Mixing
If two rotation matrices commute, do their infinitesimal generators commute too?
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Suppose that $e^A$ and $e^B$ are two rotations in $mathrm{SO}(n)$. If $e^{A}e^{B} = e^{B}e^{A}$, can we conclude that $e^{A+B}=e^Ae^B$? More importantly, can we say that $AB=BA$?
I'm particularly interested in the cases when $n=2,3,4$ because I was working on a physics problem that this question was brought up. $n=3$ is the most important case for me.
linear-algebra lie-groups lie-algebras rotations matrix-exponential
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
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show 1 more comment
$begingroup$
Suppose that $e^A$ and $e^B$ are two rotations in $mathrm{SO}(n)$. If $e^{A}e^{B} = e^{B}e^{A}$, can we conclude that $e^{A+B}=e^Ae^B$? More importantly, can we say that $AB=BA$?
I'm particularly interested in the cases when $n=2,3,4$ because I was working on a physics problem that this question was brought up. $n=3$ is the most important case for me.
linear-algebra lie-groups lie-algebras rotations matrix-exponential
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
$endgroup$
$begingroup$
For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
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– Widawensen
Jan 8 at 11:54
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@Widawensen Well, intuitively yes. That's exactly why I asked this question in the first place. But how do you conclude from $e^Ae^B=e^Be^A$ that $AB=BA$?
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– stressed out
Jan 8 at 11:57
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If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
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– Widawensen
Jan 8 at 11:59
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@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
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– stressed out
Jan 8 at 12:01
1
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Sort of obvious, but if $ain so(3)$ and $||a||=pi$, then $e^a=-Id$ which is in the center of $SO(3)$. You probably want to add a hypothesis, like the eigenvalues of $e^a$ and $e^b$ are all of multiplicity $1$.
$endgroup$
– Charlie Frohman
Jan 8 at 12:21
|
show 1 more comment
$begingroup$
Suppose that $e^A$ and $e^B$ are two rotations in $mathrm{SO}(n)$. If $e^{A}e^{B} = e^{B}e^{A}$, can we conclude that $e^{A+B}=e^Ae^B$? More importantly, can we say that $AB=BA$?
I'm particularly interested in the cases when $n=2,3,4$ because I was working on a physics problem that this question was brought up. $n=3$ is the most important case for me.
linear-algebra lie-groups lie-algebras rotations matrix-exponential
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
$endgroup$
Suppose that $e^A$ and $e^B$ are two rotations in $mathrm{SO}(n)$. If $e^{A}e^{B} = e^{B}e^{A}$, can we conclude that $e^{A+B}=e^Ae^B$? More importantly, can we say that $AB=BA$?
I'm particularly interested in the cases when $n=2,3,4$ because I was working on a physics problem that this question was brought up. $n=3$ is the most important case for me.
linear-algebra lie-groups lie-algebras rotations matrix-exponential
linear-algebra lie-groups lie-algebras rotations matrix-exponential
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
asked Jan 8 at 11:47
stressed outstressed out
4,8141634
4,8141634
$begingroup$
For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
$endgroup$
– Widawensen
Jan 8 at 11:54
$begingroup$
@Widawensen Well, intuitively yes. That's exactly why I asked this question in the first place. But how do you conclude from $e^Ae^B=e^Be^A$ that $AB=BA$?
$endgroup$
– stressed out
Jan 8 at 11:57
$begingroup$
If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
$endgroup$
– Widawensen
Jan 8 at 11:59
$begingroup$
@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
$endgroup$
– stressed out
Jan 8 at 12:01
1
$begingroup$
Sort of obvious, but if $ain so(3)$ and $||a||=pi$, then $e^a=-Id$ which is in the center of $SO(3)$. You probably want to add a hypothesis, like the eigenvalues of $e^a$ and $e^b$ are all of multiplicity $1$.
