Distributive law for matrices

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
-1
down vote

favorite












Unless I'm mistaken, matrices follow a distributive law, provided that the dimensions line up, so $X(Y+Z) = XY + XZ$. I'm struggling somewhat to prove this fact, though. Is it enough to say that because matrices represent linear transformations, we're simply applying the linearity of $X$? Or is a more rigorous derivation, likely requiring the summation representation of a matrix, required?



A hint on this would be very helpful. I'm mainly interested on the 'how' here rather than the proof itself, as it's very possible I'm misunderstanding the definition of linearity.




linear-transformations proof-explanation




share|cite|improve this question



















  • Linearity seems enough to support your statement.
    – xbh
    Jul 29 at 14:12














up vote
-1
down vote

favorite












Unless I'm mistaken, matrices follow a distributive law, provided that the dimensions line up, so $X(Y+Z) = XY + XZ$. I'm struggling somewhat to prove this fact, though. Is it enough to say that because matrices represent linear transformations, we're simply applying the linearity of $X$? Or is a more rigorous derivation, likely requiring the summation representation of a matrix, required?



A hint on this would be very helpful. I'm mainly interested on the 'how' here rather than the proof itself, as it's very possible I'm misunderstanding the definition of linearity.




linear-transformations proof-explanation




share|cite|improve this question



















  • Linearity seems enough to support your statement.
    – xbh
    Jul 29 at 14:12












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Unless I'm mistaken, matrices follow a distributive law, provided that the dimensions line up, so $X(Y+Z) = XY + XZ$. I'm struggling somewhat to prove this fact, though. Is it enough to say that because matrices represent linear transformations, we're simply applying the linearity of $X$? Or is a more rigorous derivation, likely requiring the summation representation of a matrix, required?



A hint on this would be very helpful. I'm mainly interested on the 'how' here rather than the proof itself, as it's very possible I'm misunderstanding the definition of linearity.




linear-transformations proof-explanation




share|cite|improve this question











Unless I'm mistaken, matrices follow a distributive law, provided that the dimensions line up, so $X(Y+Z) = XY + XZ$. I'm struggling somewhat to prove this fact, though. Is it enough to say that because matrices represent linear transformations, we're simply applying the linearity of $X$? Or is a more rigorous derivation, likely requiring the summation representation of a matrix, required?



A hint on this would be very helpful. I'm mainly interested on the 'how' here rather than the proof itself, as it's very possible I'm misunderstanding the definition of linearity.





linear-transformations proof-explanation





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 29 at 14:06









Matt.P

666313




666313











  • Linearity seems enough to support your statement.
    – xbh
    Jul 29 at 14:12
















  • Linearity seems enough to support your statement.
    – xbh
    Jul 29 at 14:12















Linearity seems enough to support your statement.
– xbh
Jul 29 at 14:12




Linearity seems enough to support your statement.
– xbh
Jul 29 at 14:12










2 Answers
2






active

oldest

votes

















up vote
1
down vote













If you've established:



  • the correspondence between matrices and linear transformations

  • matrix multiplication corresponds to composition of transformations

  • matrix addition corresponds to addition of transformations

  • linear transformations satisfy this law

then that is enough to conclude it for matrices as well. Explicitly, if $X$ is a matrix transformation, let $overlineX$ denote the associated linear transformation, and you have



$$ overlineX(Y+Z) = overlineX circ overline(Y+Z)
= overlineX circ left( overlineY + overlineZ right)
= overlineX circ overlineY + overlineX circ overlineZ
= overlineXY + overlineXZ
= overlineXY + XZ$$



Since the associated transformations are equal, the matrices $X(Y+Z)$ and $XY + XZ$ are equal. Observe how all of the bullet points above are used in this calculation.






share|cite|improve this answer




























    up vote
    1
    down vote













    The fact that matrices represent linear functions means that for a matrix $A$ and vectors $u, v$, we have $A(u +v) = A u + A v$. This doesn't tell us anything about what happens when we multiply matrices together.



