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Distributing a matrix

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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.






linear-algebra







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.






    linear-algebra







    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.






      linear-algebra







      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




      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









            $endgroup$













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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

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              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



























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