Definition of the joint spectrum of Hilbert space operators












3












$begingroup$


I see in this paper the following definition:



enter image description here




How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:




Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
$$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$






functional-analysis operator-theory definition operator-algebras







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I see in this paper the following definition:



    enter image description here




    How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:




    Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
    $$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$






    functional-analysis operator-theory definition operator-algebras







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I see in this paper the following definition:



      enter image description here




      How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:




      Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
      $$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$






      functional-analysis operator-theory definition operator-algebras







      share|cite|improve this question











      $endgroup$




      I see in this paper the following definition:



      enter image description here




      How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:




      Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
      $$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$






      functional-analysis operator-theory definition operator-algebras




      functional-analysis operator-theory definition operator-algebras






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 17:50







      Student

















      asked Dec 25 '18 at 8:06









      StudentStudent

      2,4562524




      2,4562524






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          The double commutant $A''$ is defined as $(A')'$, where
          $$
          A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
          $$
          and
          $$
          A''={Sin B(H): ST=TS forall Tin A'}.
          $$
          The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.



          The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
          $$
          M''=overline{M}^{rm sot}.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
            $endgroup$
            – Student
            Dec 26 '18 at 7:53










          • $begingroup$
            Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
            $endgroup$
            – Student
            Dec 26 '18 at 7:54










          • $begingroup$
            Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
            $endgroup$
            – Martin Argerami
            Dec 26 '18 at 13:10










          • $begingroup$
            Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
            $endgroup$
            – Student
            Dec 29 '18 at 11:11










          • $begingroup$
            What I said about the inclusion seems to be wrong. I have deleted that paragraph.
            $endgroup$
            – Martin Argerami
            Dec 29 '18 at 18:15











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          The double commutant $A''$ is defined as $(A')'$, where
          $$
          A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
          $$
          and
          $$
          A''={Sin B(H): ST=TS forall Tin A'}.
          $$
          The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.



          The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
          $$
          M''=overline{M}^{rm sot}.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
            $endgroup$
            – Student
            Dec 26 '18 at 7:53










          • $begingroup$
            Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
            $endgroup$
            – Student
            Dec 26 '18 at 7:54










          • $begingroup$
            Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
            $endgroup$
            – Martin Argerami
            Dec 26 '18 at 13:10










          • $begingroup$
            Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
            $endgroup$
            – Student
            Dec 29 '18 at 11:11










          • $begingroup$
            What I said about the inclusion seems to be wrong. I have deleted that paragraph.
            $endgroup$
            – Martin Argerami
            Dec 29 '18 at 18:15
















          2












          $begingroup$

          The double commutant $A''$ is defined as $(A')'$, where
          $$
          A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
          $$
          and
          $$
          A''={Sin B(H): ST=TS forall Tin A'}.
          $$
          The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.



          The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
          $$
          M''=overline{M}^{rm sot}.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
            $endgroup$
            – Student
            Dec 26 '18 at 7:53










          • $begingroup$
            Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
            $endgroup$
            – Student
            Dec 26 '18 at 7:54










          • $begingroup$
            Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
            $endgroup$
            – Martin Argerami
            Dec 26 '18 at 13:10










          • $begingroup$
            Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
            $endgroup$
            – Student
            Dec 29 '18 at 11:11










          • $begingroup$
            What I said about the inclusion seems to be wrong. I have deleted that paragraph.
            $endgroup$
            – Martin Argerami
            Dec 29 '18 at 18:15














          2












          2








          2





          $begingroup$

          The double commutant $A''$ is defined as $(A')'$, where
          $$
          A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
          $$
          and
          $$
          A''={Sin B(H): ST=TS forall Tin A'}.
          $$
          The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.



          The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
          $$
          M''=overline{M}^{rm sot}.
          $$






          share|cite|improve this answer











          $endgroup$



          The double commutant $A''$ is defined as $(A')'$, where
          $$
          A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
          $$
          and
          $$
          A''={Sin B(H): ST=TS forall Tin A'}.
          $$
          The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.



          The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
          $$
          M''=overline{M}^{rm sot}.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 18:15

























          answered Dec 26 '18 at 6:02









          Martin ArgeramiMartin Argerami

          127k1182182




          127k1182182












          • $begingroup$
            Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
            $endgroup$
            – Student
            Dec 26 '18 at 7:53










          • $begingroup$
            Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
            $endgroup$
            – Student
            Dec 26 '18 at 7:54










          • $begingroup$
            Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
            $endgroup$
            – Martin Argerami
            Dec 26 '18 at 13:10










          • $begingroup$
            Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
            $endgroup$
            – Student
            Dec 29 '18 at 11:11










          • $begingroup$
            What I said about the inclusion seems to be wrong. I have deleted that paragraph.
            $endgroup$
            – Martin Argerami
            Dec 29 '18 at 18:15


















          • $begingroup$
            Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
            $endgroup$
            – Student
            Dec 26 '18 at 7:53










          • $begingroup$
            Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
            $endgroup$
            – Student
            Dec 26 '18 at 7:54










          • $begingroup$
            Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
            $endgroup$
            – Martin Argerami
            Dec 26 '18 at 13:10










          • $begingroup$
            Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
            $endgroup$
            – Student
            Dec 29 '18 at 11:11










          • $begingroup$
            What I said about the inclusion seems to be wrong. I have deleted that paragraph.
            $endgroup$
            – Martin Argerami
            Dec 29 '18 at 18:15
















          $begingroup$
          Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
          $endgroup$
          – Student
          Dec 26 '18 at 7:53




          $begingroup$
          Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
          $endgroup$
          – Student
          Dec 26 '18 at 7:53












          $begingroup$
          Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
          $endgroup$
          – Student
          Dec 26 '18 at 7:54




          $begingroup$
          Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
          $endgroup$
          – Student
          Dec 26 '18 at 7:54












          $begingroup$
          Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
          $endgroup$
          – Martin Argerami
          Dec 26 '18 at 13:10




          $begingroup$
          Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
          $endgroup$
          – Martin Argerami
          Dec 26 '18 at 13:10












          $begingroup$
          Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
          $endgroup$
          – Student
          Dec 29 '18 at 11:11




          $begingroup$
          Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
          $endgroup$
          – Student
          Dec 29 '18 at 11:11












          $begingroup$
          What I said about the inclusion seems to be wrong. I have deleted that paragraph.
          $endgroup$
          – Martin Argerami
          Dec 29 '18 at 18:15




          $begingroup$
          What I said about the inclusion seems to be wrong. I have deleted that paragraph.
          $endgroup$
          – Martin Argerami
          Dec 29 '18 at 18:15


















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