Inequalities inside operator norm for a normal operator












1












$begingroup$


Suppose $T$ is normal, I am trying to show that



$|T^{n}|^2|T|^2 = |T^{n+1}|^2$



I know that



$|T^{n+1}|^2 = |T^{n}T|^2 leq |T^{n}|^2|T|^2$.



Is there a way to show that



$ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $?



I could be taking the wrong approach, although I am curious if it is possible to show if the second inequality is true.






operator-theory







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Suppose $T$ is normal, I am trying to show that



    $|T^{n}|^2|T|^2 = |T^{n+1}|^2$



    I know that



    $|T^{n+1}|^2 = |T^{n}T|^2 leq |T^{n}|^2|T|^2$.



    Is there a way to show that



    $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $?



    I could be taking the wrong approach, although I am curious if it is possible to show if the second inequality is true.






    operator-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose $T$ is normal, I am trying to show that



      $|T^{n}|^2|T|^2 = |T^{n+1}|^2$



      I know that



      $|T^{n+1}|^2 = |T^{n}T|^2 leq |T^{n}|^2|T|^2$.



      Is there a way to show that



      $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $?



      I could be taking the wrong approach, although I am curious if it is possible to show if the second inequality is true.






      operator-theory







      share|cite|improve this question









      $endgroup$




      Suppose $T$ is normal, I am trying to show that



      $|T^{n}|^2|T|^2 = |T^{n+1}|^2$



      I know that



      $|T^{n+1}|^2 = |T^{n}T|^2 leq |T^{n}|^2|T|^2$.



      Is there a way to show that



      $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $?



      I could be taking the wrong approach, although I am curious if it is possible to show if the second inequality is true.






      operator-theory




      operator-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 0:16









      Axion004Axion004

      314212




      314212






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          We only need to prove
          $$|T^n|=|T|^n$$
          when $T$ is a normal operator.



          For any operator $R$, we have $$|R^*R|=|R|^2.$$ So for a self-ajoint operator $S$, we have $|S^2|=|S|^2$, thus $$|S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}.$$
          For any $nin mathbb{N}$, pick a positive integer $k$ such that $n<2^k$, we have $$|S|^{2^k}=|S^{2^k}|=|S^nS^{2^k-n}|leq|S^n||S|^{2^k-n}leq |S|^{2^k},$$ hence $
          |S^n|=|S|^n$          .
          For a normal operator $T$, $|T^n|^2=|(T^n)^*T^n|=|(T^*T)^n|=|T^*T|^n=|T|^{2n}.$ Done!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To show that $|S^2|=|S|^2 implies |S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}$, you must have used induction. The base case is $|S^2|=|S|^2$ and the induction hypothesis is $|S^{2^k}|=|S|^{2^k}$. For the induction step, you must have shown that $|S^{2^{k+1}}|=|S|^{2^{k+1}}$, which doesn't appear to be a straightforward calculation.
            $endgroup$
            – Axion004
            Nov 29 '18 at 14:30












          • $begingroup$
            I just mean $|S^{2^k}|=|S^{2^{k-1}}S^{2^{k-1}}|=|S^{2^{k-1}}|^2=|S^{2^{k-2}}|^4=cdots=|S|^{2^{k}}$. And using induction seems better.
            $endgroup$
            – C.Ding
            Nov 29 '18 at 23:54





















          1












          $begingroup$

          Since $T$ is normal, you have $|T|=r(T)$, the spectral radius. So now your equality, without the squares is saying that
          $$
          max{|lambda|^n: lambdainsigma(T)},max{|lambda|: lambdainsigma(T)}=max{|lambda|^{n+1}: lambdainsigma(T)}.
          $$
          This is trivial after you notice that $max{|lambda|^n: lambdainsigma(T)}=(max{|lambda|: lambdainsigma(T)})^n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That makes sense. Is there a way to show that the second inequality is true without applying $|T|=r(T)$? Could one show that $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $ directly through properties of the norm?
            $endgroup$
            – Axion004
            Nov 29 '18 at 5:10











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          2 Answers
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          2 Answers
          2






          active

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          active

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          active

          oldest

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          1












          $begingroup$

          We only need to prove
          $$|T^n|=|T|^n$$
          when $T$ is a normal operator.



