Self-adjoint element of $C^*$-algebra has real spectrum












1












$begingroup$


I'm trying to understand the proof in $S$3.9 on p.23 of these notes: http://strung.me/karen/CStarIntroDraft.pdf. The argument is as follows:



Let $A$ be a unital $C^*$-algebra, let $a in A$ be self-adjoint ($a = a^*$), and let $lambda in sigma(a)$. We want to show that $lambda in mathbb{R}$, by showing that $e^{ilambda} in sigma(e^{ia})$, where $e^{ia}$ has the usual power series definition; since $e^{ia}$ is unitary, any $mu in sigma(e^{ia})$ has $|mu| = 1$. I understand things so far.



We now write
$$ e^{ia} - e^{ilambda} = e^{ilambda}( e^{i(a - lambda)} - 1) = e^{ilambda}(a-lambda)b $$
where $b = sum_{n=1}^{infty} i^n(a-lambda)^{n-1}/n!$ (in the notes, the sum starts from $n = 2$, but I think this is a typo). Then $b$ commutes with $a-lambda$, so $e^{ia} - e^{ilambda}$ cannot be invertible, since $a-lambda$ is not.



My understanding is that the implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is valid in a ring without zero divisors, but I don't see how it's valid here.






spectral-theory c-star-algebras







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm trying to understand the proof in $S$3.9 on p.23 of these notes: http://strung.me/karen/CStarIntroDraft.pdf. The argument is as follows:



    Let $A$ be a unital $C^*$-algebra, let $a in A$ be self-adjoint ($a = a^*$), and let $lambda in sigma(a)$. We want to show that $lambda in mathbb{R}$, by showing that $e^{ilambda} in sigma(e^{ia})$, where $e^{ia}$ has the usual power series definition; since $e^{ia}$ is unitary, any $mu in sigma(e^{ia})$ has $|mu| = 1$. I understand things so far.



    We now write
    $$ e^{ia} - e^{ilambda} = e^{ilambda}( e^{i(a - lambda)} - 1) = e^{ilambda}(a-lambda)b $$
    where $b = sum_{n=1}^{infty} i^n(a-lambda)^{n-1}/n!$ (in the notes, the sum starts from $n = 2$, but I think this is a typo). Then $b$ commutes with $a-lambda$, so $e^{ia} - e^{ilambda}$ cannot be invertible, since $a-lambda$ is not.



    My understanding is that the implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is valid in a ring without zero divisors, but I don't see how it's valid here.






    spectral-theory c-star-algebras







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm trying to understand the proof in $S$3.9 on p.23 of these notes: http://strung.me/karen/CStarIntroDraft.pdf. The argument is as follows:



      Let $A$ be a unital $C^*$-algebra, let $a in A$ be self-adjoint ($a = a^*$), and let $lambda in sigma(a)$. We want to show that $lambda in mathbb{R}$, by showing that $e^{ilambda} in sigma(e^{ia})$, where $e^{ia}$ has the usual power series definition; since $e^{ia}$ is unitary, any $mu in sigma(e^{ia})$ has $|mu| = 1$. I understand things so far.



      We now write
      $$ e^{ia} - e^{ilambda} = e^{ilambda}( e^{i(a - lambda)} - 1) = e^{ilambda}(a-lambda)b $$
      where $b = sum_{n=1}^{infty} i^n(a-lambda)^{n-1}/n!$ (in the notes, the sum starts from $n = 2$, but I think this is a typo). Then $b$ commutes with $a-lambda$, so $e^{ia} - e^{ilambda}$ cannot be invertible, since $a-lambda$ is not.



      My understanding is that the implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is valid in a ring without zero divisors, but I don't see how it's valid here.






      spectral-theory c-star-algebras







      share|cite|improve this question









      $endgroup$




      I'm trying to understand the proof in $S$3.9 on p.23 of these notes: http://strung.me/karen/CStarIntroDraft.pdf. The argument is as follows:



      Let $A$ be a unital $C^*$-algebra, let $a in A$ be self-adjoint ($a = a^*$), and let $lambda in sigma(a)$. We want to show that $lambda in mathbb{R}$, by showing that $e^{ilambda} in sigma(e^{ia})$, where $e^{ia}$ has the usual power series definition; since $e^{ia}$ is unitary, any $mu in sigma(e^{ia})$ has $|mu| = 1$. I understand things so far.



