On the commutativity of matrices and their exponentials

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












It is fairly easy to see that if $A$ and $B $, real square matrices, commute, then $A $ and $e^B $ commute. In fact,



$$Ae^B = sum_n=0^infty AfracB^nn! = sum_n=0^infty fracB^nn!A = e^BA $$



But is the reverse true as well? If $A $ and $e^B $ commute, do $A $ and $B $ commute as well? If not, can you help me finding a counter-example?



What if both $A $ and $B $ are not constant matrices but matricial functions of $t $?






linear-algebra matrix-exponential







share|cite|improve this question

























    up vote
    2
    down vote

    favorite












    It is fairly easy to see that if $A$ and $B $, real square matrices, commute, then $A $ and $e^B $ commute. In fact,



    $$Ae^B = sum_n=0^infty AfracB^nn! = sum_n=0^infty fracB^nn!A = e^BA $$



    But is the reverse true as well? If $A $ and $e^B $ commute, do $A $ and $B $ commute as well? If not, can you help me finding a counter-example?



    What if both $A $ and $B $ are not constant matrices but matricial functions of $t $?






    linear-algebra matrix-exponential







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      It is fairly easy to see that if $A$ and $B $, real square matrices, commute, then $A $ and $e^B $ commute. In fact,



      $$Ae^B = sum_n=0^infty AfracB^nn! = sum_n=0^infty fracB^nn!A = e^BA $$



      But is the reverse true as well? If $A $ and $e^B $ commute, do $A $ and $B $ commute as well? If not, can you help me finding a counter-example?



      What if both $A $ and $B $ are not constant matrices but matricial functions of $t $?






      linear-algebra matrix-exponential







      share|cite|improve this question













      It is fairly easy to see that if $A$ and $B $, real square matrices, commute, then $A $ and $e^B $ commute. In fact,



      $$Ae^B = sum_n=0^infty AfracB^nn! = sum_n=0^infty fracB^nn!A = e^BA $$



      But is the reverse true as well? If $A $ and $e^B $ commute, do $A $ and $B $ commute as well? If not, can you help me finding a counter-example?



      What if both $A $ and $B $ are not constant matrices but matricial functions of $t $?






      linear-algebra matrix-exponential




      linear-algebra matrix-exponential






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      RGS

      8,40711129




      8,40711129




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          5
          down vote













          You can cook up nontrivial real matrices $B$ with $exp(B)=I$, for instance
          $$pmatrix0&2pi\-2pi&0.$$
          There are matrices $A$ that don't commute with $B$, say
          $$A=pmatrix1&0\0&0.$$
          But $A$ commutes with $exp(B)=I$.






          share|cite|improve this answer




















          • It might be worth adding a proof that choice for $B$ satisfies $exp B=I$.
            – J.G.
            51 mins ago










          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',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f2973071%2fon-the-commutativity-of-matrices-and-their-exponentials%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          5
          down vote













          You can cook up nontrivial real matrices $B$ with $exp(B)=I$, for instance
          $$pmatrix0&2pi\-2pi&0.$$
          There are matrices $A$ that don't commute with $B$, say
          $$A=pmatrix1&0\0&0.$$
          But $A$ commutes with $exp(B)=I$.






          share|cite|improve this answer




















          • It might be worth adding a proof that choice for $B$ satisfies $exp B=I$.
            – J.G.
            51 mins ago














          up vote
          5
          down vote













          You can cook up nontrivial real matrices $B$ with $exp(B)=I$, for instance
          $$pmatrix0&2pi\-2pi&0.$$
          There are matrices $A$ that don't commute with $B$, say
          $$A=pmatrix1&0\0&0.$$
          But $A$ commutes with $exp(B)=I$.






