Question about rotation $2times 2$ rotation matrices
Clash Royale CLAN TAG#URR8PPP
How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= beginbmatrix cosx& -sinx\ sinx & cosxendbmatrix ,$$
for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.
linear-algebra rotations
edited Dec 16 at 22:51
amWhy
191k28224439
asked Dec 16 at 15:08
Tanny Sieben
30118
add a comment |
How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= beginbmatrix cosx& -sinx\ sinx & cosxendbmatrix ,$$
for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.
linear-algebra rotations
edited Dec 16 at 22:51
amWhy
191k28224439
asked Dec 16 at 15:08
Tanny Sieben
30118
add a comment |
How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= beginbmatrix cosx& -sinx\ sinx & cosxendbmatrix ,$$
for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.
linear-algebra rotations
edited Dec 16 at 22:51
amWhy
191k28224439
asked Dec 16 at 15:08
Tanny Sieben
30118
How do I prove that, if for a $2 times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= beginbmatrix cosx& -sinx\ sinx & cosxendbmatrix ,$$
for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.
linear-algebra rotations
linear-algebra rotations
edited Dec 16 at 22:51
amWhy
191k28224439
asked Dec 16 at 15:08
Tanny Sieben
30118
edited Dec 16 at 22:51
amWhy
191k28224439
asked Dec 16 at 15:08
Tanny Sieben
30118
edited Dec 16 at 22:51
amWhy
191k28224439
edited Dec 16 at 22:51
amWhy
191k28224439
edited Dec 16 at 22:51
amWhy
191k28224439
191k28224439
asked Dec 16 at 15:08
Tanny Sieben
30118
asked Dec 16 at 15:08
Tanny Sieben
30118
asked Dec 16 at 15:08
Tanny Sieben
30118
30118
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Even for the case $A^2=R$ there are many possible roots.
Look for instance at:
https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf
With the additional constraint $det(R)=1$ here you get $$operatornametr(A)A=Rpm I$$
and have to discuss according values of $x$, whether the trace is zero or not and so on.
- e.g. $R=I$ and null trace, gives $beginpmatrixa&b\c&-aendpmatrix$ with $a^2+bc=1$
[ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]
For $A^n=R$ you get even more solutions.
Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...
answered Dec 16 at 16:29
zwim
11.5k728
add a comment |
"Very obvious" or not, it's not true. If $A=beginbmatrix0&1\1&0endbmatrix$ then $A^2=I$ but $A$ is not a rotation.
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
– Tanny Sieben
Dec 16 at 15:31
2
I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_(theta+2pi k)/n$, $k=1,dots,n$.
– David C. Ullrich
Dec 16 at 15:41
I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
– Tanny Sieben
Dec 16 at 15:52
add a comment |
$DeclareMathOperatorTrTr$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:
If $R=beginpmatrix-1\&-1endpmatrix$, then $A=beginpmatrixalpha&beta\gamma&-alphaendpmatrix$ with $alpha^2+betagamma=-1$.
If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfracR+Isqrt2cosphi+2$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.
answered Dec 16 at 21:09
I like Serena
3,6721718
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Even for the case $A^2=R$ there are many possible roots.
Look for instance at:
https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf
With the additional constraint $det(R)=1$ here you get $$operatornametr(A)A=Rpm I$$
and have to discuss according values of $x$, whether the trace is zero or not and so on.
- e.g. $R=I$ and null trace, gives $beginpmatrixa&b\c&-aendpmatrix$ with $a^2+bc=1$
[ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]
For $A^n=R$ you get even more solutions.
Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...
answered Dec 16 at 16:29
zwim
11.5k728
add a comment |
Even for the case $A^2=R$ there are many possible roots.
Look for instance at:
https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf
With the additional constraint $det(R)=1$ here you get $$operatornametr(A)A=Rpm I$$
and have to discuss according values of $x$, whether the trace is zero or not and so on.
- e.g. $R=I$ and null trace, gives $beginpmatrixa&b\c&-aendpmatrix$ with $a^2+bc=1$
[ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]
For $A^n=R$ you get even more solutions.
Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...
answered Dec 16 at 16:29
zwim
11.5k728
add a comment |
Even for the case $A^2=R$ there are many possible roots.
Look for instance at:
https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf
With the additional constraint $det(R)=1$ here you get $$operatornametr(A)A=Rpm I$$
and have to discuss according values of $x$, whether the trace is zero or not and so on.
- e.g. $R=I$ and null trace, gives $beginpmatrixa&b\c&-aendpmatrix$ with $a^2+bc=1$
[ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]
For $A^n=R$ you get even more solutions.
Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...
answered Dec 16 at 16:29
zwim
11.5k728
Even for the case $A^2=R$ there are many possible roots.
Look for instance at:
https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf
With the additional constraint $det(R)=1$ here you get $$operatornametr(A)A=Rpm I$$
and have to discuss according values of $x$, whether the trace is zero or not and so on.
