Normal Operator || T^2|| = ||T||^2












2












$begingroup$


Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.

How can we show ||T$^2$|| = ||T||$^2$?



By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?






linear-algebra







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.

    How can we show ||T$^2$|| = ||T||$^2$?



    By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?






    linear-algebra







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.

      How can we show ||T$^2$|| = ||T||$^2$?



      By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?






      linear-algebra







      share|cite|improve this question









      $endgroup$




      Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.

      How can we show ||T$^2$|| = ||T||$^2$?



      By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      EricEric

      798




      798






















          1 Answer
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          $begingroup$

          If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
          =left<x,TT^*xright>=|T^*x|^2$          , so $|Tx|=|T^*x|$ (and therefore
          $|T|=|T^*|$).
          Then (replacing $x$ by $Tx$)
          $|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
          $|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
          =|T^2|$          . But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
          whenever $T$ is normal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
            $endgroup$
            – Eric
            7 mins ago










          • $begingroup$
            (ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
            $endgroup$
            – Eric
            6 mins ago












          • $begingroup$
            By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
            $endgroup$
            – Lord Shark the Unknown
            6 mins ago










          • $begingroup$
            @eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
            $endgroup$
            – Lord Shark the Unknown
            3 mins ago












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          active

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          $begingroup$

          If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
          =left<x,TT^*xright>=|T^*x|^2$          , so $|Tx|=|T^*x|$ (and therefore
          $|T|=|T^*|$).
          Then (replacing $x$ by $Tx$)
          $|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
          $|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
          =|T^2|$          . But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
          whenever $T$ is normal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
            $endgroup$
            – Eric
            7 mins ago










          • $begingroup$
            (ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
            $endgroup$
            – Eric
            6 mins ago












          • $begingroup$
            By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
            $endgroup$
            – Lord Shark the Unknown
            6 mins ago










          • $begingroup$
            @eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
            $endgroup$
            – Lord Shark the Unknown
            3 mins ago
















          5












          $begingroup$

          If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
          =left<x,TT^*xright>=|T^*x|^2$          , so $|Tx|=|T^*x|$ (and therefore
          $|T|=|T^*|$).
          Then (replacing $x$ by $Tx$)
          $|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
          $|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
          =|T^2|$          . But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
          whenever $T$ is normal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
            $endgroup$
            – Eric
            7 mins ago










          • $begingroup$
            (ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
            $endgroup$
            – Eric
            6 mins ago












          • $begingroup$
            By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
            $endgroup$
            – Lord Shark the Unknown
            6 mins ago










          • $begingroup$
            @eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
            $endgroup$
            – Lord Shark the Unknown
            3 mins ago














          5












          5








          5





          $begingroup$

          If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
          =left<x,TT^*xright>=|T^*x|^2$          , so $|Tx|=|T^*x|$ (and therefore
          $|T|=|T^*|$).
          Then (replacing $x$ by $Tx$)
          $|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
          $|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
          =|T^2|$          . But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
          whenever $T$ is normal.






          share|cite|improve this answer









          $endgroup$



          If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
          =left<x,TT^*xright>=|T^*x|^2$          , so $|Tx|=|T^*x|$ (and therefore
          $|T|=|T^*|$).
          Then (replacing $x$ by $Tx$)
          $|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
          $|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
          =|T^2|$          . But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
          whenever $T$ is normal.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 23 mins ago









          Lord Shark the UnknownLord Shark the Unknown

          109k1163136




          109k1163136












          • $begingroup$
            Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
            $endgroup$
            – Eric
            7 mins ago










          • $begingroup$
            (ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
            $endgroup$
            – Eric
            6 mins ago












          • $begingroup$
            By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
            $endgroup$
            – Lord Shark the Unknown
            6 mins ago










          • $begingroup$
            @eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
            $endgroup$
            – Lord Shark the Unknown
            3 mins ago


















          • $begingroup$
            Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
            $endgroup$
            – Eric
            7 mins ago










          • $begingroup$
            (ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
            $endgroup$
            – Eric
            6 mins ago












          • $begingroup$
            By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
            $endgroup$
            – Lord Shark the Unknown
            6 mins ago










          • $begingroup$
            @eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
            $endgroup$
            – Lord Shark the Unknown
            3 mins ago
















          $begingroup$
          Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
          $endgroup$
          – Eric
          7 mins ago




          $begingroup$
          Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
          $endgroup$
          – Eric
          7 mins ago












          $begingroup$
          (ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
          $endgroup$
          – Eric
          6 mins ago






          $begingroup$
          (ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
          $endgroup$
          – Eric
          6 mins ago














          $begingroup$
          By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
          $endgroup$
          – Lord Shark the Unknown
          6 mins ago




          $begingroup$
          By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
          $endgroup$
          – Lord Shark the Unknown
          6 mins ago












          $begingroup$
          @eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
          $endgroup$
          – Lord Shark the Unknown
          3 mins ago




          $begingroup$
          @eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
          $endgroup$
          – Lord Shark the Unknown
          3 mins ago


















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