Mixing
Counting the number of orbits for the group action
$begingroup$
I have have to find the number of orbits for the group action given by the set of all real $2times 2$ invertible matrices acting on $mathbb{R^2}$ by matrix multiplication.
I know that if a group $G$ acts on a set $X$ then for each $xin X$,
$orbit(x) = {gx: gin G}$, where $g$ runs over all the elements of the group $G$.
Thus here for any $xin mathbb{R^2}$ its orbit is given by set ${Ax : A in G}$ and I have to keep in mind that orbits partition the set $X = mathbb{R^2} $.
My confusion:
Here both group $G$ and set $X$ are of infinite order. So, how to find orbit of all the infinite elements of $mathbb{R^2}$. Do I have to look at the basis elements of $mathbb{R^2}$?
I am learning to find the orbits of a set. Any help and basic idea to solve such kind of problem would be very helpful to me.
Thank you very much for your kind consideration.
abstract-algebra group-theory
asked May 27 '14 at 11:10
srijansrijan
6,46563980
$endgroup$
add a comment |
$begingroup$
I have have to find the number of orbits for the group action given by the set of all real $2times 2$ invertible matrices acting on $mathbb{R^2}$ by matrix multiplication.
I know that if a group $G$ acts on a set $X$ then for each $xin X$,
$orbit(x) = {gx: gin G}$, where $g$ runs over all the elements of the group $G$.
Thus here for any $xin mathbb{R^2}$ its orbit is given by set ${Ax : A in G}$ and I have to keep in mind that orbits partition the set $X = mathbb{R^2} $.
My confusion:
Here both group $G$ and set $X$ are of infinite order. So, how to find orbit of all the infinite elements of $mathbb{R^2}$. Do I have to look at the basis elements of $mathbb{R^2}$?
I am learning to find the orbits of a set. Any help and basic idea to solve such kind of problem would be very helpful to me.
Thank you very much for your kind consideration.
abstract-algebra group-theory
asked May 27 '14 at 11:10
srijansrijan
6,46563980
$endgroup$
1
$begingroup$
Hint: prove that if $;x,yinBbb R;,;;x,yneq0;$ then there exists $;Ain GL(n,Bbb R);;s.t.;;Ax=y;$
$endgroup$
– DonAntonio
May 27 '14 at 11:15
$begingroup$
@DonAntonio Thanks for you comment. I don't get your point. Could you explain little more?
$endgroup$
– srijan
May 27 '14 at 11:22
$begingroup$
this is basic (=not necessarily elementary but important) linear algebra: (1) every linear transformation (and thus any matrix) is uniquely and completely determined by its action on any basis of the vector space in question, and (2) A square matrix is regular iff it maps a basis to a basis. Now, remember that $;xneq 0implies x;$ is part of a basis, so there you go...
$endgroup$
– DonAntonio
May 27 '14 at 11:39
add a comment |
$begingroup$
I have have to find the number of orbits for the group action given by the set of all real $2times 2$ invertible matrices acting on $mathbb{R^2}$ by matrix multiplication.
I know that if a group $G$ acts on a set $X$ then for each $xin X$,
$orbit(x) = {gx: gin G}$, where $g$ runs over all the elements of the group $G$.
Thus here for any $xin mathbb{R^2}$ its orbit is given by set ${Ax : A in G}$ and I have to keep in mind that orbits partition the set $X = mathbb{R^2} $.
My confusion:
Here both group $G$ and set $X$ are of infinite order. So, how to find orbit of all the infinite elements of $mathbb{R^2}$. Do I have to look at the basis elements of $mathbb{R^2}$?
I am learning to find the orbits of a set. Any help and basic idea to solve such kind of problem would be very helpful to me.
Thank you very much for your kind consideration.
abstract-algebra group-theory
asked May 27 '14 at 11:10
srijansrijan
6,46563980
$endgroup$
I have have to find the number of orbits for the group action given by the set of all real $2times 2$ invertible matrices acting on $mathbb{R^2}$ by matrix multiplication.
I know that if a group $G$ acts on a set $X$ then for each $xin X$,
$orbit(x) = {gx: gin G}$, where $g$ runs over all the elements of the group $G$.