          For any operator $R$, we have $$|R^*R|=|R|^2.$$ So for a self-ajoint operator $S$, we have $|S^2|=|S|^2$, thus $$|S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}.$$
          For any $nin mathbb{N}$, pick a positive integer $k$ such that $n<2^k$, we have $$|S|^{2^k}=|S^{2^k}|=|S^nS^{2^k-n}|leq|S^n||S|^{2^k-n}leq |S|^{2^k},$$ hence $
          |S^n|=|S|^n$          .
          For a normal operator $T$, $|T^n|^2=|(T^n)^*T^n|=|(T^*T)^n|=|T^*T|^n=|T|^{2n}.$ Done!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To show that $|S^2|=|S|^2 implies |S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}$, you must have used induction. The base case is $|S^2|=|S|^2$ and the induction hypothesis is $|S^{2^k}|=|S|^{2^k}$. For the induction step, you must have shown that $|S^{2^{k+1}}|=|S|^{2^{k+1}}$, which doesn't appear to be a straightforward calculation.
            $endgroup$
            – Axion004
            Nov 29 '18 at 14:30












          • $begingroup$
            I just mean $|S^{2^k}|=|S^{2^{k-1}}S^{2^{k-1}}|=|S^{2^{k-1}}|^2=|S^{2^{k-2}}|^4=cdots=|S|^{2^{k}}$. And using induction seems better.
            $endgroup$
            – C.Ding
            Nov 29 '18 at 23:54


















          1












          $begingroup$

          We only need to prove
          $$|T^n|=|T|^n$$
          when $T$ is a normal operator.



          For any operator $R$, we have $$|R^*R|=|R|^2.$$ So for a self-ajoint operator $S$, we have $|S^2|=|S|^2$, thus $$|S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}.$$
          For any $nin mathbb{N}$, pick a positive integer $k$ such that $n<2^k$, we have $$|S|^{2^k}=|S^{2^k}|=|S^nS^{2^k-n}|leq|S^n||S|^{2^k-n}leq |S|^{2^k},$$ hence $
          |S^n|=|S|^n$          .
          For a normal operator $T$, $|T^n|^2=|(T^n)^*T^n|=|(T^*T)^n|=|T^*T|^n=|T|^{2n}.$ Done!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To show that $|S^2|=|S|^2 implies |S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}$, you must have used induction. The base case is $|S^2|=|S|^2$ and the induction hypothesis is $|S^{2^k}|=|S|^{2^k}$. For the induction step, you must have shown that $|S^{2^{k+1}}|=|S|^{2^{k+1}}$, which doesn't appear to be a straightforward calculation.
            $endgroup$
            – Axion004
            Nov 29 '18 at 14:30












          • $begingroup$
            I just mean $|S^{2^k}|=|S^{2^{k-1}}S^{2^{k-1}}|=|S^{2^{k-1}}|^2=|S^{2^{k-2}}|^4=cdots=|S|^{2^{k}}$. And using induction seems better.
            $endgroup$
            – C.Ding
            Nov 29 '18 at 23:54
















          1












          1








          1





          $begingroup$

          We only need to prove
          $$|T^n|=|T|^n$$
          when $T$ is a normal operator.



          For any operator $R$, we have $$|R^*R|=|R|^2.$$ So for a self-ajoint operator $S$, we have $|S^2|=|S|^2$, thus $$|S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}.$$
          For any $nin mathbb{N}$, pick a positive integer $k$ such that $n<2^k$, we have $$|S|^{2^k}=|S^{2^k}|=|S^nS^{2^k-n}|leq|S^n||S|^{2^k-n}leq |S|^{2^k},$$ hence $
          |S^n|=|S|^n$          .
          For a normal operator $T$, $|T^n|^2=|(T^n)^*T^n|=|(T^*T)^n|=|T^*T|^n=|T|^{2n}.$ Done!






          share|cite|improve this answer









          $endgroup$



          We only need to prove
          $$|T^n|=|T|^n$$
          when $T$ is a normal operator.