$endgroup$
– Charlie Frohman
Jan 8 at 12:21
|
show 1 more comment
$begingroup$
For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
$endgroup$
– Widawensen
Jan 8 at 11:54
$begingroup$
@Widawensen Well, intuitively yes. That's exactly why I asked this question in the first place. But how do you conclude from $e^Ae^B=e^Be^A$ that $AB=BA$?
$endgroup$
– stressed out
Jan 8 at 11:57
$begingroup$
If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
$endgroup$
– Widawensen
Jan 8 at 11:59
$begingroup$
@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
$endgroup$
– stressed out
Jan 8 at 12:01
1
$begingroup$
Sort of obvious, but if $ain so(3)$ and $||a||=pi$, then $e^a=-Id$ which is in the center of $SO(3)$. You probably want to add a hypothesis, like the eigenvalues of $e^a$ and $e^b$ are all of multiplicity $1$.
$endgroup$
– Charlie Frohman
Jan 8 at 12:21
$begingroup$
For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
$endgroup$
– Widawensen
Jan 8 at 11:54
$begingroup$
For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
$endgroup$
– Widawensen
Jan 8 at 11:54
$begingroup$
@Widawensen Well, intuitively yes. That's exactly why I asked this question in the first place. But how do you conclude from $e^Ae^B=e^Be^A$ that $AB=BA$?
$endgroup$
– stressed out
Jan 8 at 11:57
$begingroup$
@Widawensen Well, intuitively yes. That's exactly why I asked this question in the first place. But how do you conclude from $e^Ae^B=e^Be^A$ that $AB=BA$?
$endgroup$
– stressed out
Jan 8 at 11:57
$begingroup$
If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
$endgroup$
– Widawensen
Jan 8 at 11:59
$begingroup$
If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
$endgroup$
– Widawensen
Jan 8 at 11:59
$begingroup$
@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
$endgroup$
– stressed out
Jan 8 at 12:01
$begingroup$
@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
$endgroup$
– stressed out
Jan 8 at 12:01
1
1
$begingroup$
Sort of obvious, but if $ain so(3)$ and $||a||=pi$, then $e^a=-Id$ which is in the center of $SO(3)$. You probably want to add a hypothesis, like the eigenvalues of $e^a$ and $e^b$ are all of multiplicity $1$.
$endgroup$
– Charlie Frohman
Jan 8 at 12:21
$begingroup$
Sort of obvious, but if $ain so(3)$ and $||a||=pi$, then $e^a=-Id$ which is in the center of $SO(3)$. You probably want to add a hypothesis, like the eigenvalues of $e^a$ and $e^b$ are all of multiplicity $1$.
$endgroup$
– Charlie Frohman
Jan 8 at 12:21
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Similar questions --- without the requirement that $e^A,e^Bin SO(n)$ --- have been asked multiple times on this site before. See If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? for instance.
The answer is still "no" even if you require that $A,B$ are skew-symmetric and $e^A,e^Bin SO(n,mathbb R)$ when $nge3$. Consider
$$
A=pmatrix{0&-2pi&0\ 2pi&0&0\ 0&0&0}, B=pmatrix{0&0&0\ 0&0&-2pi\ 0&2pi&0}.
$$
Then $e^A$ commutes with $e^B$ because both matrix exponentials are equal to $I_3$, but
$$
AB-BA=pmatrix{0&0&4pi^2\ 0&0&0\ -4pi^2&0&0}ne0
$$
and the eigenvalues of $A+B$ are $0$ and $pmsqrt{8}pi i$, so that $e^{A+B}$ is similar to $operatorname{diag}(1,e^{sqrt{8}pi i},e^{-sqrt{8}pi i})$ and it cannot possibly be equal to $e^Ae^B=I_3$.
However, in a "generic" case, the answer to your question is "yes". More specifically, if the spectra of $A$ and $B$ are $2pi i$-congruence free (this assumption does not hold in the counterexample above), then $e^A$ commutes with $e^B$ if and only if $A$ and $B$ commute. Therefore, we also have $e^{A+B}=e^Ae^B$ in this case. See Is $exp:overline{mathbb{M}}_ntomathbb{M}_n$ injective? (and loup blanc's answer in particular) for more details.
