For bounded linear operators $S$ and $T$ such that $ST-TS=I$, show that $|T^n|=0$ for some $n in mathbb{N}$












1












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I found in the first part of the question that $ST^{n+1}-T^{n+1}S=(n+1)T^n$ for any non negative integer $n$, any clues on how to proceed?



Thanks.






operator-theory linear-transformations







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    1












    $begingroup$


    I found in the first part of the question that $ST^{n+1}-T^{n+1}S=(n+1)T^n$ for any non negative integer $n$, any clues on how to proceed?



    Thanks.






    operator-theory linear-transformations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I found in the first part of the question that $ST^{n+1}-T^{n+1}S=(n+1)T^n$ for any non negative integer $n$, any clues on how to proceed?



      Thanks.






      operator-theory linear-transformations







      share|cite|improve this question









      $endgroup$




      I found in the first part of the question that $ST^{n+1}-T^{n+1}S=(n+1)T^n$ for any non negative integer $n$, any clues on how to proceed?



      Thanks.






      operator-theory linear-transformations




      operator-theory linear-transformations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 31 at 12:07









      MarmosetMarmoset

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          $begingroup$

          From $ST^{n+1}-T^{n+1}S=(n+1)T^n$ we get



          $(n+1)||T^n|| le 2||S||cdot||T^{n+1}||le 2||S||cdot ||T|| cdot||T^n||.$



          Now suppose that $||T^n|| ne 0$ for all $n$. Then we get



          $n+1 le 2||S||cdot ||T||$ for all $n$, which is absurd.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe for completeness you should include the fact that the relation is impossible for bounded operators. If $T^n=0$, then $T^{n-1}=tfrac1n(ST^n-T^nS)=0$; iterate this until $n=2$, to get $T=0$.
            $endgroup$
            – Martin Argerami
            Jan 31 at 17:27












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          $begingroup$

          From $ST^{n+1}-T^{n+1}S=(n+1)T^n$ we get



          $(n+1)||T^n|| le 2||S||cdot||T^{n+1}||le 2||S||cdot ||T|| cdot||T^n||.$



          Now suppose that $||T^n|| ne 0$ for all $n$. Then we get



          $n+1 le 2||S||cdot ||T||$ for all $n$, which is absurd.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe for completeness you should include the fact that the relation is impossible for bounded operators. If $T^n=0$, then $T^{n-1}=tfrac1n(ST^n-T^nS)=0$; iterate this until $n=2$, to get $T=0$.
            $endgroup$
            – Martin Argerami
            Jan 31 at 17:27
















          4












          $begingroup$

          From $ST^{n+1}-T^{n+1}S=(n+1)T^n$ we get



          $(n+1)||T^n|| le 2||S||cdot||T^{n+1}||le 2||S||cdot ||T|| cdot||T^n||.$



          Now suppose that $||T^n|| ne 0$ for all $n$. Then we get



          $n+1 le 2||S||cdot ||T||$ for all $n$, which is absurd.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe for completeness you should include the fact that the relation is impossible for bounded operators. If $T^n=0$, then $T^{n-1}=tfrac1n(ST^n-T^nS)=0$; iterate this until $n=2$, to get $T=0$.
            $endgroup$
            – Martin Argerami
            Jan 31 at 17:27














          4












          4








          4





          $begingroup$

          From $ST^{n+1}-T^{n+1}S=(n+1)T^n$ we get



          $(n+1)||T^n|| le 2||S||cdot||T^{n+1}||le 2||S||cdot ||T|| cdot||T^n||.$



          Now suppose that $||T^n|| ne 0$ for all $n$. Then we get



          $n+1 le 2||S||cdot ||T||$ for all $n$, which is absurd.






          share|cite|improve this answer









          $endgroup$



          From $ST^{n+1}-T^{n+1}S=(n+1)T^n$ we get



          $(n+1)||T^n|| le 2||S||cdot||T^{n+1}||le 2||S||cdot ||T|| cdot||T^n||.$



          Now suppose that $||T^n|| ne 0$ for all $n$. Then we get



          $n+1 le 2||S||cdot ||T||$ for all $n$, which is absurd.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 31 at 12:18









          FredFred

          48.6k11849




          48.6k11849












          • $begingroup$
            Maybe for completeness you should include the fact that the relation is impossible for bounded operators. If $T^n=0$, then $T^{n-1}=tfrac1n(ST^n-T^nS)=0$; iterate this until $n=2$, to get $T=0$.
            $endgroup$
            – Martin Argerami
            Jan 31 at 17:27


















          • $begingroup$
            Maybe for completeness you should include the fact that the relation is impossible for bounded operators. If $T^n=0$, then $T^{n-1}=tfrac1n(ST^n-T^nS)=0$; iterate this until $n=2$, to get $T=0$.
            $endgroup$
            – Martin Argerami
            Jan 31 at 17:27
















          $begingroup$
          Maybe for completeness you should include the fact that the relation is impossible for bounded operators. If $T^n=0$, then $T^{n-1}=tfrac1n(ST^n-T^nS)=0$; iterate this until $n=2$, to get $T=0$.
          $endgroup$
          – Martin Argerami
          Jan 31 at 17:27




          $begingroup$
          Maybe for completeness you should include the fact that the relation is impossible for bounded operators. If $T^n=0$, then $T^{n-1}=tfrac1n(ST^n-T^nS)=0$; iterate this until $n=2$, to get $T=0$.
          $endgroup$
          – Martin Argerami
          Jan 31 at 17:27


















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