      We now write
      $$ e^{ia} - e^{ilambda} = e^{ilambda}( e^{i(a - lambda)} - 1) = e^{ilambda}(a-lambda)b $$
      where $b = sum_{n=1}^{infty} i^n(a-lambda)^{n-1}/n!$ (in the notes, the sum starts from $n = 2$, but I think this is a typo). Then $b$ commutes with $a-lambda$, so $e^{ia} - e^{ilambda}$ cannot be invertible, since $a-lambda$ is not.



      My understanding is that the implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is valid in a ring without zero divisors, but I don't see how it's valid here.






      spectral-theory c-star-algebras




      spectral-theory c-star-algebras






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 12 at 23:00









      Kyle MacDonaldKyle MacDonald

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          $begingroup$

          The implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is true in any ring (with unity).



          Indeed, you have
          $$x (y (xy)^{-1}) =1$$



          It is a bit trickier, but we can also show that
          $$(y (xy)^{-1})x =1$$



          Indeed, let
          $$t:= (y (xy)^{-1})x$$



          Then
          $$ty=(y (xy)^{-1})xy=y \
          (t-1)y=0\
          (t-1)yx=0 \
          (t-1)xy=0\
          (t-1)xy(xy)^{-1}=0\
          t-1=0
          $$



          Note I am not sure but I think that the claim is not true if you don't know that $x,y$ commute, there is probably where you need the no zero divisors.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            The implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is true in any ring (with unity).



            Indeed, you have
            $$x (y (xy)^{-1}) =1$$



            It is a bit trickier, but we can also show that
            $$(y (xy)^{-1})x =1$$



            Indeed, let
            $$t:= (y (xy)^{-1})x$$



            Then
            $$ty=(y (xy)^{-1})xy=y \
            (t-1)y=0\
            (t-1)yx=0 \
            (t-1)xy=0\
            (t-1)xy(xy)^{-1}=0\
            t-1=0
            $$



            Note I am not sure but I think that the claim is not true if you don't know that $x,y$ commute, there is probably where you need the no zero divisors.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is true in any ring (with unity).



              Indeed, you have
              $$x (y (xy)^{-1}) =1$$



              It is a bit trickier, but we can also show that
              $$(y (xy)^{-1})x =1$$



              Indeed, let
              $$t:= (y (xy)^{-1})x$$



              Then
              $$ty=(y (xy)^{-1})xy=y \
              (t-1)y=0\
              (t-1)yx=0 \
              (t-1)xy=0\
              (t-1)xy(xy)^{-1}=0\
              t-1=0
              $$



              Note I am not sure but I think that the claim is not true if you don't know that $x,y$ commute, there is probably where you need the no zero divisors.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is true in any ring (with unity).



                Indeed, you have
                $$x (y (xy)^{-1}) =1$$



                It is a bit trickier, but we can also show that
                $$(y (xy)^{-1})x =1$$



                Indeed, let
                $$t:= (y (xy)^{-1})x$$



                Then
                $$ty=(y (xy)^{-1})xy=y \
                (t-1)y=0\
                (t-1)yx=0 \
                (t-1)xy=0\
                (t-1)xy(xy)^{-1}=0\
                t-1=0
                $$



                Note I am not sure but I think that the claim is not true if you don't know that $x,y$ commute, there is probably where you need the no zero divisors.






                share|cite|improve this answer









                $endgroup$



                The implication $(xy = yx text{ and } (xy)^{-1} text{ exists}) implies (x^{-1} text{ and } y^{-1} text{ exist})$ is true in any ring (with unity).



                Indeed, you have
                $$x (y (xy)^{-1}) =1$$



                It is a bit trickier, but we can also show that
                $$(y (xy)^{-1})x =1$$



                Indeed, let
                $$t:= (y (xy)^{-1})x$$



                Then
                $$ty=(y (xy)^{-1})xy=y \
                (t-1)y=0\
                (t-1)yx=0 \
                (t-1)xy=0\
                (t-1)xy(xy)^{-1}=0\
                t-1=0
                $$



                Note I am not sure but I think that the claim is not true if you don't know that $x,y$ commute, there is probably where you need the no zero divisors.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 12 at 23:12









                N. S.N. S.

                103k6111208




                103k6111208






























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