Given the direct system $$\mathbb{Z}^2 \xrightarrow{A} \mathbb{Z}^2 \xrightarrow{A}\mathbb{Z}^2 \xrightarrow{A}\cdots$$ with $$A = \begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix},$$ the direct limit should be $\mathbb{Z} \oplus \mathbb{Z}[\frac{1}{2}]$.
How is this done? Well, the eigenvalues of $A$ are $-1$ and $2$ with eigenvectors $(-1,2)^T$ and $(1,1)^T$ respectively. The best I can do right now is to show that the limit contains a subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}[\frac{1}{2}]$...
The reason why I am asking is because by replacing $A$ above with $A' = \begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}$, then even though $A'$ has eigenvalues $-2$ and $3$, the direct limit in this case cannot be the direct sum of $\mathbb{Z}[\frac{1}{2}]$ and $\mathbb{Z}[\frac{1}{3}]$.
Hint: The direct limit of this system is isomorphic to the direct limit of $$ \Bbb Z^2\xrightarrow{S}\Bbb Z^2\xrightarrow{S}\Bbb Z^2\xrightarrow{S}\dotsb $$ where $S$ is the Smith normal form of $A$. In this case, the Smith form of $A$ is $$ S= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} $$ and we have $UAV=S$ where \begin{align*} U &=\begin{pmatrix}1&0\\ 2&-1\end{pmatrix} & V &=\begin{pmatrix}1 & -1\\ 0 & 1\end{pmatrix} \end{align*}