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












3












$begingroup$


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







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    3












    $begingroup$


    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











    $endgroup$















      3












      3








      3





      $begingroup$


      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











      $endgroup$




      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

      119k639112




      119k639112










      asked Nov 28 '18 at 18:22









      MPB94MPB94

      27017




      27017






















          1 Answer
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          $begingroup$

          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









          $endgroup$













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










          • $begingroup$
            @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 :)
            $endgroup$
            – Botond
            Nov 28 '18 at 19:43










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











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          active

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          3












          $begingroup$

          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









          $endgroup$













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










          • $begingroup$
            @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 :)
            $endgroup$
            – Botond
            Nov 28 '18 at 19:43










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
















          3












          $begingroup$

          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









          $endgroup$













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










          • $begingroup$
            @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 :)
            $endgroup$
            – Botond
            Nov 28 '18 at 19:43










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














          3












          3








          3





          $begingroup$

          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









          $endgroup$



          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

          5,6032732




          5,6032732












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










          • $begingroup$
            @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 :)
            $endgroup$
            – Botond
            Nov 28 '18 at 19:43










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


















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










          • $begingroup$
            @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 :)
            $endgroup$
            – Botond
            Nov 28 '18 at 19:43










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
















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




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












          $begingroup$
          @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 :)
          $endgroup$
          – Botond
          Nov 28 '18 at 19:43




          $begingroup$
          @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 :)
          $endgroup$
          – Botond
          Nov 28 '18 at 19:43












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




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


















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