Let $A, B$ be $2\times 2 $ matrices satisfying:
- The eigenvalues $\lambda,\mu$ of $A$ satisfy $|\lambda|<1<|\mu|$.
- The eigenvalues $\lambda',\mu'$ of $B$ satisfy $|\lambda'|<1<|\mu'|$.
- There exists a homeomorphism $h:\mathbb{R}^2 \to \mathbb{R}^2$ such that $$h(A x) = B h(x), \ \forall \ x \in \mathbb{R}^2.$$
I'm reading the paper "Generic Singularities of 3D Piecewise Smooth Dynamical Systems", and the author "kinda says" that that 1) + 2) + 3) implies that $$\frac{\log(\lambda)}{\log(\mu)} =\frac{\log(\lambda')}{\log(\mu')}. $$
My question: Does someone know if 1)+2)+3) $\Rightarrow$ $\frac{\log(\lambda)}{\log(\mu)} =\frac{\log(\lambda')}{\log(\mu')} $?
Just some commentaries.
I might be confusing something. However, my assumption is based on this phrase
The references listed on the picture above are:
Moreover, on Proposition 9, the author uses (in my view) the fact of existing a conjugation between two diffeomorphisms $\phi$ and $\phi_0$, to conclude that $P(\phi) = P(\phi_0)$ and then construct a homeomorphism.
Can anyone help me?
EDIT: the author at no time says that $1) + 2) +3) \Rightarrow \frac{\log(\lambda)}{\log(\mu)} =\frac{\log(\lambda')}{\log(\mu')}$, I understood what was written in the wrong way, it was my mistake.
This is not true. Consider the linear maps $A(x,y)=(\frac12x,\,2y),\ B(x,y)=(\frac12x,\,8y)$ and the function $h(x,y)=(x,y^3)$ defined on $\mathbb R^2$. Then $h$ is a homeomorphism and $$ h(A(x,y))=h\left(\frac12x,\,2y\right)=\left(\frac12x,\,8y^3\right) =B(x,\,y^3)=B(h(x,y)), $$ but $$ \frac{\log(\lambda)}{\log(\mu)}=\frac{\log(\frac12)}{\log(2)} \ne\frac{\log(\frac12)}{\log(8)}=\frac{\log(\lambda')}{\log(\mu')}. $$