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5

0

$begingroup$

Let $A, B$ be $2times 2 $ matrices satisfying:

1. The eigenvalues $lambda,mu$ of $A$ satisfy $|lambda|<1<|mu|$.


2. The eigenvalues $lambda',mu'$ of $B$ satisfy $|lambda'|<1<|mu'|$.


3. 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.

matrices dynamical-systems diffeomorphism

share|cite|improve this question

edited Mar 11 at 18:53

Matheus Manzatto

asked Feb 13 at 0:22

Matheus ManzattoMatheus Manzatto

1,3771625

$endgroup$

add a comment | 

5

0

$begingroup$

Let $A, B$ be $2times 2 $ matrices satisfying:

1. The eigenvalues $lambda,mu$ of $A$ satisfy $|lambda|<1<|mu|$.


2. The eigenvalues $lambda',mu'$ of $B$ satisfy $|lambda'|<1<|mu'|$.


3. 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.

matrices dynamical-systems diffeomorphism

share|cite|improve this question

edited Mar 11 at 18:53

Matheus Manzatto

asked Feb 13 at 0:22

Matheus ManzattoMatheus Manzatto

1,3771625

$endgroup$

add a comment | 

$begingroup$

Let $A, B$ be $2times 2 $ matrices satisfying:

1. The eigenvalues $lambda,mu$ of $A$ satisfy $|lambda|<1<|mu|$.


2. The eigenvalues $lambda',mu'$ of $B$ satisfy $|lambda'|<1<|mu'|$.


3. 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.

matrices dynamical-systems diffeomorphism

share|cite|improve this question

edited Mar 11 at 18:53

Matheus Manzatto

asked Feb 13 at 0:22

Matheus ManzattoMatheus Manzatto

1,3771625

$endgroup$

share|cite|improve this question

edited Mar 11 at 18:53

Matheus Manzatto

asked Feb 13 at 0:22

Matheus ManzattoMatheus Manzatto

1,3771625

share|cite|improve this question

edited Mar 11 at 18:53

Matheus Manzatto

asked Feb 13 at 0:22

Matheus ManzattoMatheus Manzatto

1,3771625

asked Feb 13 at 0:22

Matheus ManzattoMatheus Manzatto

1,3771625

asked Feb 13 at 0:22

Matheus ManzattoMatheus Manzatto

1,3771625

1 Answer
1

active

oldest

votes

2

$begingroup$

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
$$
h(A(x,y))=hleft(frac12x,,2yright)=left(frac12x,,8y^3right)
=B(x,,y^3)=B(h(x,y)),
$$
but
$$
frac{log(lambda)}{log(mu)}=frac{log(frac12)}{log(2)}
nefrac{log(frac12)}{log(8)}=frac{log(lambda')}{log(mu')}.
$$

share|cite|improve this answer

answered Feb 13 at 19:43

user1551user1551

73.7k566129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect counterexample.
  $endgroup$
  – Matheus Manzatto
  Feb 13 at 19:48
  

  
  
  
  

add a comment | 

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2

$begingroup$

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
$$
h(A(x,y))=hleft(frac12x,,2yright)=left(frac12x,,8y^3right)
=B(x,,y^3)=B(h(x,y)),
$$
but
$$
frac{log(lambda)}{log(mu)}=frac{log(frac12)}{log(2)}
nefrac{log(frac12)}{log(8)}=frac{log(lambda')}{log(mu')}.
$$

share|cite|improve this answer

answered Feb 13 at 19:43

user1551user1551

73.7k566129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect counterexample.
  $endgroup$
  – Matheus Manzatto
  Feb 13 at 19:48
  

  
  
  
  

add a comment | 

2

$begingroup$

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
$$
h(A(x,y))=hleft(frac12x,,2yright)=left(frac12x,,8y^3right)
=B(x,,y^3)=B(h(x,y)),
$$
but
$$
frac{log(lambda)}{log(mu)}=frac{log(frac12)}{log(2)}
nefrac{log(frac12)}{log(8)}=frac{log(lambda')}{log(mu')}.
$$

share|cite|improve this answer

answered Feb 13 at 19:43

user1551user1551

73.7k566129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect counterexample.
  $endgroup$
  – Matheus Manzatto
  Feb 13 at 19:48
  

  
  
  
  

add a comment | 

$begingroup$

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
$$
h(A(x,y))=hleft(frac12x,,2yright)=left(frac12x,,8y^3right)
=B(x,,y^3)=B(h(x,y)),
$$
but
$$
frac{log(lambda)}{log(mu)}=frac{log(frac12)}{log(2)}
nefrac{log(frac12)}{log(8)}=frac{log(lambda')}{log(mu')}.
$$

share|cite|improve this answer

answered Feb 13 at 19:43

user1551user1551

73.7k566129

$endgroup$

share|cite|improve this answer

answered Feb 13 at 19:43

user1551user1551

73.7k566129

answered Feb 13 at 19:43

user1551user1551

73.7k566129

answered Feb 13 at 19:43

user1551user1551

73.7k566129