When $n=2$ and $A,B$ are skew-symmetric, the answer is certainly yes because all skew-symmetric matrices commute in this case.
answered Jan 8 at 13:00
user1551user1551
72.5k566127
$endgroup$
$begingroup$
Wait... Aren't all elements in $mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $exp(cdot): mathrm{Skew}(n) to mathrm{SO}(n)$ is not surjective in general?
$endgroup$
– stressed out
Jan 8 at 15:31
$begingroup$
@stressedout The mapping is surjective. Is there anything wrong with that?
$endgroup$
– user1551
Jan 8 at 15:49
2
$begingroup$
@stressedout Close but not exactly. I mean $exp(cdot):M_n(mathbb R)to GL_n(mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^Xin SO_n(mathbb R)$, such as $X=pmatrix{0&-2pi\ 2k^2pi&0}$ where $k$ is any integer strictly greater than $1$.
$endgroup$
– user1551
Jan 8 at 16:59
2
$begingroup$
@stressedout We don't write $exp:M_n(mathbb R)to SO_n(mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $exppmatrix{0&1\ 0&0}=pmatrix{1&1\ 0&1}notin SO(2)$). As for learning, sorry, I haven't any suggestions.
$endgroup$
– user1551
Jan 8 at 17:09
1
$begingroup$
OK. Thank you for answering all my questions patiently. I appreciate it lot. :)
$endgroup$
– stressed out
Jan 8 at 17:17
|
show 9 more comments
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If $A in mathrm{SO}(n)$, then $|lambda|=1$ for each eigenvalue $lambda$ of $A$. Hence the set of eigenvalues of $A$ is $2 pi i - $ congruence - free.
Now take a look in http://www.math.kit.edu/iana3/~schmoeger/seite/publikationen/media/normexpproc.pdf
answered Jan 8 at 12:06
FredFred
45.3k1847
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Similar questions --- without the requirement that $e^A,e^Bin SO(n)$ --- have been asked multiple times on this site before. See If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? for instance.
The answer is still "no" even if you require that $A,B$ are skew-symmetric and $e^A,e^Bin SO(n,mathbb R)$ when $nge3$. Consider
$$
A=pmatrix{0&-2pi&0\ 2pi&0&0\ 0&0&0}, B=pmatrix{0&0&0\ 0&0&-2pi\ 0&2pi&0}.
$$
Then $e^A$ commutes with $e^B$ because both matrix exponentials are equal to $I_3$, but
$$
AB-BA=pmatrix{0&0&4pi^2\ 0&0&0\ -4pi^2&0&0}ne0
$$
and the eigenvalues of $A+B$ are $0$ and $pmsqrt{8}pi i$, so that $e^{A+B}$ is similar to $operatorname{diag}(1,e^{sqrt{8}pi i},e^{-sqrt{8}pi i})$ and it cannot possibly be equal to $e^Ae^B=I_3$.
However, in a "generic" case, the answer to your question is "yes". More specifically, if the spectra of $A$ and $B$ are $2pi i$-congruence free (this assumption does not hold in the counterexample above), then $e^A$ commutes with $e^B$ if and only if $A$ and $B$ commute. Therefore, we also have $e^{A+B}=e^Ae^B$ in this case. See Is $exp:overline{mathbb{M}}_ntomathbb{M}_n$ injective? (and loup blanc's answer in particular) for more details.
When $n=2$ and $A,B$ are skew-symmetric, the answer is certainly yes because all skew-symmetric matrices commute in this case.
answered Jan 8 at 13:00
user1551user1551
72.5k566127
$endgroup$
$begingroup$
Wait... Aren't all elements in $mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $exp(cdot): mathrm{Skew}(n) to mathrm{SO}(n)$ is not surjective in general?