- e.g. $R=I$ and null trace, gives $beginpmatrixa&b\c&-aendpmatrix$ with $a^2+bc=1$
[ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]
For $A^n=R$ you get even more solutions.
Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...
answered Dec 16 at 16:29
zwim
11.5k728
edited Dec 16 at 16:35
answered Dec 16 at 16:29
zwim
11.5k728
answered Dec 16 at 16:29
zwim
11.5k728
answered Dec 16 at 16:29
zwim
11.5k728
11.5k728
add a comment |
add a comment |
"Very obvious" or not, it's not true. If $A=beginbmatrix0&1\1&0endbmatrix$ then $A^2=I$ but $A$ is not a rotation.
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
– Tanny Sieben
Dec 16 at 15:31
2
I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_(theta+2pi k)/n$, $k=1,dots,n$.
– David C. Ullrich
Dec 16 at 15:41
I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
– Tanny Sieben
Dec 16 at 15:52
add a comment |
"Very obvious" or not, it's not true. If $A=beginbmatrix0&1\1&0endbmatrix$ then $A^2=I$ but $A$ is not a rotation.
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
– Tanny Sieben
Dec 16 at 15:31
2
I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_(theta+2pi k)/n$, $k=1,dots,n$.
– David C. Ullrich
Dec 16 at 15:41
I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
– Tanny Sieben
Dec 16 at 15:52
add a comment |
"Very obvious" or not, it's not true. If $A=beginbmatrix0&1\1&0endbmatrix$ then $A^2=I$ but $A$ is not a rotation.
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
"Very obvious" or not, it's not true. If $A=beginbmatrix0&1\1&0endbmatrix$ then $A^2=I$ but $A$ is not a rotation.
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
answered Dec 16 at 15:17
David C. Ullrich
58.1k43891
58.1k43891
I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
– Tanny Sieben
Dec 16 at 15:31
2
I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_(theta+2pi k)/n$, $k=1,dots,n$.
– David C. Ullrich
Dec 16 at 15:41
I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
– Tanny Sieben
Dec 16 at 15:52
add a comment |
I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
– Tanny Sieben
Dec 16 at 15:31
2
I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_(theta+2pi k)/n$, $k=1,dots,n$.
– David C. Ullrich
Dec 16 at 15:41
I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
– Tanny Sieben
Dec 16 at 15:52
I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
– Tanny Sieben
Dec 16 at 15:31
I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix?
– Tanny Sieben
Dec 16 at 15:31
2
2
I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_(theta+2pi k)/n$, $k=1,dots,n$.
– David C. Ullrich
Dec 16 at 15:41
I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_theta$ is the rotation through the angle $theta$; then $B^n=A_theta$ if $B=A_(theta+2pi k)/n$, $k=1,dots,n$.
– David C. Ullrich
Dec 16 at 15:41
I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
– Tanny Sieben
Dec 16 at 15:52
I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix?
– Tanny Sieben
Dec 16 at 15:52
add a comment |
$DeclareMathOperatorTrTr$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:
If $R=beginpmatrix-1\&-1endpmatrix$, then $A=beginpmatrixalpha&beta\gamma&-alphaendpmatrix$ with $alpha^2+betagamma=-1$.
If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfracR+Isqrt2cosphi+2$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.
answered Dec 16 at 21:09
I like Serena
3,6721718
add a comment |
$DeclareMathOperatorTrTr$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:
If $R=beginpmatrix-1\&-1endpmatrix$, then $A=beginpmatrixalpha&beta\gamma&-alphaendpmatrix$ with $alpha^2+betagamma=-1$.
If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfracR+Isqrt2cosphi+2$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.
answered Dec 16 at 21:09
I like Serena
3,6721718
add a comment |
$DeclareMathOperatorTrTr$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:
If $R=beginpmatrix-1\&-1endpmatrix$, then $A=beginpmatrixalpha&beta\gamma&-alphaendpmatrix$ with $alpha^2+betagamma=-1$.
If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfracR+Isqrt2cosphi+2$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.
answered Dec 16 at 21:09
I like Serena
3,6721718
$DeclareMathOperatorTrTr$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:
If $R=beginpmatrix-1\&-1endpmatrix$, then $A=beginpmatrixalpha&beta\gamma&-alphaendpmatrix$ with $alpha^2+betagamma=-1$.
If $R$ is any other rotation over an angle $phi$, then $displaystyle A=pmfracR+Isqrt2cosphi+2$. Working through the possibilities, it means that $A$ is a rotation matrix over either $frac 12phi$ or $frac 12phi + pi$.
answered Dec 16 at 21:09
I like Serena
3,6721718
answered Dec 16 at 21:09
I like Serena
3,6721718
answered Dec 16 at 21:09
I like Serena
3,6721718
answered Dec 16 at 21:09
I like Serena
3,6721718
3,6721718
add a comment |
add a comment |
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