Spectrum of a pair of commuting operators












2














Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$




Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$











share|cite|improve this question





























    2














    Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
    $(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
    $$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$




    Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
    $$sigma_H(I,A)={(1,1);(1,-1)}.$$











    share|cite|improve this question



























      2












      2








      2







      Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
      $(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
      $$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$




      Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
      $$sigma_H(I,A)={(1,1);(1,-1)}.$$











      share|cite|improve this question















      Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
      $(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
      $$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$




      Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
      $$sigma_H(I,A)={(1,1);(1,-1)}.$$








      functional-analysis operator-theory hilbert-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 '18 at 15:11

























      asked Nov 28 '18 at 12:57









      Student

      2,5202524




      2,5202524






















          1 Answer
          1






          active

          oldest

          votes


















          1














          First a remark:



          If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.



          This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.



          So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.






          share|cite|improve this answer





















          • Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
            – Student
            Dec 2 '18 at 11:12











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017110%2fspectrum-of-a-pair-of-commuting-operators%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          First a remark:



          If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.



          This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.



          So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.






          share|cite|improve this answer





















          • Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
            – Student
            Dec 2 '18 at 11:12
















          1














          First a remark:



          If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.



          This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.



          So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.






          share|cite|improve this answer





















          • Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
            – Student
            Dec 2 '18 at 11:12














          1












          1








          1






          First a remark:



          If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.



          This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.



          So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.






          share|cite|improve this answer












          First a remark:



          If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.



          This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.



          So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 16:59









          s.harp

          8,42312049




          8,42312049












          • Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
            – Student
            Dec 2 '18 at 11:12


















          • Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
            – Student
            Dec 2 '18 at 11:12
















          Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
          – Student
          Dec 2 '18 at 11:12




          Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
          – Student
          Dec 2 '18 at 11:12


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017110%2fspectrum-of-a-pair-of-commuting-operators%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Aardman Animations

          Are they similar matrix

          “minimization” problem in Euclidean space related to orthonormal basis