$endgroup$
– stressed out
Jan 8 at 15:31
$begingroup$
@stressedout The mapping is surjective. Is there anything wrong with that?
$endgroup$
– user1551
Jan 8 at 15:49
2
$begingroup$
@stressedout Close but not exactly. I mean $exp(cdot):M_n(mathbb R)to GL_n(mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^Xin SO_n(mathbb R)$, such as $X=pmatrix{0&-2pi\ 2k^2pi&0}$ where $k$ is any integer strictly greater than $1$.
$endgroup$
– user1551
Jan 8 at 16:59
2
$begingroup$
@stressedout We don't write $exp:M_n(mathbb R)to SO_n(mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $exppmatrix{0&1\ 0&0}=pmatrix{1&1\ 0&1}notin SO(2)$). As for learning, sorry, I haven't any suggestions.
$endgroup$
– user1551
Jan 8 at 17:09
1
$begingroup$
OK. Thank you for answering all my questions patiently. I appreciate it lot. :)
$endgroup$
– stressed out
Jan 8 at 17:17
|
show 9 more comments
$begingroup$
Similar questions --- without the requirement that $e^A,e^Bin SO(n)$ --- have been asked multiple times on this site before. See If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? for instance.
The answer is still "no" even if you require that $A,B$ are skew-symmetric and $e^A,e^Bin SO(n,mathbb R)$ when $nge3$. Consider
$$
A=pmatrix{0&-2pi&0\ 2pi&0&0\ 0&0&0}, B=pmatrix{0&0&0\ 0&0&-2pi\ 0&2pi&0}.
$$
Then $e^A$ commutes with $e^B$ because both matrix exponentials are equal to $I_3$, but
$$
AB-BA=pmatrix{0&0&4pi^2\ 0&0&0\ -4pi^2&0&0}ne0
$$
and the eigenvalues of $A+B$ are $0$ and $pmsqrt{8}pi i$, so that $e^{A+B}$ is similar to $operatorname{diag}(1,e^{sqrt{8}pi i},e^{-sqrt{8}pi i})$ and it cannot possibly be equal to $e^Ae^B=I_3$.
However, in a "generic" case, the answer to your question is "yes". More specifically, if the spectra of $A$ and $B$ are $2pi i$-congruence free (this assumption does not hold in the counterexample above), then $e^A$ commutes with $e^B$ if and only if $A$ and $B$ commute. Therefore, we also have $e^{A+B}=e^Ae^B$ in this case. See Is $exp:overline{mathbb{M}}_ntomathbb{M}_n$ injective? (and loup blanc's answer in particular) for more details.
When $n=2$ and $A,B$ are skew-symmetric, the answer is certainly yes because all skew-symmetric matrices commute in this case.
answered Jan 8 at 13:00
user1551user1551
72.5k566127
$endgroup$
$begingroup$
Wait... Aren't all elements in $mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $exp(cdot): mathrm{Skew}(n) to mathrm{SO}(n)$ is not surjective in general?
$endgroup$
– stressed out
Jan 8 at 15:31
$begingroup$
@stressedout The mapping is surjective. Is there anything wrong with that?
$endgroup$
– user1551
Jan 8 at 15:49
2
$begingroup$
@stressedout Close but not exactly. I mean $exp(cdot):M_n(mathbb R)to GL_n(mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^Xin SO_n(mathbb R)$, such as $X=pmatrix{0&-2pi\ 2k^2pi&0}$ where $k$ is any integer strictly greater than $1$.
$endgroup$
– user1551
Jan 8 at 16:59
2
$begingroup$
@stressedout We don't write $exp:M_n(mathbb R)to SO_n(mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $exppmatrix{0&1\ 0&0}=pmatrix{1&1\ 0&1}notin SO(2)$). As for learning, sorry, I haven't any suggestions.
