Distributing a matrix The 2019 Stack Overflow Developer Survey Results Are InOn multiplying quaternion matricesWhen is matrix multiplication commutative?Matrix multiplicationWhy aren't all matrices diagonalisable?Linear Transformation vs Matrixhow many ways is there to factor matrix?Can an arbitrary matrix represent any linear map just by changing the basis?Inverse matrix confusionA question matrix multiplication commutative?Joint Matrices Factorization

Why did Acorn's A3000 have red function keys?

slides for 30min~1hr skype tenure track application interview

How to notate time signature switching consistently every measure

Why isn't airport relocation done gradually?

Can a rogue use sneak attack with weapons that have the thrown property even if they are not thrown?

How technical should a Scrum Master be to effectively remove impediments?

Why can Shazam fly?

Can you compress metal and what would be the consequences?

How are circuits which use complex ICs normally simulated?

Output the Arecibo Message

For what reasons would an animal species NOT cross a *horizontal* land bridge?

Return to UK after having been refused entry years ago

What are the motivations for publishing new editions of an existing textbook, beyond new discoveries in a field?

Should I use my personal e-mail address, or my workplace one, when registering to external websites for work purposes?

Why isn't the circumferential light around the M87 black hole's event horizon symmetric?

What is the most effective way of iterating a std::vector and why?

Does the shape of a die affect the probability of a number being rolled?

How to deal with fear of taking dependencies

How to check whether the reindex working or not in Magento?

Why hard-Brexiteers don't insist on a hard border to prevent illegal immigration after Brexit?

Looking for Correct Greek Translation for Heraclitus

If a Druid sees an animal’s corpse, can they Wild Shape into that animal?

Distributing a matrix

Button changing it's text & action. Good or terrible?



Distributing a matrix



The 2019 Stack Overflow Developer Survey Results Are InOn multiplying quaternion matricesWhen is matrix multiplication commutative?Matrix multiplicationWhy aren't all matrices diagonalisable?Linear Transformation vs Matrixhow many ways is there to factor matrix?Can an arbitrary matrix represent any linear map just by changing the basis?Inverse matrix confusionA question matrix multiplication commutative?Joint Matrices Factorization










3












$begingroup$


Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.



In particular, if I want to distribute



$$((I - A) + A)(I - A)^-1,$$



would it become



$$(I - A)(I - A)^-1 + A(I - A)^-1 $$



OR would it be



$$(I - A)^-1(I - A) + (I - A)^-1A?$$



How do I know which side it goes on? I think the first one is correct.










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.



    In particular, if I want to distribute



    $$((I - A) + A)(I - A)^-1,$$



    would it become



    $$(I - A)(I - A)^-1 + A(I - A)^-1 $$



    OR would it be



    $$(I - A)^-1(I - A) + (I - A)^-1A?$$



    How do I know which side it goes on? I think the first one is correct.










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.



      In particular, if I want to distribute



      $$((I - A) + A)(I - A)^-1,$$



      would it become



      $$(I - A)(I - A)^-1 + A(I - A)^-1 $$



      OR would it be



      $$(I - A)^-1(I - A) + (I - A)^-1A?$$



      How do I know which side it goes on? I think the first one is correct.










      share|cite|improve this question









      $endgroup$




      Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.



      In particular, if I want to distribute



      $$((I - A) + A)(I - A)^-1,$$



      would it become



      $$(I - A)(I - A)^-1 + A(I - A)^-1 $$



      OR would it be



      $$(I - A)^-1(I - A) + (I - A)^-1A?$$



      How do I know which side it goes on? I think the first one is correct.







      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      redblacktreesredblacktrees

      424




      424




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Your first answer is correct. There are two distributive laws for matrices,
          $$A(B+C)=AB+ACquadhboxandquad (A+B)C=AC+BC ,$$
          but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.




            Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by



            $$a cdot (b+c) = acdot b + a cdot c$$



            Similarly, right-distributivity is given by



            $$(b+c)cdot a = bcdot a + ccdot a$$



            Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.



            In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).




            So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have



            $$(B+C)A = BA + CA$$



            Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.






            share|cite|improve this answer









            $endgroup$













              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%2f3183231%2fdistributing-a-matrix%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Your first answer is correct. There are two distributive laws for matrices,
              $$A(B+C)=AB+ACquadhboxandquad (A+B)C=AC+BC ,$$
              but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Your first answer is correct. There are two distributive laws for matrices,
                $$A(B+C)=AB+ACquadhboxandquad (A+B)C=AC+BC ,$$
                but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Your first answer is correct. There are two distributive laws for matrices,
                  $$A(B+C)=AB+ACquadhboxandquad (A+B)C=AC+BC ,$$
                  but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....






                  share|cite|improve this answer









                  $endgroup$



                  Your first answer is correct. There are two distributive laws for matrices,
                  $$A(B+C)=AB+ACquadhboxandquad (A+B)C=AC+BC ,$$
                  but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  DavidDavid

                  69.8k668131




                  69.8k668131





















                      1












                      $begingroup$

                      In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.




                      Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by



                      $$a cdot (b+c) = acdot b + a cdot c$$



                      Similarly, right-distributivity is given by



                      $$(b+c)cdot a = bcdot a + ccdot a$$



                      Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.



                      In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).




                      So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have



                      $$(B+C)A = BA + CA$$



                      Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.




                        Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by



                        $$a cdot (b+c) = acdot b + a cdot c$$



                        Similarly, right-distributivity is given by



                        $$(b+c)cdot a = bcdot a + ccdot a$$



                        Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.



                        In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).




                        So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have



                        $$(B+C)A = BA + CA$$



                        Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.




                          Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by



                          $$a cdot (b+c) = acdot b + a cdot c$$



                          Similarly, right-distributivity is given by



                          $$(b+c)cdot a = bcdot a + ccdot a$$



                          Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.



                          In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).




                          So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have



                          $$(B+C)A = BA + CA$$



                          Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.






                          share|cite|improve this answer









                          $endgroup$



                          In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.




                          Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by



                          $$a cdot (b+c) = acdot b + a cdot c$$



                          Similarly, right-distributivity is given by



                          $$(b+c)cdot a = bcdot a + ccdot a$$



                          Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.



                          In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).




                          So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have



                          $$(B+C)A = BA + CA$$



                          Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 4 hours ago









                          Eevee TrainerEevee Trainer

                          10.4k31742




                          10.4k31742



























                              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.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3183231%2fdistributing-a-matrix%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

                              Möglingen Índice Localización Historia Demografía Referencias Enlaces externos Menú de navegación48°53′18″N 9°07′45″E / 48.888333333333, 9.129166666666748°53′18″N 9°07′45″E / 48.888333333333, 9.1291666666667Sitio web oficial Mapa de Möglingen«Gemeinden in Deutschland nach Fläche, Bevölkerung und Postleitzahl am 30.09.2016»Möglingen

                              Virtualbox - Configuration error: Querying “UUID” failed (VERR_CFGM_VALUE_NOT_FOUND)“VERR_SUPLIB_WORLD_WRITABLE” error when trying to installing OS in virtualboxVirtual Box Kernel errorFailed to open a seesion for the virtual machineFailed to open a session for the virtual machineUbuntu 14.04 LTS Virtualbox errorcan't use VM VirtualBoxusing virtualboxI can't run Linux-64 Bit on VirtualBoxUnable to insert the virtual optical disk (VBoxguestaddition) in virtual machine for ubuntu server in win 10VirtuaBox in Ubuntu 18.04 Issues with Win10.ISO Installation

                              Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org