          For any operator $R$, we have $$|R^*R|=|R|^2.$$ So for a self-ajoint operator $S$, we have $|S^2|=|S|^2$, thus $$|S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}.$$
          For any $nin mathbb{N}$, pick a positive integer $k$ such that $n<2^k$, we have $$|S|^{2^k}=|S^{2^k}|=|S^nS^{2^k-n}|leq|S^n||S|^{2^k-n}leq |S|^{2^k},$$ hence $
          |S^n|=|S|^n$          .
          For a normal operator $T$, $|T^n|^2=|(T^n)^*T^n|=|(T^*T)^n|=|T^*T|^n=|T|^{2n}.$ Done!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 10:18









          C.DingC.Ding

          1,3931321




          1,3931321












          • $begingroup$
            To show that $|S^2|=|S|^2 implies |S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}$, you must have used induction. The base case is $|S^2|=|S|^2$ and the induction hypothesis is $|S^{2^k}|=|S|^{2^k}$. For the induction step, you must have shown that $|S^{2^{k+1}}|=|S|^{2^{k+1}}$, which doesn't appear to be a straightforward calculation.
            $endgroup$
            – Axion004
            Nov 29 '18 at 14:30












          • $begingroup$
            I just mean $|S^{2^k}|=|S^{2^{k-1}}S^{2^{k-1}}|=|S^{2^{k-1}}|^2=|S^{2^{k-2}}|^4=cdots=|S|^{2^{k}}$. And using induction seems better.
            $endgroup$
            – C.Ding
            Nov 29 '18 at 23:54




















          • $begingroup$
            To show that $|S^2|=|S|^2 implies |S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}$, you must have used induction. The base case is $|S^2|=|S|^2$ and the induction hypothesis is $|S^{2^k}|=|S|^{2^k}$. For the induction step, you must have shown that $|S^{2^{k+1}}|=|S|^{2^{k+1}}$, which doesn't appear to be a straightforward calculation.
            $endgroup$
            – Axion004
            Nov 29 '18 at 14:30












          • $begingroup$
            I just mean $|S^{2^k}|=|S^{2^{k-1}}S^{2^{k-1}}|=|S^{2^{k-1}}|^2=|S^{2^{k-2}}|^4=cdots=|S|^{2^{k}}$. And using induction seems better.
            $endgroup$
            – C.Ding
            Nov 29 '18 at 23:54


















          $begingroup$
          To show that $|S^2|=|S|^2 implies |S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}$, you must have used induction. The base case is $|S^2|=|S|^2$ and the induction hypothesis is $|S^{2^k}|=|S|^{2^k}$. For the induction step, you must have shown that $|S^{2^{k+1}}|=|S|^{2^{k+1}}$, which doesn't appear to be a straightforward calculation.
          $endgroup$
          – Axion004
          Nov 29 '18 at 14:30






          $begingroup$
          To show that $|S^2|=|S|^2 implies |S^{2^k}|=|S|^{2^k} ~forall kin mathbb{N}$, you must have used induction. The base case is $|S^2|=|S|^2$ and the induction hypothesis is $|S^{2^k}|=|S|^{2^k}$. For the induction step, you must have shown that $|S^{2^{k+1}}|=|S|^{2^{k+1}}$, which doesn't appear to be a straightforward calculation.
          $endgroup$
          – Axion004
          Nov 29 '18 at 14:30














          $begingroup$
          I just mean $|S^{2^k}|=|S^{2^{k-1}}S^{2^{k-1}}|=|S^{2^{k-1}}|^2=|S^{2^{k-2}}|^4=cdots=|S|^{2^{k}}$. And using induction seems better.
          $endgroup$
          – C.Ding
          Nov 29 '18 at 23:54






