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

A poker game description that does not feel gimmicky

If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?

What does Linus Torvalds mean when he says that Git "never ever" tracks a file?

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

Apparent duplicates between Haynes service instructions and MOT

Can you compress metal and what would be the consequences?

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

How to answer pointed "are you quitting" questioning when I don't want them to suspect

Falsification in Math vs Science

FPGA - DIY Programming

Why didn't the Event Horizon Telescope team mention Sagittarius A*?

Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?

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

Why was M87 targetted for the Event Horizon Telescope instead of Sagittarius A*?

How to support a colleague who finds meetings extremely tiring?

How to deal with fear of taking dependencies

What is the meaning of Triage in Cybersec world?

Have you ever entered Singapore using a different passport or name?

Why do UK politicians seemingly ignore opinion polls on Brexit?

Is this app Icon Browser Safe/Legit?

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

How to manage monthly salary

Origin of "cooter" meaning "vagina"

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



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

                              Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

                              2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

                              Metrô de Los Teques Índice Linhas | Estações | Ver também | Referências Ligações externas | Menu de navegação«INSTITUCIÓN»«Mapa de rutas»originalMetrô de Los TequesC.A. Metro Los Teques |Alcaldía de Guaicaipuro – Sitio OficialGobernacion de Mirandaeeeeeee