$endgroup$
– user1551
Jan 8 at 17:09
1
$begingroup$
OK. Thank you for answering all my questions patiently. I appreciate it lot. :)
$endgroup$
– stressed out
Jan 8 at 17:17
|
show 9 more comments
$begingroup$
Similar questions --- without the requirement that $e^A,e^Bin SO(n)$ --- have been asked multiple times on this site before. See If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? for instance.
The answer is still "no" even if you require that $A,B$ are skew-symmetric and $e^A,e^Bin SO(n,mathbb R)$ when $nge3$. Consider
$$
A=pmatrix{0&-2pi&0\ 2pi&0&0\ 0&0&0}, B=pmatrix{0&0&0\ 0&0&-2pi\ 0&2pi&0}.
$$
Then $e^A$ commutes with $e^B$ because both matrix exponentials are equal to $I_3$, but
$$
AB-BA=pmatrix{0&0&4pi^2\ 0&0&0\ -4pi^2&0&0}ne0
$$
and the eigenvalues of $A+B$ are $0$ and $pmsqrt{8}pi i$, so that $e^{A+B}$ is similar to $operatorname{diag}(1,e^{sqrt{8}pi i},e^{-sqrt{8}pi i})$ and it cannot possibly be equal to $e^Ae^B=I_3$.
However, in a "generic" case, the answer to your question is "yes". More specifically, if the spectra of $A$ and $B$ are $2pi i$-congruence free (this assumption does not hold in the counterexample above), then $e^A$ commutes with $e^B$ if and only if $A$ and $B$ commute. Therefore, we also have $e^{A+B}=e^Ae^B$ in this case. See Is $exp:overline{mathbb{M}}_ntomathbb{M}_n$ injective? (and loup blanc's answer in particular) for more details.
When $n=2$ and $A,B$ are skew-symmetric, the answer is certainly yes because all skew-symmetric matrices commute in this case.
answered Jan 8 at 13:00
user1551user1551
72.5k566127
$endgroup$
Similar questions --- without the requirement that $e^A,e^Bin SO(n)$ --- have been asked multiple times on this site before. See If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? for instance.
The answer is still "no" even if you require that $A,B$ are skew-symmetric and $e^A,e^Bin SO(n,mathbb R)$ when $nge3$. Consider
$$
A=pmatrix{0&-2pi&0\ 2pi&0&0\ 0&0&0}, B=pmatrix{0&0&0\ 0&0&-2pi\ 0&2pi&0}.
$$
Then $e^A$ commutes with $e^B$ because both matrix exponentials are equal to $I_3$, but
$$
AB-BA=pmatrix{0&0&4pi^2\ 0&0&0\ -4pi^2&0&0}ne0
$$
and the eigenvalues of $A+B$ are $0$ and $pmsqrt{8}pi i$, so that $e^{A+B}$ is similar to $operatorname{diag}(1,e^{sqrt{8}pi i},e^{-sqrt{8}pi i})$ and it cannot possibly be equal to $e^Ae^B=I_3$.
However, in a "generic" case, the answer to your question is "yes". More specifically, if the spectra of $A$ and $B$ are $2pi i$-congruence free (this assumption does not hold in the counterexample above), then $e^A$ commutes with $e^B$ if and only if $A$ and $B$ commute. Therefore, we also have $e^{A+B}=e^Ae^B$ in this case. See Is $exp:overline{mathbb{M}}_ntomathbb{M}_n$ injective? (and loup blanc's answer in particular) for more details.
When $n=2$ and $A,B$ are skew-symmetric, the answer is certainly yes because all skew-symmetric matrices commute in this case.
answered Jan 8 at 13:00
user1551user1551
72.5k566127
edited Jan 8 at 14:19
answered Jan 8 at 13:00
user1551user1551
72.5k566127
answered Jan 8 at 13:00
user1551user1551
72.5k566127
answered Jan 8 at 13:00
user1551user1551
72.5k566127
72.5k566127
$begingroup$
Wait... Aren't all elements in $mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $exp(cdot): mathrm{Skew}(n) to mathrm{SO}(n)$ is not surjective in general?