          share|cite|improve this answer




















          • It might be worth adding a proof that choice for $B$ satisfies $exp B=I$.
            – J.G.
            51 mins ago












          up vote
          5
          down vote










          up vote
          5
          down vote









          You can cook up nontrivial real matrices $B$ with $exp(B)=I$, for instance
          $$pmatrix0&2pi\-2pi&0.$$
          There are matrices $A$ that don't commute with $B$, say
          $$A=pmatrix1&0\0&0.$$
          But $A$ commutes with $exp(B)=I$.






          share|cite|improve this answer












          You can cook up nontrivial real matrices $B$ with $exp(B)=I$, for instance
          $$pmatrix0&2pi\-2pi&0.$$
          There are matrices $A$ that don't commute with $B$, say
          $$A=pmatrix1&0\0&0.$$
          But $A$ commutes with $exp(B)=I$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Lord Shark the Unknown

          94.1k956122




          94.1k956122











          • It might be worth adding a proof that choice for $B$ satisfies $exp B=I$.
            – J.G.
            51 mins ago
















          • It might be worth adding a proof that choice for $B$ satisfies $exp B=I$.
            – J.G.
            51 mins ago















          It might be worth adding a proof that choice for $B$ satisfies $exp B=I$.
          – J.G.
          51 mins ago




          It might be worth adding a proof that choice for $B$ satisfies $exp B=I$.
          – J.G.
          51 mins ago

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2973071%2fon-the-commutativity-of-matrices-and-their-exponentials%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          White Anglo-Saxon Protestant

          Image
          White Anglo-Saxon Protestants ( WASPs ) are a social group of wealthy and well-connected white Americans, of Protestant and predominantly British ancestry, who trace their ancestry to the American colonial period. Until at least the 1960s, this group has dominated American society and culture and dominated in the leadership of the Whig and Republican parties. They usually are very well placed in major financial, business, legal and academic institutions and had close to a monopoly of elite society due to intermarriage and nepotism. [1] During the latter half of the twentieth century, outsider ethnic and racial groups grew in influence and WASP dominance weakened. Americans are increasingly criticizing the WASP hegemony and disparaging WASPs as the epitome of "the Establishment". The Random House Unabridged Dictionary (1998) says the term is "Sometimes Disparaging and Offensive". [2] [3] The term WASP is often used as a pejorative to classify their historical dom...
          Read more

          BuddyTV

          BuddyTV Owner David Niu, Andy Liu Created by Various contributors Website www.buddytv.com Launched July 31, 2007 Current status Offline BuddyTV was an entertainment-based website based in Seattle, Washington, which generated content about television programs and sporting events. The website published information about celebrity and related entertainment news through a series of articles, entertainment profiles, actor biographies and user forums. [1] On 31 December 2014, Smart TV manufacturer VIZIO acquired BuddyTV's parent Advanced Media Research Group, Inc., in order to expand content and service offerings. [2] The site was shut down on 22 May 2018. [3] Contents 1 History 2 Business model 3 Impact 4 See also 5 References 6 External links History BuddyTV was co-founded in 2005 by Andy Liu and David Niu. The pair graduated from the University of Pennsylvania with a Master of Business Administration. They sold their previous company, NetConversions, to aQuantive in 2004...
          Read more

          Conflict (narrative)

          Image
          Conflict in narrative comes in many forms. "Man versus man", such as is depicted here in the battle between King Arthur and Mordred, is particularly common in traditional literature, fairy tales and myths. [1] In works of narrative, conflict is the challenge main characters need to solve to achieve their goals. Traditionally, conflict is a major literary element that creates challenges in a story by adding uncertainty to if the goal would be achieved. A narrative is not limited to a single conflict. While conflicts may not always resolve in narrative, the resolution of a conflict creates closure or fulfillment, which may or may not occur at a story's end. Contents 1 Basic nature 2 Classification 2.1 Man against man 2.2 Man against nature 2.3 Man against self 2.4 Man against society 3 History 4 See also 5 References 6 External links Basic nature Conflict in literature refers to the different drives of the characters or forces involved. Conflict may be internal o...
          Read more