Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$ $A$ is a matrix with elements
$$\begin{pmatrix}
\frac{2}{3} & -\frac{2}{3}& -\frac{1}{3} \\
-\frac{2}{3}& 1 & \frac{2}{3} \\
\frac{4}{3} & 0 & -\frac{1}{3}
\end{pmatrix}$$
Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $$\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$$

A's eigenvalues are $-\frac{1}{3}$,1 and $\frac{2}{3}$ and eigenvectors are $\left(
  \begin{array}{c}
    0 \\
    -\frac{1}{2} \\
    1
  \end{array}
\right),\left(
  \begin{array}{c}
    1 \\
    -1 \\
    1
  \end{array}
\right),\left(
  \begin{array}{c}
    \frac{3}{4} \\
    -\frac{1}{2} \\
    1
  \end{array}
\right)$ for each.
I tried to get $A^n$ at first then got $$\begin{pmatrix}
(\frac{2}{3})^n & -2+\frac{2^{n+1}}{3}& 0 \\
-\frac{2+2^{n+1}}{3^{n+1}}& \frac{1-2^{n+2}}{3^{n+1}}-2 & \frac{2}{3^n} \\
\frac{4+2^{n+2}}{3^{n+1}} & \frac{2-2^{n+3}}{3^{n+1}} & -\frac{1}{3^n}
\end{pmatrix}$$
Then I tried find $$\displaystyle \lim_{ n \to \infty }(\frac{2}{3})^nx_1+(-2+\frac{2^{n+1}}{3})x_2=x_1 \\\displaystyle \lim_{ n \to \infty }-\frac{2+2^{n+1}}{3^{n+1}}x_1+(\frac{1-2^{n+2}}{3^{n+1}}-2)x_2+\frac{2}{3^n}x_3=x_2\\\ \displaystyle \lim_{ n \to \infty }\frac{4+2^{n+2}}{3^{n+1}}x_1+\frac{2-2^{n+3}}{3^{n+1}}x_2-\frac{1}{3^n}x_3=x_3$$
But I could not lead meaningful conclusion from these equations but $(0,0,0)$.
Where did I get wrong?
 A: Let $e_1, e_2, e_3$ be three linearly independent eigenvectors which are of unit norm. (You have shown that these exist.)
Suppose that they are corresponding to the eigenvalues $1, -1/3, 2/3$, respectively.
Let $\mathbf{x} \in \Bbb R^3$ be any such with the property you stated. Since $\{e_1, e_2, e_3\}$ form a basis, we can write $$\mathbf{x} = a_1e_1 + a_2e_2 + a_3e_3$$ for some reals $a_1, a_2, a_3$.
Then, applying $A^n$ to both sides, we get $$A^n\mathbf{x} = 1^na_1e_1 + \left(-\dfrac{1}{3}\right)^n a_2e_2 + \left(\dfrac{2}{3}\right)^na_3e_3.$$
As $n \to \infty$, the last two terms on the right tend to $0$ and thus, we get $$\lim_{n \to \infty}A^n\mathbf{x} = a_1e_1.$$
Thus, $$\mathbf{x} = a_1e_1.$$ Thus, we get that $a_2 = a_3 = 0.$
Conversely, it is easy to check that any $\mathbf{x} \in \operatorname{span}\{e_1\}$ does have the property you want.
Thus, we conclude that $\operatorname{span}\{e_1\}$ is precisely the set of all such vectors.
A: The answer of @AryamanMaithani explains very well how you can solve this question, but it uses the fact that the matrix is diagonalizable. I just would like to add a small precision, that you can prove directly and more easily that :
for every matrix $A \in \mathcal{M}_n(\mathbb{R})$ (diagonalizable or not),
$$\forall x \in \mathbb{R}^n, \text{ } \lim_{n\rightarrow +\infty} A^nx =x \text{ } \Longleftrightarrow \text{ }Ax=x$$
(so the set $\lbrace x \in \mathbb{R}^n \text{ }|\text{ } \lim_{n\rightarrow +\infty} A^nx =x \rbrace$ is always equal to the eigenspace associated to the eigenvalue $1$)
Indeed, if $A^nx \rightarrow x$, then you have $A^{n+1}x \rightarrow Ax$ and $A^{n+1}x \rightarrow x$, so $Ax=x$, so $x$ is an eigenvector associated to the eigenvalue $1$. Conversely, if $Ax=x$, it is immediate that $A^nx=x$ for all $n$, and then $A^nx \rightarrow x$ when $n$ tends to $+\infty$.
A: As the Eigenvalues of $A$ are $-\dfrac13,1,\dfrac23$, when you take the $n^{th}$ power of any vector, only the Eigenvector associated to $\lambda=1$ will remain.
Hence $\mathbb x$ are the multiple of the second Eigenvector.