$endgroup$
– stressed out
Jan 8 at 15:31
$begingroup$
@stressedout The mapping is surjective. Is there anything wrong with that?
$endgroup$
– user1551
Jan 8 at 15:49
2
$begingroup$
@stressedout Close but not exactly. I mean $exp(cdot):M_n(mathbb R)to GL_n(mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^Xin SO_n(mathbb R)$, such as $X=pmatrix{0&-2pi\ 2k^2pi&0}$ where $k$ is any integer strictly greater than $1$.
$endgroup$
– user1551
Jan 8 at 16:59
2
$begingroup$
@stressedout We don't write $exp:M_n(mathbb R)to SO_n(mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $exppmatrix{0&1\ 0&0}=pmatrix{1&1\ 0&1}notin SO(2)$). As for learning, sorry, I haven't any suggestions.
$endgroup$
– user1551
Jan 8 at 17:09
1
$begingroup$
OK. Thank you for answering all my questions patiently. I appreciate it lot. :)
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– stressed out
Jan 8 at 17:17
|
show 9 more comments
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Wait... Aren't all elements in $mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $exp(cdot): mathrm{Skew}(n) to mathrm{SO}(n)$ is not surjective in general?
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– stressed out
Jan 8 at 15:31
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@stressedout The mapping is surjective. Is there anything wrong with that?
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– user1551
Jan 8 at 15:49
2
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@stressedout Close but not exactly. I mean $exp(cdot):M_n(mathbb R)to GL_n(mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^Xin SO_n(mathbb R)$, such as $X=pmatrix{0&-2pi\ 2k^2pi&0}$ where $k$ is any integer strictly greater than $1$.
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– user1551
Jan 8 at 16:59
2
$begingroup$
@stressedout We don't write $exp:M_n(mathbb R)to SO_n(mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $exppmatrix{0&1\ 0&0}=pmatrix{1&1\ 0&1}notin SO(2)$). As for learning, sorry, I haven't any suggestions.
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– user1551
Jan 8 at 17:09
1
$begingroup$
OK. Thank you for answering all my questions patiently. I appreciate it lot. :)
$endgroup$
– stressed out
Jan 8 at 17:17
$begingroup$
Wait... Aren't all elements in $mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $exp(cdot): mathrm{Skew}(n) to mathrm{SO}(n)$ is not surjective in general?
$endgroup$
– stressed out
Jan 8 at 15:31
$begingroup$
Wait... Aren't all elements in $mathrm{SO}(n)$, the exponential of some skew-symmetric matrix? :( at least for $n=2,3$? :( So, $exp(cdot): mathrm{Skew}(n) to mathrm{SO}(n)$ is not surjective in general?
$endgroup$
– stressed out
Jan 8 at 15:31
$begingroup$
@stressedout The mapping is surjective. Is there anything wrong with that?
$endgroup$
– user1551
Jan 8 at 15:49
$begingroup$
@stressedout The mapping is surjective. Is there anything wrong with that?
$endgroup$
– user1551
Jan 8 at 15:49
2
2
$begingroup$
@stressedout Close but not exactly. I mean $exp(cdot):M_n(mathbb R)to GL_n(mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^Xin SO_n(mathbb R)$, such as $X=pmatrix{0&-2pi\ 2k^2pi&0}$ where $k$ is any integer strictly greater than $1$.
$endgroup$
– user1551
Jan 8 at 16:59
$begingroup$
@stressedout Close but not exactly. I mean $exp(cdot):M_n(mathbb R)to GL_n(mathbb R)$ is not injective. In particular, there exists some $X$ that is not skew-symmetric such that $e^Xin SO_n(mathbb R)$, such as $X=pmatrix{0&-2pi\ 2k^2pi&0}$ where $k$ is any integer strictly greater than $1$.
