it towards $\left( 2,1\right) ^{\prime}$. Since $b$ is on the other side of the orange line, repeated multiplication of $b$ by $A$ will take it towards $\left( -2,-1\right) ^{\prime}$. But what happens if a vector is EXACTLY on the orange line?
Consider vector $b$. Is it possible that after multiplying by $A$, the new vector happens to coincide with the second eigenvector? That is, $$ A^{k}b=u_{2}% $$ for some positive integer $k$. I assume the answer is no, but what exactly assures that this does not happen?
