Positive operator is symmetric?












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If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?



However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?



Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.



Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed



begin{align*}
0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
end{align*}
This implies that
$$langle Tx,yrangle = -langle Ty,xrangle.$$



Exchanging $x$ with $ix$ yields



$$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
why
$$langle Tx,yrangle = langle Ty,xrangle$$
all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.



Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:



We have that



$$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.






matrices functional-analysis operator-theory







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    4












    $begingroup$


    If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?



    However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?



    Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.



    Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed



    begin{align*}
    0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
    end{align*}
    This implies that
    $$langle Tx,yrangle = -langle Ty,xrangle.$$



    Exchanging $x$ with $ix$ yields



    $$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
    why
    $$langle Tx,yrangle = langle Ty,xrangle$$
    all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.



    Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:



    We have that



    $$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
    for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.






    matrices functional-analysis operator-theory







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?



      However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?



      Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.



      Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed



      begin{align*}
      0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
      end{align*}
      This implies that
      $$langle Tx,yrangle = -langle Ty,xrangle.$$



      Exchanging $x$ with $ix$ yields



      $$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
      why
      $$langle Tx,yrangle = langle Ty,xrangle$$
      all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.



      Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:



      We have that



      $$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
      for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.






      matrices functional-analysis operator-theory







      share|cite|improve this question









      $endgroup$




      If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?



      However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?



      Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.



      Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed



      begin{align*}
      0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
      end{align*}
      This implies that
      $$langle Tx,yrangle = -langle Ty,xrangle.$$



      Exchanging $x$ with $ix$ yields



      $$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
      why
      $$langle Tx,yrangle = langle Ty,xrangle$$
      all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.



      Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:



      We have that



      $$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
      for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.






      matrices functional-analysis operator-theory




      matrices functional-analysis operator-theory






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      asked Jan 3 at 11:50









      Olof RubinOlof Rubin

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          In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.






          share|cite|improve this answer











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          • $begingroup$
            Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
            $endgroup$
            – Olof Rubin
            Jan 3 at 11:57












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          $begingroup$

          In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
            $endgroup$
            – Olof Rubin
            Jan 3 at 11:57
















          3












          $begingroup$

          In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
            $endgroup$
            – Olof Rubin
            Jan 3 at 11:57














          3












          3








          3





          $begingroup$

          In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.






          share|cite|improve this answer











          $endgroup$



          In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 11:58

























          answered Jan 3 at 11:55









          Kavi Rama MurthyKavi Rama Murthy

          71.6k53170




          71.6k53170












          • $begingroup$
            Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
            $endgroup$
            – Olof Rubin
            Jan 3 at 11:57


















          • $begingroup$
            Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
            $endgroup$
            – Olof Rubin
            Jan 3 at 11:57
















          $begingroup$
          Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
          $endgroup$
          – Olof Rubin
          Jan 3 at 11:57




          $begingroup$
          Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
          $endgroup$
          – Olof Rubin
          Jan 3 at 11:57


















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