We are given the matrix $A=\begin{bmatrix}0.9&0.5\\0.1&0.5\end{bmatrix}$ and any initial vector $X^{(0)}=\begin{bmatrix}a\\b\end{bmatrix}$.
The matrix $A$ has the following eigensystem:
$\qquad\text{Eigenvalues}: 1,\,0.4\\\qquad\text{Eigenvectors: }E_1=\begin{bmatrix}5&1\end{bmatrix}^T, \,E_2=\begin{bmatrix}-1&1\end{bmatrix}^T$
Given in order of occurring eigenvalue.
We can write $A^nE_1=1^nE_1=E$, and also $A^nE_2=(0.4)^nE_2$.
Since $E_1$ and $E_2$ form a basis in $\mathbb{R}^2$ we can now write our initial value vector $X^{(0)}$ as $x_1'E_1+x_2'E_2$. However, since $X^{(n)}=A^nX^{(0)}$ and because of the linear property of our matrix we get the following equation:
$$A^nX^{(0)}=x_1'A^nE_1+x_2'E_2 \iff X^{(n)}=x_1'E_1+x_2'(0.4)^nE_2$$
Because $0.4^n \rightarrow 0$ when $n\rightarrow\infty \iff X^{(n)}\rightarrow x_1'E_1$ when $n\rightarrow\infty$. As such, $X^{(n)}$ always converges to the component of $X^{(0)}$ in the direction of $E_1$.
Trying this out, say we have the initial vector $X$ where $a=30,\,b=20$.
Note that $a+b=50$
$\text{Proj}_{E_1}X^{(0)} \approx (32.6, 6.5)$
And now, the proportions are right, as $\frac{32.6}{32.6+6.5}\approx\frac{5}{6}$.
Note that $32.6+6.5$ is not near $50$.
HOWEVER, when I let Wolfram calculate this manually, i.e. $$X^{(n)}=A^nX^{(0)}\rightarrow (41.6667,\,8.3333) \text{ when } n\rightarrow \infty$$
Note that $41.6667\ldots+8.3333\ldots = 50$
- Why does this differ from the projection? Where is my theory wrong?
Because you are trying to rewrite your vector using a non-orthogonal basis, you cannot rely on projections to find the decomposition coefficients. The identity $v = \Sigma \text{ Proj}_{e_i}(v)$ only works if the $e_i$ form an orthogonal basis. To find the decomposition in the general case, to write $v = \Sigma a_i w_i$ where the $w_i$ make up a possibly non-orthogonal basis, you must solve the system $[w_1,w_2...w_n][a_1...a_n]^T = v$. Here, that amounts to solving $\begin{bmatrix}5&-1\\1&1\end{bmatrix}\begin{bmatrix}a_1\\a_2\end{bmatrix} = \begin{bmatrix}30\\20\end{bmatrix}$, giving $a_1 = \frac{25}{3}$ and $a_2 = \frac{65}{3}$.
So $X^{(0)}=\frac{25}{3}E_1 + \frac{65}{3}E_2$, and $\lim_{n\to\infty}A^nX^{(0)} = \frac{25}{3}E_1 = [\frac{125}{3}, \frac{25}{3}]^T$ as desired.