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Suppose we have the following sequence of $\mathbb{Z}$-modules $$G\,\,\overset{M}{\longrightarrow}\,\, G\,\,\overset{M}{\longrightarrow}\,\, G\,\,\overset{M}{\longrightarrow}\,\, \cdots,$$

where each $G:=(\frac{\mathbb{Z}}{2\mathbb{Z}})^5\oplus\mathbb{Z}^5$, and the bonding map $M$ is given by the square matrix \begin{eqnarray*} M:=\left( \begin{array}{cccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 3 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 1 & 2 \\ \end{array} \right). \end{eqnarray*} For example, if $t=(t_1,\ldots,t_{10})\in \mathbb{Z}^{10}$, then the element $([t_1]_2,\ldots,[t_5]_2,t_6,\ldots,t_{10})\in G$ is mapped to $([(Mt)_1]_2,\ldots,[(Mt)_5]_2,(Mt)_6,\ldots,(Mt)_{10})$. What is the direct limit of such sequence?

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Can you say anything interesting bout the lower right quarter submatrix of $M$? –  Hagen von Eitzen Nov 20 '12 at 18:52
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Where does this problem come from? It looks a little arbitrary... Anyway, the matrix in the lower right corner is diagonalizable with eigenvalues $6,1,1,1,1$., where the eigenspace of $6$ is spanned by $[1,0,1,1,1]$. And in general, if you denote the matrices in the upper right and lower right corner with $A$ and $B$, then $M^n$ has the same form with $AB^{n-1}$ and $B^n$ instead of $A$ and $B$, respectively. That should help, I guess. –  Lukas Geyer Nov 21 '12 at 0:39
    
For each $i\in\mathbb{N}$ we let $G_i=G$ and view $M$ as a map from $G_i$ to $G_{i+1}$. Then the direct limit is $\oplus G_i/\sim$ where the equivalence relation $\sim$ is generated by declaring that $x\sim Mx$ for $x\in G_i$ for any $i$ (Note that here $Mx\in G_{i+1}$. Let $K_i=\mathbb{Z}/2\mathbb{Z}^5\subseteq G_i$. Then $M$ annihilates $K_i$ so for $x\in K_i$, $x\sim Mx = 0$. Hence the image of $K_i$ in the limit is 0. Essentially this should mean that we can ignore the summand $\mathbb{Z}/2\mathbb{Z}^5$ and just take the limit of $\mathbb{Z}^5$. –  Jeff Tolliver Nov 21 '12 at 2:26
    
If the lower right corner of your matrix is diagonalizable as Lukas suggests, then it is not difficult to see that your direct limit is isomorphic to $\mathbb Z[1/6]\oplus \mathbb Z^4$. Use the fact that the direct limit of a system of the form $\mathbb Z\overset{n}{\to}\mathbb Z\overset{n}{\to}\dots \overset{n}{\to}\mathbb Z\overset{n}{\to}\dots$ is isomorphic to $\mathbb Z[1/n]$, where $\mathbb Z[1/n]$ is the subring of $\mathbb Q$ generated by $1/n$. –  Simone Dec 3 '12 at 8:25

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