          $begingroup$
          I just mean $|S^{2^k}|=|S^{2^{k-1}}S^{2^{k-1}}|=|S^{2^{k-1}}|^2=|S^{2^{k-2}}|^4=cdots=|S|^{2^{k}}$. And using induction seems better.
          $endgroup$
          – C.Ding
          Nov 29 '18 at 23:54













          1












          $begingroup$

          Since $T$ is normal, you have $|T|=r(T)$, the spectral radius. So now your equality, without the squares is saying that
          $$
          max{|lambda|^n: lambdainsigma(T)},max{|lambda|: lambdainsigma(T)}=max{|lambda|^{n+1}: lambdainsigma(T)}.
          $$
          This is trivial after you notice that $max{|lambda|^n: lambdainsigma(T)}=(max{|lambda|: lambdainsigma(T)})^n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That makes sense. Is there a way to show that the second inequality is true without applying $|T|=r(T)$? Could one show that $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $ directly through properties of the norm?
            $endgroup$
            – Axion004
            Nov 29 '18 at 5:10
















          1












          $begingroup$

          Since $T$ is normal, you have $|T|=r(T)$, the spectral radius. So now your equality, without the squares is saying that
          $$
          max{|lambda|^n: lambdainsigma(T)},max{|lambda|: lambdainsigma(T)}=max{|lambda|^{n+1}: lambdainsigma(T)}.
          $$
          This is trivial after you notice that $max{|lambda|^n: lambdainsigma(T)}=(max{|lambda|: lambdainsigma(T)})^n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That makes sense. Is there a way to show that the second inequality is true without applying $|T|=r(T)$? Could one show that $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $ directly through properties of the norm?
            $endgroup$
            – Axion004
            Nov 29 '18 at 5:10














          1












          1








          1





          $begingroup$

          Since $T$ is normal, you have $|T|=r(T)$, the spectral radius. So now your equality, without the squares is saying that
          $$
          max{|lambda|^n: lambdainsigma(T)},max{|lambda|: lambdainsigma(T)}=max{|lambda|^{n+1}: lambdainsigma(T)}.
          $$
          This is trivial after you notice that $max{|lambda|^n: lambdainsigma(T)}=(max{|lambda|: lambdainsigma(T)})^n$.






          share|cite|improve this answer









          $endgroup$



          Since $T$ is normal, you have $|T|=r(T)$, the spectral radius. So now your equality, without the squares is saying that
          $$
          max{|lambda|^n: lambdainsigma(T)},max{|lambda|: lambdainsigma(T)}=max{|lambda|^{n+1}: lambdainsigma(T)}.
          $$
          This is trivial after you notice that $max{|lambda|^n: lambdainsigma(T)}=(max{|lambda|: lambdainsigma(T)})^n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 1:25









          Martin ArgeramiMartin Argerami

          124k1177176




          124k1177176












          • $begingroup$
            That makes sense. Is there a way to show that the second inequality is true without applying $|T|=r(T)$? Could one show that $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $ directly through properties of the norm?
            $endgroup$
            – Axion004
            Nov 29 '18 at 5:10


















          • $begingroup$
            That makes sense. Is there a way to show that the second inequality is true without applying $|T|=r(T)$? Could one show that $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $ directly through properties of the norm?
            $endgroup$
            – Axion004
            Nov 29 '18 at 5:10
















          $begingroup$
          That makes sense. Is there a way to show that the second inequality is true without applying $|T|=r(T)$? Could one show that $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $ directly through properties of the norm?
          $endgroup$
          – Axion004
          Nov 29 '18 at 5:10




          $begingroup$
          That makes sense. Is there a way to show that the second inequality is true without applying $|T|=r(T)$? Could one show that $ |T^{n}|^2|T|^2 leq |T^{n+1}|^2 $ directly through properties of the norm?
          $endgroup$
          – Axion004
          Nov 29 '18 at 5:10


















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