Commutative property of linear operators











up vote
0
down vote

favorite












I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.



But my question is about linear operators.



If :
$$ L1: V rightarrow V$$
$$ L2: V rightarrow V$$
$$L1[aX+bY]=aL1[X]+bL1[Y]$$



$$L2[aX+bY]=aL2[X]+bL2[Y]$$



Does the below equality hold? if yes what is the proof?



$$L2[L1[X]]=L1[L2[X]]$$










share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
    We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.



    But my question is about linear operators.



    If :
    $$ L1: V rightarrow V$$
    $$ L2: V rightarrow V$$
    $$L1[aX+bY]=aL1[X]+bL1[Y]$$



    $$L2[aX+bY]=aL2[X]+bL2[Y]$$



    Does the below equality hold? if yes what is the proof?



    $$L2[L1[X]]=L1[L2[X]]$$










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
      We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.



      But my question is about linear operators.



      If :
      $$ L1: V rightarrow V$$
      $$ L2: V rightarrow V$$
      $$L1[aX+bY]=aL1[X]+bL1[Y]$$



      $$L2[aX+bY]=aL2[X]+bL2[Y]$$



      Does the below equality hold? if yes what is the proof?



      $$L2[L1[X]]=L1[L2[X]]$$










      share|cite|improve this question















      I know that if we have to cascade linear system we can substitute the place of each system and get the same result.
      We can simply proof this property by writing the output of each system as convolution of impulse response and combine integrals.



      But my question is about linear operators.



      If :
      $$ L1: V rightarrow V$$
      $$ L2: V rightarrow V$$
      $$L1[aX+bY]=aL1[X]+bL1[Y]$$



      $$L2[aX+bY]=aL2[X]+bL2[Y]$$



      Does the below equality hold? if yes what is the proof?



      $$L2[L1[X]]=L1[L2[X]]$$







      linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 15 at 19:02

























      asked Nov 15 at 18:54









      alireza

      73




      73



























          active

          oldest

          votes











          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',
          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%2f3000121%2fcommutative-property-of-linear-operators%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown






























          active

          oldest

          votes













          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000121%2fcommutative-property-of-linear-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

          Plaza Victoria

          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

          How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...