How do I work out the generator for $GL(2,Bbb{R})$?












2












$begingroup$


Where $G$ is a group $$G=GL(2,mathbb{R})$$



$$x=begin{bmatrix}
cos(pi/4) & sin(pi/4)\
-sin(pi/4)& cos(pi/4)
end{bmatrix}$$



How do I work out the generator $langle xrangle$... Quite new to group theory so any help would help, thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    That's a rotation matrix. What are its powers?
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:33










  • $begingroup$
    What do you mean by rotation matrix?
    $endgroup$
    – Reety
    Dec 29 '18 at 11:33












  • $begingroup$
    In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. To learn more: en.wikipedia.org/wiki/Rotation_matrix
    $endgroup$
    – toric_actions
    Dec 29 '18 at 11:44
















2












$begingroup$


Where $G$ is a group $$G=GL(2,mathbb{R})$$



$$x=begin{bmatrix}
cos(pi/4) & sin(pi/4)\
-sin(pi/4)& cos(pi/4)
end{bmatrix}$$



How do I work out the generator $langle xrangle$... Quite new to group theory so any help would help, thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    That's a rotation matrix. What are its powers?
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:33










  • $begingroup$
    What do you mean by rotation matrix?
    $endgroup$
    – Reety
    Dec 29 '18 at 11:33












  • $begingroup$
    In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. To learn more: en.wikipedia.org/wiki/Rotation_matrix
    $endgroup$
    – toric_actions
    Dec 29 '18 at 11:44














2












2








2


1



$begingroup$


Where $G$ is a group $$G=GL(2,mathbb{R})$$



$$x=begin{bmatrix}
cos(pi/4) & sin(pi/4)\
-sin(pi/4)& cos(pi/4)
end{bmatrix}$$



How do I work out the generator $langle xrangle$... Quite new to group theory so any help would help, thanks.










share|cite|improve this question











$endgroup$




Where $G$ is a group $$G=GL(2,mathbb{R})$$



$$x=begin{bmatrix}
cos(pi/4) & sin(pi/4)\
-sin(pi/4)& cos(pi/4)
end{bmatrix}$$



How do I work out the generator $langle xrangle$... Quite new to group theory so any help would help, thanks.







linear-algebra group-theory trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 12:11









mechanodroid

27.8k62447




27.8k62447










asked Dec 29 '18 at 11:31









ReetyReety

15311




15311












  • $begingroup$
    That's a rotation matrix. What are its powers?
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:33










  • $begingroup$
    What do you mean by rotation matrix?
    $endgroup$
    – Reety
    Dec 29 '18 at 11:33












  • $begingroup$
    In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. To learn more: en.wikipedia.org/wiki/Rotation_matrix
    $endgroup$
    – toric_actions
    Dec 29 '18 at 11:44


















  • $begingroup$
    That's a rotation matrix. What are its powers?
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:33










  • $begingroup$
    What do you mean by rotation matrix?
    $endgroup$
    – Reety
    Dec 29 '18 at 11:33












  • $begingroup$
    In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. To learn more: en.wikipedia.org/wiki/Rotation_matrix
    $endgroup$
    – toric_actions
    Dec 29 '18 at 11:44
















$begingroup$
That's a rotation matrix. What are its powers?
$endgroup$
– Lord Shark the Unknown
Dec 29 '18 at 11:33




$begingroup$
That's a rotation matrix. What are its powers?
$endgroup$
– Lord Shark the Unknown
Dec 29 '18 at 11:33












$begingroup$
What do you mean by rotation matrix?
$endgroup$
– Reety
Dec 29 '18 at 11:33






$begingroup$
What do you mean by rotation matrix?
$endgroup$
– Reety
Dec 29 '18 at 11:33














$begingroup$
In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. To learn more: en.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– toric_actions
Dec 29 '18 at 11:44




$begingroup$
In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. To learn more: en.wikipedia.org/wiki/Rotation_matrix
$endgroup$
– toric_actions
Dec 29 '18 at 11:44










1 Answer
1






active

oldest

votes


















3












$begingroup$

Your matrix $x$ represents a rotation in $mathbb{R}^2$ around the origin by the angle $frac{pi}4$ clockwise.



