Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, if [A,[A,B]]=0$?












3














Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?



I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:



$$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
$$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$



But now I do not see how to use $[A,[A,B]] = 0$.



Does anyone has hints for this or a hint how to advance?






linear-algebra matrices







share|cite|improve this question





























    3














    Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?



    I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:



    $$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
    $$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$



    But now I do not see how to use $[A,[A,B]] = 0$.



    Does anyone has hints for this or a hint how to advance?






    linear-algebra matrices







    share|cite|improve this question



























      3












      3








      3







      Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?



      I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:



      $$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
      $$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$



      But now I do not see how to use $[A,[A,B]] = 0$.



      Does anyone has hints for this or a hint how to advance?






      linear-algebra matrices







      share|cite|improve this question















      Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?



      I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:



      $$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
      $$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$



      But now I do not see how to use $[A,[A,B]] = 0$.



      Does anyone has hints for this or a hint how to advance?






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 '18 at 19:29









      Bernard

      118k639112




      118k639112










      asked Nov 28 '18 at 18:22









      MPB94MPB94

      27016




      27016






















          1 Answer
          1






          active

          oldest

          votes


















          3














          From
          $$[A,[A,B]]=0$$
          we have that
          $$A[A,B]=[A,B]A$$
          And
          $$A[A,B]^n=[A,B]^nA$$
          So using your work:
          $$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
          which is what you wanted to prove.






          share|cite|improve this answer





















          • Thanks, I looked for such an identity and somehow did not see it. :)
            – MPB94
            Nov 28 '18 at 19:37










          • @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
            – Botond
            Nov 28 '18 at 19:43










          • $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
            – MPB94
            Nov 28 '18 at 19:47











          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%2f3017507%2fwhy-is-operatornametra-bm-operatornametrab-a-bm-1-if%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          From
          $$[A,[A,B]]=0$$
          we have that
          $$A[A,B]=[A,B]A$$
          And
          $$A[A,B]^n=[A,B]^nA$$
          So using your work:
          $$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
          which is what you wanted to prove.






          share|cite|improve this answer





















          • Thanks, I looked for such an identity and somehow did not see it. :)
            – MPB94
            Nov 28 '18 at 19:37










          • @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
            – Botond
            Nov 28 '18 at 19:43










          • $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
            – MPB94
            Nov 28 '18 at 19:47
















          3














          From
          $$[A,[A,B]]=0$$
          we have that
          $$A[A,B]=[A,B]A$$
          And
          $$A[A,B]^n=[A,B]^nA$$
          So using your work:
          $$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
          which is what you wanted to prove.






          share|cite|improve this answer





















          • Thanks, I looked for such an identity and somehow did not see it. :)
            – MPB94
            Nov 28 '18 at 19:37










          • @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
            – Botond
            Nov 28 '18 at 19:43










          • $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
            – MPB94
            Nov 28 '18 at 19:47














          3












          3








          3






          From
          $$[A,[A,B]]=0$$
          we have that
          $$A[A,B]=[A,B]A$$
          And
          $$A[A,B]^n=[A,B]^nA$$
          So using your work:
          $$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
          which is what you wanted to prove.






          share|cite|improve this answer












          From
          $$[A,[A,B]]=0$$
          we have that
          $$A[A,B]=[A,B]A$$
          And
          $$A[A,B]^n=[A,B]^nA$$
          So using your work:
          $$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
          which is what you wanted to prove.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 19:12









          BotondBotond

          5,5882732




          5,5882732












          • Thanks, I looked for such an identity and somehow did not see it. :)
            – MPB94
            Nov 28 '18 at 19:37










          • @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
            – Botond
            Nov 28 '18 at 19:43










          • $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
            – MPB94
            Nov 28 '18 at 19:47


















          • Thanks, I looked for such an identity and somehow did not see it. :)
            – MPB94
            Nov 28 '18 at 19:37










          • @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
            – Botond
            Nov 28 '18 at 19:43










          • $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
            – MPB94
            Nov 28 '18 at 19:47
















          Thanks, I looked for such an identity and somehow did not see it. :)
          – MPB94
          Nov 28 '18 at 19:37




          Thanks, I looked for such an identity and somehow did not see it. :)
          – MPB94
          Nov 28 '18 at 19:37












          @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
          – Botond
          Nov 28 '18 at 19:43




          @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
          – Botond
          Nov 28 '18 at 19:43












          $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
          – MPB94
          Nov 28 '18 at 19:47




          $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
          – MPB94
          Nov 28 '18 at 19:47


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3017507%2fwhy-is-operatornametra-bm-operatornametrab-a-bm-1-if%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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more