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Let $A$ and $B$ be $2\times 2$ matrices with $\mathrm{tr}(A)>2$, $\mathrm{tr}(B)>2$ and $\det(A)=1$, $\det(B)=1$.

My Question : there exists an bijetive application $F:\mathbb{R}^{2}\to\mathbb{R}^{2}$ such that $F(\mathbb{Z}^2)=\mathbb{Z}^2$ and $A\circ F=F\circ B$??

EDIT 1: Think $A$, $B$ and $C$ as applications of $\mathbb{R}^2\to \mathbb{R}^2$

EDIT2: $A$ and $B$ has integer entries.

I apologize for the careless drafting the question

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Does $tr(A) = tr(B)$? –  Owen Biesel Oct 8 '12 at 22:33
    
No! $tr(A)$ can be diferent of $tr(B)$ –  user27456 Oct 8 '12 at 22:35
    
Could you clarify your notation? $A,B$ are matrices and $F$ is a function. What do $A\circ F$ and $F\circ B$ mean? Are you first applying $F$ then multiplying by $A$ and for the latter are you multiplying by $B$ and then applying $F$? –  Alex R. Oct 8 '12 at 22:53
    
I tried to clarify my question. if it has not been clear can ask more questions –  user27456 Oct 8 '12 at 22:57
    
What have you done so far? –  Alexander Gruber Oct 8 '12 at 23:02
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2 Answers 2

If your $F$ exists and is linear, then $A=F^{-1}BF$, so $A$ and $B$ would have the same eigenvalues. This means that any $A$, $B$ with different eigenvalues give a counterexample. For instance, take $$ A=\begin{bmatrix}3&1\\ 2 &1\end{bmatrix}, \ \ \ B=\begin{bmatrix}4&1\\3&1\end{bmatrix}. $$ Then for any $F$ like you want, $AF$ has eigenvalues $2\pm\sqrt3$, while $FB$ has eigenvalues $(5\pm\sqrt{21})/2$, so they cannot be equal.

Note that in this example $F$ does not exist even without the requirement of integer entries.

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Are you assuming $F$ is linear? That's not stated in the question. –  Gerry Myerson Oct 9 '12 at 1:09
    
Good point, although then there is little point in seeing $A,B$ as matrices. I'll edit accordingly. –  Martin Argerami Oct 9 '12 at 2:21
    
I had already thought of that approach, but as you found it, it is very restrictive, requires that $A$ and $B$ have the same eigenvalues​​, and I do not have it as a hypothesis. –  user27456 Oct 9 '12 at 13:59
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A parcial solution : I could only show that

$$F|_{\mathbb{Z}^2}\circ A|_{\mathbb{Z}^2}=B|_{\mathbb{Z}^2}\circ F|_{\mathbb{Z}^2} ~~~~~~(*)$$

but this is enough for my purposes.

In fact, consider in $\mathbb{Z}^2$ the equivalence relation $\sim_A$ $$\underline{x} \sim_A \underline{y}~~~~ \text{if only if} ~~~~Orb_A(\underline{x})=Orb_A(\underline{y})$$

and similarly define the equivalence relation $\sim_B.$ Let $G_A:=\mathbb{Z}^2 /\sim_A$ and $G_B:=\mathbb{Z}^2 /\sim_B$.

$G_A$ and $G_B$ are infinite enumerable sets, then there is a bijection $\Lambda$ between them.

Now we show how to construct $ F $ over each orbit of $ A $. Let be, $Orb_A(\underline{x})$ and $Orb_B(\underline{y})$ such that

$$\Lambda(Orb_A(\underline{x}))=Orb_B(\underline{y})~~~~$$ define $$~~F(\underline{x})=\underline{y}, ~~F(A\underline{x})=B\underline{y}, \ldots F(A^k\underline{x})=B^k\underline{y}, \text{etc.}$$

See, $ F $ is bijetive along each orbit, since neither nor $ A $ nor $ B $ have eigenvalues ​​which are the roots of unity

It is easy to see that $ F $ so defined satisfies $(*).$

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