In general $R(phi) = begin{bmatrix} cosphi & sinphi \ -sinphi & cosphiend{bmatrix}$ does the same thing with the angle $phi$.



Clearly if you rotate by $phi$ and then by $psi$, it is the same as rotating by $phi + psi$. Hence $R(phi)R(psi) = R(phi + psi)$. You can also check this algebraically, of course.



Also $R(phi)^{-1} = R(-phi)$ as the inverse of a rotation by $phi$ is a rotation by $-phi$.



Therefore



$$langle xrangle = {x^n : n in mathbb{Z}} = left{Rleft(frac{pi}4right)^n : n in mathbb{Z}right} = left{Rleft(frac{npi}4right) : n in mathbb{Z}right}$$



Hence your group consists of all rotations by an angle which is a multiple of $fracpi4$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a great explanation thank you, although the book I'm reading says the answer is {e, x, x^2, x^3} how can I convert this notation? - @mechanodroid
    $endgroup$
    – Reety
    Dec 29 '18 at 12:16












  • $begingroup$
    @Reety Are you sure about that? Notice that $x^8 = R(pi/4)^8 = R(2pi) = e$ and the smaller powers are not equal to $e$ so the group should be $$langle xrangle = {e,x,x^2,x^3,x^4,x^5,x^6,x^7}$$
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:27










  • $begingroup$
    @Reety: Which book are you using?
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:30










  • $begingroup$
    @Shaun I don't know what it's called since It's a pdf file and it doesn't have the author on it and the names generic.
    $endgroup$
    – Reety
    Dec 29 '18 at 12:36










  • $begingroup$
    @Reety: If you found it online, then copy & paste a long sentence from it into a search engine, with speechmarks both sides; it might show up.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:40













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Your matrix $x$ represents a rotation in $mathbb{R}^2$ around the origin by the angle $frac{pi}4$ clockwise.



In general $R(phi) = begin{bmatrix} cosphi & sinphi \ -sinphi & cosphiend{bmatrix}$ does the same thing with the angle $phi$.



Clearly if you rotate by $phi$ and then by $psi$, it is the same as rotating by $phi + psi$. Hence $R(phi)R(psi) = R(phi + psi)$. You can also check this algebraically, of course.



Also $R(phi)^{-1} = R(-phi)$ as the inverse of a rotation by $phi$ is a rotation by $-phi$.



Therefore



$$langle xrangle = {x^n : n in mathbb{Z}} = left{Rleft(frac{pi}4right)^n : n in mathbb{Z}right} = left{Rleft(frac{npi}4right) : n in mathbb{Z}right}$$



Hence your group consists of all rotations by an angle which is a multiple of $fracpi4$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a great explanation thank you, although the book I'm reading says the answer is {e, x, x^2, x^3} how can I convert this notation? - @mechanodroid
    $endgroup$
    – Reety
    Dec 29 '18 at 12:16












  • $begingroup$
    @Reety Are you sure about that? Notice that $x^8 = R(pi/4)^8 = R(2pi) = e$ and the smaller powers are not equal to $e$ so the group should be $$langle xrangle = {e,x,x^2,x^3,x^4,x^5,x^6,x^7}$$
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:27










  • $begingroup$
    @Reety: Which book are you using?
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:30










  • $begingroup$
    @Shaun I don't know what it's called since It's a pdf file and it doesn't have the author on it and the names generic.
    $endgroup$
    – Reety
    Dec 29 '18 at 12:36










  • $begingroup$
    @Reety: If you found it online, then copy & paste a long sentence from it into a search engine, with speechmarks both sides; it might show up.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:40


















3












$begingroup$

Your matrix $x$ represents a rotation in $mathbb{R}^2$ around the origin by the angle $frac{pi}4$ clockwise.