Thus here for any $xin mathbb{R^2}$ its orbit is given by set ${Ax : A in G}$ and I have to keep in mind that orbits partition the set $X = mathbb{R^2} $.
My confusion:
Here both group $G$ and set $X$ are of infinite order. So, how to find orbit of all the infinite elements of $mathbb{R^2}$. Do I have to look at the basis elements of $mathbb{R^2}$?
I am learning to find the orbits of a set. Any help and basic idea to solve such kind of problem would be very helpful to me.
Thank you very much for your kind consideration.
abstract-algebra group-theory
abstract-algebra group-theory
asked May 27 '14 at 11:10
srijansrijan
6,46563980
asked May 27 '14 at 11:10
srijansrijan
6,46563980
asked May 27 '14 at 11:10
srijansrijan
6,46563980
asked May 27 '14 at 11:10
srijansrijan
6,46563980
asked May 27 '14 at 11:10
srijansrijan
6,46563980
6,46563980
1
$begingroup$
Hint: prove that if $;x,yinBbb R;,;;x,yneq0;$ then there exists $;Ain GL(n,Bbb R);;s.t.;;Ax=y;$
$endgroup$
– DonAntonio
May 27 '14 at 11:15
$begingroup$
@DonAntonio Thanks for you comment. I don't get your point. Could you explain little more?
$endgroup$
– srijan
May 27 '14 at 11:22
$begingroup$
this is basic (=not necessarily elementary but important) linear algebra: (1) every linear transformation (and thus any matrix) is uniquely and completely determined by its action on any basis of the vector space in question, and (2) A square matrix is regular iff it maps a basis to a basis. Now, remember that $;xneq 0implies x;$ is part of a basis, so there you go...
$endgroup$
– DonAntonio
May 27 '14 at 11:39
add a comment |
1
$begingroup$
Hint: prove that if $;x,yinBbb R;,;;x,yneq0;$ then there exists $;Ain GL(n,Bbb R);;s.t.;;Ax=y;$
$endgroup$
– DonAntonio
May 27 '14 at 11:15
$begingroup$
@DonAntonio Thanks for you comment. I don't get your point. Could you explain little more?
$endgroup$
– srijan
May 27 '14 at 11:22
$begingroup$
this is basic (=not necessarily elementary but important) linear algebra: (1) every linear transformation (and thus any matrix) is uniquely and completely determined by its action on any basis of the vector space in question, and (2) A square matrix is regular iff it maps a basis to a basis. Now, remember that $;xneq 0implies x;$ is part of a basis, so there you go...
$endgroup$
– DonAntonio
May 27 '14 at 11:39
1
1
$begingroup$
Hint: prove that if $;x,yinBbb R;,;;x,yneq0;$ then there exists $;Ain GL(n,Bbb R);;s.t.;;Ax=y;$
$endgroup$
– DonAntonio
May 27 '14 at 11:15
$begingroup$
Hint: prove that if $;x,yinBbb R;,;;x,yneq0;$ then there exists $;Ain GL(n,Bbb R);;s.t.;;Ax=y;$
$endgroup$
– DonAntonio
May 27 '14 at 11:15
$begingroup$
@DonAntonio Thanks for you comment. I don't get your point. Could you explain little more?
$endgroup$
– srijan
May 27 '14 at 11:22
$begingroup$
@DonAntonio Thanks for you comment. I don't get your point. Could you explain little more?
$endgroup$
– srijan
May 27 '14 at 11:22
$begingroup$
this is basic (=not necessarily elementary but important) linear algebra: (1) every linear transformation (and thus any matrix) is uniquely and completely determined by its action on any basis of the vector space in question, and (2) A square matrix is regular iff it maps a basis to a basis. Now, remember that $;xneq 0implies x;$ is part of a basis, so there you go...
$endgroup$
– DonAntonio
May 27 '14 at 11:39
$begingroup$
this is basic (=not necessarily elementary but important) linear algebra: (1) every linear transformation (and thus any matrix) is uniquely and completely determined by its action on any basis of the vector space in question, and (2) A square matrix is regular iff it maps a basis to a basis. Now, remember that $;xneq 0implies x;$ is part of a basis, so there you go...
