Normal Operator || T^2|| = ||T||^2 Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)If $S$ and $T$ are commuting, normal operators, then $ST$ is normal$T^2=I$ implies that $T$ is a normal operatorProving an operator is Self-adjoint using the Spectral TheoremEigenvalues of adjoint operator [General Case]diagonalizability implies existence of an inner product wrt an operator is normalNormal operator over real inner product spaceNormal matrix over real inner product space with real eigenvalues is Hermitianpolar form of unitary operatorWe have a linear operator T. Show $T^2=Id$ implies $T=T^*$Some property of Normal Operator

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



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)If $S$ and $T$ are commuting, normal operators, then $ST$ is normal$T^2=I$ implies that $T$ is a normal operatorProving an operator is Self-adjoint using the Spectral TheoremEigenvalues of adjoint operator [General Case]diagonalizability implies existence of an inner product wrt an operator is normalNormal operator over real inner product spaceNormal matrix over real inner product space with real eigenvalues is Hermitianpolar form of unitary operatorWe have a linear operator T. Show $T^2=Id$ implies $T=T^*$Some property of Normal Operator










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 $frac$ and ||T$^2$|| = sup $frac$. 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?










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 $frac$ and ||T$^2$|| = sup $frac$. 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?










    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 $frac$ and ||T$^2$|| = sup $frac$. 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?










      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 $frac$ and ||T$^2$|| = sup $frac$. 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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      EricEric

      798




      798




















          1 Answer
          1






          active

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          4












          $begingroup$

          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
            just now











          Your Answer








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          1 Answer
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          1 Answer
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          active

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          active

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          4












          $begingroup$

          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
            just now















          4












          $begingroup$

          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
            just now













          4












          4








          4





          $begingroup$

          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 16 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
            just now
















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















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




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

















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