If two rotation matrices commute, do their infinitesimal generators commute too?












6














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







share|cite|improve this question






















  • For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
    – Widawensen
    2 days ago












  • @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 $AB=BA$?
    – stressed out
    2 days ago










  • If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
    – Widawensen
    2 days ago










  • @Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
    – stressed out
    2 days ago






  • 1




    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 $1$.
    – Charlie Frohman
    2 days ago


















6














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







share|cite|improve this question






















  • For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
    – Widawensen
    2 days ago












  • @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 $AB=BA$?
    – stressed out
    2 days ago










  • If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
    – Widawensen
    2 days ago










  • @Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
    – stressed out
    2 days ago






  • 1




    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 $1$.
    – Charlie Frohman
    2 days ago
















6












6








6







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









stressed outstressed out

4,0811533




4,0811533












  • For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
    – Widawensen
    2 days ago












  • @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 $AB=BA$?
    – stressed out
    2 days ago










  • If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
    – Widawensen
    2 days ago










  • @Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
    – stressed out
    2 days ago






  • 1




    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 $1$.
    – Charlie Frohman
    2 days ago




















  • For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
    – Widawensen
    2 days ago












  • @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 $AB=BA$?
    – stressed out
    2 days ago










  • If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
    – Widawensen
    2 days ago










  • @Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
    – stressed out
    2 days ago






  • 1




    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 $1$.
    – Charlie Frohman
    2 days ago


















For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
– Widawensen
2 days ago






For $n=3$ matrices certainly commute because they are about the same axis but by different angles, $R=e^{theta S(v)}$ so $A=theta S(v)$
– Widawensen
2 days ago














@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 $AB=BA$?
– stressed out
2 days ago




@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 $AB=BA$?
– stressed out
2 days ago












If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
– Widawensen
2 days ago




If general form of both matrices is $A=theta_1 S(v)$ and $B=theta_2 S(v)$ they must commute.
– Widawensen
2 days ago












@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
– stressed out
2 days ago




@Widawensen No, I mean how do you conclude from $e^Ae^B=e^Be^A$ that they have the same axis?
– stressed out
2 days ago




1




1




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 $1$.
– Charlie Frohman
2 days ago






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 $1$.
– Charlie Frohman
2 days ago












2 Answers
2






active

oldest

votes


















10














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.






share|cite|improve this answer























  • 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?
    – stressed out
    2 days ago












  • @stressedout The mapping is surjective. Is there anything wrong with that?
    – user1551
    2 days ago










  • Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric?
    – stressed out
    2 days ago






  • 2




    @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 $1$.
    – user1551
    2 days ago








  • 2




    @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.
    – user1551
    2 days ago



















3














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






share|cite|improve this answer





















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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    10














    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.






    share|cite|improve this answer























    • 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?
      – stressed out
      2 days ago












    • @stressedout The mapping is surjective. Is there anything wrong with that?
      – user1551
      2 days ago










    • Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric?
      – stressed out
      2 days ago






    • 2




      @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 $1$.
      – user1551
      2 days ago








    • 2




      @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.
      – user1551
      2 days ago
















    10














    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.






    share|cite|improve this answer























    • 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?
      – stressed out
      2 days ago












    • @stressedout The mapping is surjective. Is there anything wrong with that?
      – user1551
      2 days ago










    • Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric?
      – stressed out
      2 days ago






    • 2




      @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 $1$.
      – user1551
      2 days ago








    • 2




      @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.
      – user1551
      2 days ago














    10












    10








    10






    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    user1551user1551

    71.9k566126




    71.9k566126












    • 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?
      – stressed out
      2 days ago












    • @stressedout The mapping is surjective. Is there anything wrong with that?
      – user1551
      2 days ago










    • Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric?
      – stressed out
      2 days ago






    • 2




      @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 $1$.
      – user1551
      2 days ago








    • 2




      @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.
      – user1551
      2 days ago


















    • 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?
      – stressed out
      2 days ago












    • @stressedout The mapping is surjective. Is there anything wrong with that?
      – user1551
      2 days ago










    • Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric?
      – stressed out
      2 days ago






    • 2




      @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 $1$.
      – user1551
      2 days ago








    • 2




      @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.
      – user1551
      2 days ago
















    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?
    – stressed out
    2 days ago






    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?
    – stressed out
    2 days ago














    @stressedout The mapping is surjective. Is there anything wrong with that?
    – user1551
    2 days ago




    @stressedout The mapping is surjective. Is there anything wrong with that?
    – user1551
    2 days ago












    Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric?
    – stressed out
    2 days ago




    Then if it's surjective, how can $e^A$ be a rotation but $A$ not skew-symmetric?
    – stressed out
    2 days ago




    2




    2




    @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 $1$.
    – user1551
    2 days ago






    @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 $1$.
    – user1551
    2 days ago






    2




    2




    @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.
    – user1551
    2 days ago




    @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.
    – user1551
    2 days ago











    3














    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






    share|cite|improve this answer


























      3














      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






      share|cite|improve this answer
























        3












        3








        3






        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






        share|cite|improve this answer












        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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        FredFred

        44.4k1845




        44.4k1845






























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