$endgroup$
– DonAntonio
May 27 '14 at 11:39
add a comment |
1 Answer
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$begingroup$
Take a point $p = (a,b) in mathbb{R}^2$. The claim is that taking any point $q =(x,y)inmathbb{R}^2$, we find a linear transformation (i.e. matrix) $A$ such that $A p^T = q^T$ (1).
Denote $A = begin{pmatrix} alpha & beta \ gamma & delta end{pmatrix}$. The equation (1) corresponds to the system $alpha a + beta b = x ,& , gamma a + delta b = y$, which has in general infinitely many solutions; adding the condition of regularity of $A$, i.e. the zero determinant, will cancel some of them.
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
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add a comment |
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$begingroup$
Take a point $p = (a,b) in mathbb{R}^2$. The claim is that taking any point $q =(x,y)inmathbb{R}^2$, we find a linear transformation (i.e. matrix) $A$ such that $A p^T = q^T$ (1).
Denote $A = begin{pmatrix} alpha & beta \ gamma & delta end{pmatrix}$. The equation (1) corresponds to the system $alpha a + beta b = x ,& , gamma a + delta b = y$, which has in general infinitely many solutions; adding the condition of regularity of $A$, i.e. the zero determinant, will cancel some of them.
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
$endgroup$
add a comment |
$begingroup$
Take a point $p = (a,b) in mathbb{R}^2$. The claim is that taking any point $q =(x,y)inmathbb{R}^2$, we find a linear transformation (i.e. matrix) $A$ such that $A p^T = q^T$ (1).
Denote $A = begin{pmatrix} alpha & beta \ gamma & delta end{pmatrix}$. The equation (1) corresponds to the system $alpha a + beta b = x ,& , gamma a + delta b = y$, which has in general infinitely many solutions; adding the condition of regularity of $A$, i.e. the zero determinant, will cancel some of them.
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
$endgroup$
add a comment |
$begingroup$
Take a point $p = (a,b) in mathbb{R}^2$. The claim is that taking any point $q =(x,y)inmathbb{R}^2$, we find a linear transformation (i.e. matrix) $A$ such that $A p^T = q^T$ (1).
Denote $A = begin{pmatrix} alpha & beta \ gamma & delta end{pmatrix}$. The equation (1) corresponds to the system $alpha a + beta b = x ,& , gamma a + delta b = y$, which has in general infinitely many solutions; adding the condition of regularity of $A$, i.e. the zero determinant, will cancel some of them.
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
$endgroup$
Take a point $p = (a,b) in mathbb{R}^2$. The claim is that taking any point $q =(x,y)inmathbb{R}^2$, we find a linear transformation (i.e. matrix) $A$ such that $A p^T = q^T$ (1).
Denote $A = begin{pmatrix} alpha & beta \ gamma & delta end{pmatrix}$. The equation (1) corresponds to the system $alpha a + beta b = x ,& , gamma a + delta b = y$, which has in general infinitely many solutions; adding the condition of regularity of $A$, i.e. the zero determinant, will cancel some of them.
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
answered Oct 23 '17 at 11:13
pepa.dvorakpepa.dvorak
839410
839410
add a comment |
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$begingroup$
Hint: prove that if $;x,yinBbb R;,;;x,yneq0;$ then there exists $;Ain GL(n,Bbb R);;s.t.;;Ax=y;$
$endgroup$
– DonAntonio
May 27 '14 at 11:15
$begingroup$
@DonAntonio Thanks for you comment. I don't get your point. Could you explain little more?
$endgroup$
– srijan
May 27 '14 at 11:22
$begingroup$
this is basic (=not necessarily elementary but important) linear algebra: (1) every linear transformation (and thus any matrix) is uniquely and completely determined by its action on any basis of the vector space in question, and (2) A square matrix is regular iff it maps a basis to a basis. Now, remember that $;xneq 0implies x;$ is part of a basis, so there you go...
$endgroup$
– DonAntonio
May 27 '14 at 11:39