$endgroup$
– user1551
Jan 8 at 16:59
2
2
$begingroup$
@stressedout We don't write $exp:M_n(mathbb R)to SO_n(mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $exppmatrix{0&1\ 0&0}=pmatrix{1&1\ 0&1}notin SO(2)$). As for learning, sorry, I haven't any suggestions.
$endgroup$
– user1551
Jan 8 at 17:09
$begingroup$
@stressedout We don't write $exp:M_n(mathbb R)to SO_n(mathbb R)$ because the exponential of a general real square matrix is not necessarily a member of the special orthogonal group (consider e.g. $exppmatrix{0&1\ 0&0}=pmatrix{1&1\ 0&1}notin SO(2)$). As for learning, sorry, I haven't any suggestions.
$endgroup$
– user1551
Jan 8 at 17:09
1
1
$begingroup$
OK. Thank you for answering all my questions patiently. I appreciate it lot. :)
$endgroup$
– stressed out
Jan 8 at 17:17
$begingroup$
OK. Thank you for answering all my questions patiently. I appreciate it lot. :)
$endgroup$
– stressed out
Jan 8 at 17:17
|
show 9 more comments
$begingroup$
If $A in mathrm{SO}(n)$, then $|lambda|=1$ for each eigenvalue $lambda$ of $A$. Hence the set of eigenvalues of $A$ is $2 pi i - $ congruence - free.
Now take a look in http://www.math.kit.edu/iana3/~schmoeger/seite/publikationen/media/normexpproc.pdf
answered Jan 8 at 12:06
FredFred
45.3k1847
$endgroup$
add a comment |
$begingroup$
If $A in mathrm{SO}(n)$, then $|lambda|=1$ for each eigenvalue $lambda$ of $A$. Hence the set of eigenvalues of $A$ is $2 pi i - $ congruence - free.
Now take a look in http://www.math.kit.edu/iana3/~schmoeger/seite/publikationen/media/normexpproc.pdf
answered Jan 8 at 12:06
FredFred
45.3k1847
$endgroup$
add a comment |
$begingroup$
If $A in mathrm{SO}(n)$, then $|lambda|=1$ for each eigenvalue $lambda$ of $A$. Hence the set of eigenvalues of $A$ is $2 pi i - $ congruence - free.
Now take a look in http://www.math.kit.edu/iana3/~schmoeger/seite/publikationen/media/normexpproc.pdf
answered Jan 8 at 12:06
FredFred
45.3k1847
$endgroup$
If $A in mathrm{SO}(n)$, then $|lambda|=1$ for each eigenvalue $lambda$ of $A$. Hence the set of eigenvalues of $A$ is $2 pi i - $ congruence - free.
Now take a look in http://www.math.kit.edu/iana3/~schmoeger/seite/publikationen/media/normexpproc.pdf
answered Jan 8 at 12:06
FredFred
45.3k1847
answered Jan 8 at 12:06
FredFred
45.3k1847
answered Jan 8 at 12:06
FredFred
45.3k1847
answered Jan 8 at 12:06
FredFred
45.3k1847
45.3k1847
add a comment |
add a comment |
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$begingroup$
For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
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– Widawensen
Jan 8 at 11:54
$begingroup$
@Widawensen Well, intuitively yes. That's exactly why I asked this question in the first place. But how do you conclude from $e^Ae^B=e^Be^A$ that $AB=BA$?
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– stressed out
Jan 8 at 11:57
$begingroup$
If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
$endgroup$
– Widawensen
Jan 8 at 11:59
$begingroup$
@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
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– stressed out
Jan 8 at 12:01
1
$begingroup$
Sort of obvious, but if $ain so(3)$ and $||a||=pi$, then $e^a=-Id$ which is in the center of $SO(3)$. You probably want to add a hypothesis, like the eigenvalues of $e^a$ and $e^b$ are all of multiplicity $1$.
$endgroup$
– Charlie Frohman
Jan 8 at 12:21