Distributing a matrix












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



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



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



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



    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



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



      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



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



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






      linear-algebra




      linear-algebra






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      share|cite|improve this question











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      asked 4 hours ago









      redblacktreesredblacktrees

      424




      424






















          2 Answers
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          2












          $begingroup$

          Your first answer is correct. There are two distributive laws for matrices,
          $$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
          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
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              2 Answers
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              2












              $begingroup$

              Your first answer is correct. There are two distributive laws for matrices,
              $$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
              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,
                $$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
                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,
                  $$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
                  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,
                  $$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
                  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

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