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If $||x_0-x_1||=||y_0-y_1||$, there exists an affine isometry such that: $f(x_0)=y_0 , f(x_1)=y_1$

prove that there is a function $g:mathbb Rrightarrow mathbb R$ such that $forall x,yin mathbb R^2space ;space f(x,y)=g(x+cy)$.Continuous mapping on cartesian product of metric spacesGiven $a,b in S^n$, then there exists an isometry $f: S^n rightarrow S^n$ such that $f(a) = b$Continuous function and densityIf $f'(x):mathbb R^n to mathbb R^n$ is a isometry, then $f(x)=T(x)+a$.Isometry on open setsProof verification: non-continuity of $f(x,y)=frac2xyx^2+y^2, f(0,0)=0$ at $(0,0)$show that $inf1-=inffrac 12(x+y)$When can we extend distance-preserving maps between finite sets of points to isometries?Lipschitzness of derivatives

1

$begingroup$

We define:

Affine transfomation: $f:mathbbR^ntomathbbR^n$ as follows: $f(x)=Ax+b$ as $Ain mathbbR_nxn$ and $binmathbbR^n$.

Isometry: $forall x,yinmathbbR^n: ||f(x)-f(y)||=||x-y||$ as $||cdot||$ denotes the euclideaמ norm.

Given $x_0,x_1,y_0,y_1 in mathbbR^n$ staisfying $||x_0-x_1||=||y_0-y_1||$.

Prove: There exists an affine isometry $f:mathbbR^ntomathbbR^n$ such that:

$f(x_0)=y_0 , f(x_1)=y_1$

I have already showed that an affine transformation is an isometry iff $A$ is orthogonal.

How can I deduce the above claim?

Any help would be appreciated.

real-analysis calculus differential-geometry isometry

share|cite|improve this question

edited Mar 22 at 16:34

Um Shmum

asked Mar 22 at 16:26

Um ShmumUm Shmum

1519

$endgroup$

add a comment | 

1

$begingroup$

We define:

Affine transfomation: $f:mathbbR^ntomathbbR^n$ as follows: $f(x)=Ax+b$ as $Ain mathbbR_nxn$ and $binmathbbR^n$.

Isometry: $forall x,yinmathbbR^n: ||f(x)-f(y)||=||x-y||$ as $||cdot||$ denotes the euclideaמ norm.

Given $x_0,x_1,y_0,y_1 in mathbbR^n$ staisfying $||x_0-x_1||=||y_0-y_1||$.

Prove: There exists an affine isometry $f:mathbbR^ntomathbbR^n$ such that:

$f(x_0)=y_0 , f(x_1)=y_1$

I have already showed that an affine transformation is an isometry iff $A$ is orthogonal.

How can I deduce the above claim?

Any help would be appreciated.

real-analysis calculus differential-geometry isometry

share|cite|improve this question

edited Mar 22 at 16:34

Um Shmum

asked Mar 22 at 16:26

Um ShmumUm Shmum

1519

$endgroup$

add a comment | 

$begingroup$

We define:

Affine transfomation: $f:mathbbR^ntomathbbR^n$ as follows: $f(x)=Ax+b$ as $Ain mathbbR_nxn$ and $binmathbbR^n$.

Isometry: $forall x,yinmathbbR^n: ||f(x)-f(y)||=||x-y||$ as $||cdot||$ denotes the euclideaמ norm.

Given $x_0,x_1,y_0,y_1 in mathbbR^n$ staisfying $||x_0-x_1||=||y_0-y_1||$.

Prove: There exists an affine isometry $f:mathbbR^ntomathbbR^n$ such that:

$f(x_0)=y_0 , f(x_1)=y_1$

I have already showed that an affine transformation is an isometry iff $A$ is orthogonal.

How can I deduce the above claim?

Any help would be appreciated.

real-analysis calculus differential-geometry isometry

share|cite|improve this question

edited Mar 22 at 16:34

Um Shmum

asked Mar 22 at 16:26

Um ShmumUm Shmum

1519

$endgroup$

share|cite|improve this question

edited Mar 22 at 16:34

Um Shmum

asked Mar 22 at 16:26

Um ShmumUm Shmum

1519

share|cite|improve this question

edited Mar 22 at 16:34

Um Shmum

asked Mar 22 at 16:26

Um ShmumUm Shmum

1519

asked Mar 22 at 16:26

Um ShmumUm Shmum

1519

asked Mar 22 at 16:26

Um ShmumUm Shmum

1519

2 Answers
2

active

oldest

votes

1

$begingroup$

1) In the trivial case where $x_0=x_1$ and $y_0=y_1$, a translation over the vector $vecx_0y_0$ does the job.

