I'm trying to understand the following example from my lecture notes:
Define $A_n=M_{F_n}(\Bbb{C})\oplus M_{F_{n+1}}(\Bbb{C})$ where $F_n$ defined by $F_1=1, \ F_2=2, \ F_{n+2}=F_{n}+F_{n+1}$, i.e., Fibonacci sequence. The connecting maps defined by $\phi_{n+1,n}:A_n\to A_{n+1}$, $$\phi_n(a,b)=(b, \begin{bmatrix} a&0\\0&b \end{bmatrix}). $$ The inductive limit is a simple unital AF algebra. Lets compute $K_0(A)$.
At each stage the group $G_n:=K_0(A_n)$ is $(\Bbb{Z}^2,\Bbb{N}^2)$ and the connecting maps are $(a,b)\to (b,a+b)$. So it is multiplication by $T=\begin{bmatrix} 0&1\\1&1 \end{bmatrix}$ which is invertible and thus a simple argument shows that $K_0(A) \cong \Bbb{Z}^2$.
Up to this I'm fine. Now, we try to compute the positive part.
Define embeddings $\alpha_n:G_n\to \Bbb{Z}^2$ by $\alpha_n(a,b)=T^{-n}(a,b)^T$. So those are consistent
(If I understand it correctly, it meants that by the universal property we can see that they define the same direct limit, up to an isomorphism). So $\Bbb{Z}^2=\cup \alpha_n(G_n)$ and the positive part is $\cup \alpha_n(G_{n+})$.(why?)
Then we compute $T^{-1}= \begin{bmatrix} -1&1\\1&0 \end{bmatrix}$, $T^{-2}= \begin{bmatrix} 2&-1\\-1&1 \end{bmatrix}$ and $T^{-3}= \begin{bmatrix} -3&2\\2&-1 \end{bmatrix}$ and by them we somehow got this image: 
and concluded that the positive part ends up being everything above a line of a certain slope, and the group is totally oredered.
I don't understand the way we found the positive part. Any clarifications would be greatly appreciated.
Thank you for your time.
Well, I think I have an answer to myself...
The natural maps $\phi^n:K_0(A_n)\to K_0(A)$ are compatible with $T^{-n}:\Bbb{Z}^2 \to \Bbb{Z}^2$, which are isomorphisms. Thus, $\Bbb{Z}^2 \cong K_0(A)=\cup \phi^n (K_0(A_n)) \cong \cup T^{-n}(\Bbb{Z}^2)$. From continuity of $K_0(A)_+$ we get:
$K_0(A)_+=\varinjlim K_0(A_n)_+ =\cup \phi^n(K_0(A_n)_+)\cong \cup T^{-n}(\Bbb{Z}^2_+)=\cup T^{-n}(\Bbb{N}^2)$.
Now, we compute $T^{-n}$ for different values of $n\in \Bbb{N}$ on $\Bbb{N}^2$, but in order to describe $T^{-n}(\Bbb{N}^2)$ we should simply compute it on the span vectors, i.e. $(1,0) , \ (0,1)$ and understand the general picture.
If someone could check me, I'd be happy.