Given matrix $$A = \begin{pmatrix} 0 & 28 & 48\\ 0.25 & 0 & 0\\ 0 & 0.5 & 0 \end{pmatrix}$$
I already know eigenvalues and corresponding eigenvectors subspaces of this matrix.
$$\lambda_1 = -1, \;\;\;v_1 = a \cdot \begin{pmatrix}8\\-2\\1\end{pmatrix}, a\in \mathbb{R}$$ $$\lambda_2 = -2, \;\;\;v_2 = a \cdot \begin{pmatrix}32\\-4\\1\end{pmatrix}, a\in \mathbb{R}$$ $$\lambda_3 = 3, \;\;\;v_3 = a \cdot \begin{pmatrix}72\\6\\1\end{pmatrix}, a\in \mathbb{R}$$
$v = \begin{pmatrix}\hat{x}\\\hat{y}\\\hat{z}\end{pmatrix}$ such that $\hat{x} + \hat{y} + \hat{z} = 1$ and it corresponds to $\lambda_3$.
So, I could see that $v = \begin{pmatrix}\frac{72}{79}\\\frac{6}{79}\\\frac{1}{79}\end{pmatrix}$.
The first complexity that I faced solving this problim is this:
Real numbers $x_0$, $y_0$, $z_0$ are such that $\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}$ is not in sum of subspaces of eigenvectors corresponding to $\lambda_1, \lambda_2$.
Also $x_n, y_n, z_n$ are such that
$$\begin{pmatrix}x_n\\y_n\\z_n\end{pmatrix} = A^n\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}$$
Prove that
$$\lim_{n \rightarrow \infty} \frac{x_n}{x_n + y_n + z_n} = \hat{x}_n$$
And, to be honest, I don't have much ideas on how to use fact that $\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}$ is not in sum of subspaces corresponding to $\lambda_1$ and $\lambda_2$. Yes, we understand that it cannot be linear combination of vectors that are basis of this subspaces sum but we definitely see that it can be represented as linear combination of $v_1, v_2$ and $v_3$ since they are basis of $\mathbb{R}^3$.
So, any help on understanding what to try here? Thanks in advance!