2) Now assume $x_0neq x_1$ and $y_0neq y_1$. Let $v_1:=vecx_0x_1/|vecx_0x_1|$ and extend it to an orthonormal basis $v_1,ldots,v_n$ of $mathbbR^n$. Similarly, let $w_1:=vecy_0y_1/|vecy_0y_1|$ and extend to an orthonormal basis $w_1,ldots,w_n$ of $mathbbR^n$. Let $A$ be the unique linear transformation of $mathbbR^n$ satisfying
begincases
A(v_1)=w_1\
hspace1.3cmvdots\
A(v_n)=w_n
endcases
Put $b:=y_0-A(x_0)$. Now define $$f:mathbbR^nrightarrowmathbbR^n:xmapsto A(x)+b.$$
I will leave it to you to check that $A$ is orthogonal, and that $f(x_0)=y_0$ and $f(x_1)=y_1$.

share|cite|improve this answer

answered Mar 22 at 16:46

studiosusstudiosus

2,294815

$endgroup$

add a comment | 

0

$begingroup$

Let us consider the vectors $x_1-x_0$ and $y_1-y_0$. Since they have the same norm there exists an isometry $A$ (actually infinitely many if $n>1$) of $mathbbR^n$ such that $(y_1-y_0)=A(x_1-x_0)$. Then, to find (one of) the affine transformation you're after, just put
$b=y_0-Ax_0$. This works because $$Ax_0+b=Ax_0-Ax_0+y_0=y_0$$ and $$Ax_1+b=A(x_1-x_0)+y_0=(y_1-y_0)+y_0=y_1$$

share|cite|improve this answer

edited Mar 22 at 16:48

answered Mar 22 at 16:41

Leo163Leo163

1,795512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it true that there exists an isometry satisfying $(y_1-y_0)=A(x_1-x_0)$?
  $endgroup$
  – Um Shmum
  Mar 22 at 16:46
  

  
  
  
  

add a comment | 

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1

$begingroup$

1) In the trivial case where $x_0=x_1$ and $y_0=y_1$, a translation over the vector $vecx_0y_0$ does the job.

2) Now assume $x_0neq x_1$ and $y_0neq y_1$. Let $v_1:=vecx_0x_1/|vecx_0x_1|$ and extend it to an orthonormal basis $v_1,ldots,v_n$ of $mathbbR^n$. Similarly, let $w_1:=vecy_0y_1/|vecy_0y_1|$ and extend to an orthonormal basis $w_1,ldots,w_n$ of $mathbbR^n$. Let $A$ be the unique linear transformation of $mathbbR^n$ satisfying
begincases
A(v_1)=w_1\
hspace1.3cmvdots\
A(v_n)=w_n
endcases
Put $b:=y_0-A(x_0)$. Now define $$f:mathbbR^nrightarrowmathbbR^n:xmapsto A(x)+b.$$
I will leave it to you to check that $A$ is orthogonal, and that $f(x_0)=y_0$ and $f(x_1)=y_1$.

share|cite|improve this answer

answered Mar 22 at 16:46

studiosusstudiosus

2,294815

$endgroup$

add a comment | 

1

$begingroup$

1) In the trivial case where $x_0=x_1$ and $y_0=y_1$, a translation over the vector $vecx_0y_0$ does the job.

2) Now assume $x_0neq x_1$ and $y_0neq y_1$. Let $v_1:=vecx_0x_1/|vecx_0x_1|$ and extend it to an orthonormal basis $v_1,ldots,v_n$ of $mathbbR^n$. Similarly, let $w_1:=vecy_0y_1/|vecy_0y_1|$ and extend to an orthonormal basis $w_1,ldots,w_n$ of $mathbbR^n$. Let $A$ be the unique linear transformation of $mathbbR^n$ satisfying
begincases
A(v_1)=w_1\
hspace1.3cmvdots\
A(v_n)=w_n
endcases
Put $b:=y_0-A(x_0)$. Now define $$f:mathbbR^nrightarrowmathbbR^n:xmapsto A(x)+b.$$
I will leave it to you to check that $A$ is orthogonal, and that $f(x_0)=y_0$ and $f(x_1)=y_1$.

share|cite|improve this answer

answered Mar 22 at 16:46

studiosusstudiosus

2,294815

$endgroup$

add a comment | 

$begingroup$

1) In the trivial case where $x_0=x_1$ and $y_0=y_1$, a translation over the vector $vecx_0y_0$ does the job.