In general $R(phi) = begin{bmatrix} cosphi & sinphi \ -sinphi & cosphiend{bmatrix}$ does the same thing with the angle $phi$.



Clearly if you rotate by $phi$ and then by $psi$, it is the same as rotating by $phi + psi$. Hence $R(phi)R(psi) = R(phi + psi)$. You can also check this algebraically, of course.



Also $R(phi)^{-1} = R(-phi)$ as the inverse of a rotation by $phi$ is a rotation by $-phi$.



Therefore



$$langle xrangle = {x^n : n in mathbb{Z}} = left{Rleft(frac{pi}4right)^n : n in mathbb{Z}right} = left{Rleft(frac{npi}4right) : n in mathbb{Z}right}$$



Hence your group consists of all rotations by an angle which is a multiple of $fracpi4$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a great explanation thank you, although the book I'm reading says the answer is {e, x, x^2, x^3} how can I convert this notation? - @mechanodroid
    $endgroup$
    – Reety
    Dec 29 '18 at 12:16












  • $begingroup$
    @Reety Are you sure about that? Notice that $x^8 = R(pi/4)^8 = R(2pi) = e$ and the smaller powers are not equal to $e$ so the group should be $$langle xrangle = {e,x,x^2,x^3,x^4,x^5,x^6,x^7}$$
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:27










  • $begingroup$
    @Reety: Which book are you using?
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:30










  • $begingroup$
    @Shaun I don't know what it's called since It's a pdf file and it doesn't have the author on it and the names generic.
    $endgroup$
    – Reety
    Dec 29 '18 at 12:36










  • $begingroup$
    @Reety: If you found it online, then copy & paste a long sentence from it into a search engine, with speechmarks both sides; it might show up.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:40
















3












3








3





$begingroup$

Your matrix $x$ represents a rotation in $mathbb{R}^2$ around the origin by the angle $frac{pi}4$ clockwise.



In general $R(phi) = begin{bmatrix} cosphi & sinphi \ -sinphi & cosphiend{bmatrix}$ does the same thing with the angle $phi$.



Clearly if you rotate by $phi$ and then by $psi$, it is the same as rotating by $phi + psi$. Hence $R(phi)R(psi) = R(phi + psi)$. You can also check this algebraically, of course.



Also $R(phi)^{-1} = R(-phi)$ as the inverse of a rotation by $phi$ is a rotation by $-phi$.



Therefore



$$langle xrangle = {x^n : n in mathbb{Z}} = left{Rleft(frac{pi}4right)^n : n in mathbb{Z}right} = left{Rleft(frac{npi}4right) : n in mathbb{Z}right}$$



Hence your group consists of all rotations by an angle which is a multiple of $fracpi4$.






share|cite|improve this answer









$endgroup$



Your matrix $x$ represents a rotation in $mathbb{R}^2$ around the origin by the angle $frac{pi}4$ clockwise.



In general $R(phi) = begin{bmatrix} cosphi & sinphi \ -sinphi & cosphiend{bmatrix}$ does the same thing with the angle $phi$.



Clearly if you rotate by $phi$ and then by $psi$, it is the same as rotating by $phi + psi$. Hence $R(phi)R(psi) = R(phi + psi)$. You can also check this algebraically, of course.



Also $R(phi)^{-1} = R(-phi)$ as the inverse of a rotation by $phi$ is a rotation by $-phi$.



