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Not sure how to do vectors using this so I apologise for the bad notation in advance ;-).

How do you go about finding the orbits of a given vector. I can find the orbits of eigen vectors but other than that it feels very laborious to check other vectors and even then you cannot check all vectors, obviously.

Eg let $H$ be the subgroup $\begin{pmatrix}a&b\\b&a\end{pmatrix}$with $a^2-b^2=1$ them $H$ acts on $R^2$ by matrix multiplication. Investigate the orbits of this action.

The eigen vectors are the lines $y=x$ and $y=-x$. These get scaled by their respective eigen values. What else can be said of the orbits?

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It's actually $SL_2(\mathbf R)$, since the conditions imply that the determinant is $1$. –  tomasz Jun 20 '12 at 12:49
    
You can write matrices as follows - writing $\text{\begin{pmatrix}a&b\\b&a\end{pmatrix}}$ produces: \begin{pmatrix}a&b\\b&a\end{pmatrix} –  Matt Pressland Jun 20 '12 at 12:53

2 Answers 2

Put $v_1=(1,1)^T,v_2=(1,-1)^T,\lambda=a+b$. Notice that $a-b=1/\lambda$.

Then for any $v=\alpha v_1+\beta v_2$ we have $A^n(v)=\alpha\lambda^nv_1+\beta\lambda^{-n}v_2$. Notice that for any power of $A$, the product of the $v_1$ coefficient and $v_2$ coefficient remains constant.

If $v_1$ and $v_2$ were the standard basis vectors, the set of points with a constant fixed nonzero product $\alpha\beta$ would be a hyperbola defined by the equation $xy=\alpha\beta$, equivalently $y=\alpha\beta/x$.

So the orbit of $v$ lies on the graph of the function $f(x)=\alpha\beta/x$ turned by $45^\circ$ (that is the transformation we need to apply to change from the standard basis vectors), and $A$ "slides" $v$ on the graph (possibly reflecting along the way if $\lambda<0$), which is what I think you were looking for.

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My phone won't let me comment for some reason. In response to tomasz.

So because the eigen vectors span the space we can express any vector in that basis and use the fact that $A(v+u)=c_1v+c_2u$ and we can find the function how exactly?

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I elaborated a little. Should be more clear now. –  tomasz Jun 20 '12 at 13:13

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