2) Now assume $x_0neq x_1$ and $y_0neq y_1$. Let $v_1:=vecx_0x_1/|vecx_0x_1|$ and extend it to an orthonormal basis $v_1,ldots,v_n$ of $mathbbR^n$. Similarly, let $w_1:=vecy_0y_1/|vecy_0y_1|$ and extend to an orthonormal basis $w_1,ldots,w_n$ of $mathbbR^n$. Let $A$ be the unique linear transformation of $mathbbR^n$ satisfying
begincases
A(v_1)=w_1\
hspace1.3cmvdots\
A(v_n)=w_n
endcases
Put $b:=y_0-A(x_0)$. Now define $$f:mathbbR^nrightarrowmathbbR^n:xmapsto A(x)+b.$$
I will leave it to you to check that $A$ is orthogonal, and that $f(x_0)=y_0$ and $f(x_1)=y_1$.

share|cite|improve this answer

answered Mar 22 at 16:46

studiosusstudiosus

2,294815

$endgroup$

share|cite|improve this answer

answered Mar 22 at 16:46

studiosusstudiosus

2,294815

answered Mar 22 at 16:46

studiosusstudiosus

2,294815

answered Mar 22 at 16:46

studiosusstudiosus

2,294815

0

$begingroup$

Let us consider the vectors $x_1-x_0$ and $y_1-y_0$. Since they have the same norm there exists an isometry $A$ (actually infinitely many if $n>1$) of $mathbbR^n$ such that $(y_1-y_0)=A(x_1-x_0)$. Then, to find (one of) the affine transformation you're after, just put
$b=y_0-Ax_0$. This works because $$Ax_0+b=Ax_0-Ax_0+y_0=y_0$$ and $$Ax_1+b=A(x_1-x_0)+y_0=(y_1-y_0)+y_0=y_1$$

share|cite|improve this answer

edited Mar 22 at 16:48

answered Mar 22 at 16:41

Leo163Leo163

1,795512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it true that there exists an isometry satisfying $(y_1-y_0)=A(x_1-x_0)$?
  $endgroup$
  – Um Shmum
  Mar 22 at 16:46
  

  
  
  
  

add a comment | 

0

$begingroup$

Let us consider the vectors $x_1-x_0$ and $y_1-y_0$. Since they have the same norm there exists an isometry $A$ (actually infinitely many if $n>1$) of $mathbbR^n$ such that $(y_1-y_0)=A(x_1-x_0)$. Then, to find (one of) the affine transformation you're after, just put
$b=y_0-Ax_0$. This works because $$Ax_0+b=Ax_0-Ax_0+y_0=y_0$$ and $$Ax_1+b=A(x_1-x_0)+y_0=(y_1-y_0)+y_0=y_1$$

share|cite|improve this answer

edited Mar 22 at 16:48

answered Mar 22 at 16:41

Leo163Leo163

1,795512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it true that there exists an isometry satisfying $(y_1-y_0)=A(x_1-x_0)$?
  $endgroup$
  – Um Shmum
  Mar 22 at 16:46
  

  
  
  
  

add a comment | 

$begingroup$

Let us consider the vectors $x_1-x_0$ and $y_1-y_0$. Since they have the same norm there exists an isometry $A$ (actually infinitely many if $n>1$) of $mathbbR^n$ such that $(y_1-y_0)=A(x_1-x_0)$. Then, to find (one of) the affine transformation you're after, just put
$b=y_0-Ax_0$. This works because $$Ax_0+b=Ax_0-Ax_0+y_0=y_0$$ and $$Ax_1+b=A(x_1-x_0)+y_0=(y_1-y_0)+y_0=y_1$$

share|cite|improve this answer

edited Mar 22 at 16:48

answered Mar 22 at 16:41

Leo163Leo163

1,795512

$endgroup$

share|cite|improve this answer

edited Mar 22 at 16:48

answered Mar 22 at 16:41

Leo163Leo163

1,795512

answered Mar 22 at 16:41

Leo163Leo163

1,795512

answered Mar 22 at 16:41

Leo163Leo163

1,795512