Therefore



$$langle xrangle = {x^n : n in mathbb{Z}} = left{Rleft(frac{pi}4right)^n : n in mathbb{Z}right} = left{Rleft(frac{npi}4right) : n in mathbb{Z}right}$$



Hence your group consists of all rotations by an angle which is a multiple of $fracpi4$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 12:10









mechanodroidmechanodroid

27.8k62447




27.8k62447












  • $begingroup$
    This is a great explanation thank you, although the book I'm reading says the answer is {e, x, x^2, x^3} how can I convert this notation? - @mechanodroid
    $endgroup$
    – Reety
    Dec 29 '18 at 12:16












  • $begingroup$
    @Reety Are you sure about that? Notice that $x^8 = R(pi/4)^8 = R(2pi) = e$ and the smaller powers are not equal to $e$ so the group should be $$langle xrangle = {e,x,x^2,x^3,x^4,x^5,x^6,x^7}$$
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:27










  • $begingroup$
    @Reety: Which book are you using?
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:30










  • $begingroup$
    @Shaun I don't know what it's called since It's a pdf file and it doesn't have the author on it and the names generic.
    $endgroup$
    – Reety
    Dec 29 '18 at 12:36










  • $begingroup$
    @Reety: If you found it online, then copy & paste a long sentence from it into a search engine, with speechmarks both sides; it might show up.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:40




















  • $begingroup$
    This is a great explanation thank you, although the book I'm reading says the answer is {e, x, x^2, x^3} how can I convert this notation? - @mechanodroid
    $endgroup$
    – Reety
    Dec 29 '18 at 12:16












  • $begingroup$
    @Reety Are you sure about that? Notice that $x^8 = R(pi/4)^8 = R(2pi) = e$ and the smaller powers are not equal to $e$ so the group should be $$langle xrangle = {e,x,x^2,x^3,x^4,x^5,x^6,x^7}$$
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 12:27










  • $begingroup$
    @Reety: Which book are you using?
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:30










  • $begingroup$
    @Shaun I don't know what it's called since It's a pdf file and it doesn't have the author on it and the names generic.
    $endgroup$
    – Reety
    Dec 29 '18 at 12:36










  • $begingroup$
    @Reety: If you found it online, then copy & paste a long sentence from it into a search engine, with speechmarks both sides; it might show up.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:40


















$begingroup$
This is a great explanation thank you, although the book I'm reading says the answer is {e, x, x^2, x^3} how can I convert this notation? - @mechanodroid
$endgroup$
– Reety
Dec 29 '18 at 12:16






$begingroup$
This is a great explanation thank you, although the book I'm reading says the answer is {e, x, x^2, x^3} how can I convert this notation? - @mechanodroid
$endgroup$
– Reety
Dec 29 '18 at 12:16














$begingroup$
@Reety Are you sure about that? Notice that $x^8 = R(pi/4)^8 = R(2pi) = e$ and the smaller powers are not equal to $e$ so the group should be $$langle xrangle = {e,x,x^2,x^3,x^4,x^5,x^6,x^7}$$
$endgroup$
– mechanodroid
Dec 29 '18 at 12:27




$begingroup$
@Reety Are you sure about that? Notice that $x^8 = R(pi/4)^8 = R(2pi) = e$ and the smaller powers are not equal to $e$ so the group should be $$langle xrangle = {e,x,x^2,x^3,x^4,x^5,x^6,x^7}$$
$endgroup$
– mechanodroid
Dec 29 '18 at 12:27












$begingroup$
@Reety: Which book are you using?
$endgroup$
– Shaun
Dec 29 '18 at 12:30




$begingroup$
@Reety: Which book are you using?
$endgroup$
– Shaun
Dec 29 '18 at 12:30












$begingroup$
@Shaun I don't know what it's called since It's a pdf file and it doesn't have the author on it and the names generic.
$endgroup$
– Reety
Dec 29 '18 at 12:36




$begingroup$
@Shaun I don't know what it's called since It's a pdf file and it doesn't have the author on it and the names generic.
$endgroup$
– Reety
Dec 29 '18 at 12:36












$begingroup$
@Reety: If you found it online, then copy & paste a long sentence from it into a search engine, with speechmarks both sides; it might show up.
$endgroup$
– Shaun
Dec 29 '18 at 12:40






$begingroup$
@Reety: If you found it online, then copy & paste a long sentence from it into a search engine, with speechmarks both sides; it might show up.
$endgroup$
– Shaun
Dec 29 '18 at 12:40




















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