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.










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.










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.










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






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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066076%2fif-two-rotation-matrices-commute-do-their-infinitesimal-generators-commute-too%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066076%2fif-two-rotation-matrices-commute-do-their-infinitesimal-generators-commute-too%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            1300-talet

            1300-talet

            Display a custom attribute below product name in the front-end Magento 1.9.3.8