    So yes, to prove distributivity you will need something more. If you consider the summation representation the proof is quite straightforward though.






    share|cite|improve this answer























      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: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      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%2f2866109%2fdistributive-law-for-matrices%23new-answer', 'question_page');

      );

      Post as a guest

















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote













      If you've established:



      • the correspondence between matrices and linear transformations

      • matrix multiplication corresponds to composition of transformations

      • matrix addition corresponds to addition of transformations

      • linear transformations satisfy this law

      then that is enough to conclude it for matrices as well. Explicitly, if $X$ is a matrix transformation, let $overlineX$ denote the associated linear transformation, and you have



      $$ overlineX(Y+Z) = overlineX circ overline(Y+Z)
      = overlineX circ left( overlineY + overlineZ right)
      = overlineX circ overlineY + overlineX circ overlineZ
      = overlineXY + overlineXZ
      = overlineXY + XZ$$



      Since the associated transformations are equal, the matrices $X(Y+Z)$ and $XY + XZ$ are equal. Observe how all of the bullet points above are used in this calculation.






      share|cite|improve this answer

























        up vote
        1
        down vote













        If you've established:



        • the correspondence between matrices and linear transformations

        • matrix multiplication corresponds to composition of transformations

        • matrix addition corresponds to addition of transformations

        • linear transformations satisfy this law

        then that is enough to conclude it for matrices as well. Explicitly, if $X$ is a matrix transformation, let $overlineX$ denote the associated linear transformation, and you have



        $$ overlineX(Y+Z) = overlineX circ overline(Y+Z)
        = overlineX circ left( overlineY + overlineZ right)
        = overlineX circ overlineY + overlineX circ overlineZ
        = overlineXY + overlineXZ
        = overlineXY + XZ$$



        Since the associated transformations are equal, the matrices $X(Y+Z)$ and $XY + XZ$ are equal. Observe how all of the bullet points above are used in this calculation.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          If you've established:



          • the correspondence between matrices and linear transformations

          • matrix multiplication corresponds to composition of transformations

          • matrix addition corresponds to addition of transformations

          • linear transformations satisfy this law

          then that is enough to conclude it for matrices as well. Explicitly, if $X$ is a matrix transformation, let $overlineX$ denote the associated linear transformation, and you have



          $$ overlineX(Y+Z) = overlineX circ overline(Y+Z)
          = overlineX circ left( overlineY + overlineZ right)
          = overlineX circ overlineY + overlineX circ overlineZ
          = overlineXY + overlineXZ
          = overlineXY + XZ$$



          Since the associated transformations are equal, the matrices $X(Y+Z)$ and $XY + XZ$ are equal. Observe how all of the bullet points above are used in this calculation.






          share|cite|improve this answer













          If you've established:



          • the correspondence between matrices and linear transformations

          • matrix multiplication corresponds to composition of transformations

          • matrix addition corresponds to addition of transformations

          • linear transformations satisfy this law

          then that is enough to conclude it for matrices as well. Explicitly, if $X$ is a matrix transformation, let $overlineX$ denote the associated linear transformation, and you have



          $$ overlineX(Y+Z) = overlineX circ overline(Y+Z)
          = overlineX circ left( overlineY + overlineZ right)
          = overlineX circ overlineY + overlineX circ overlineZ
          = overlineXY + overlineXZ
          = overlineXY + XZ$$



          Since the associated transformations are equal, the matrices $X(Y+Z)$ and $XY + XZ$ are equal. Observe how all of the bullet points above are used in this calculation.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 29 at 14:18









          Hurkyl

          107k9112253




          107k9112253




















              up vote
              1
              down vote













              The fact that matrices represent linear functions means that for a matrix $A$ and vectors $u, v$, we have $A(u +v) = A u + A v$. This doesn't tell us anything about what happens when we multiply matrices together.



              So yes, to prove distributivity you will need something more. If you consider the summation representation the proof is quite straightforward though.






              share|cite|improve this answer



























                up vote
                1
                down vote













                The fact that matrices represent linear functions means that for a matrix $A$ and vectors $u, v$, we have $A(u +v) = A u + A v$. This doesn't tell us anything about what happens when we multiply matrices together.



                So yes, to prove distributivity you will need something more. If you consider the summation representation the proof is quite straightforward though.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  The fact that matrices represent linear functions means that for a matrix $A$ and vectors $u, v$, we have $A(u +v) = A u + A v$. This doesn't tell us anything about what happens when we multiply matrices together.



                  So yes, to prove distributivity you will need something more. If you consider the summation representation the proof is quite straightforward though.






                  share|cite|improve this answer















                  The fact that matrices represent linear functions means that for a matrix $A$ and vectors $u, v$, we have $A(u +v) = A u + A v$. This doesn't tell us anything about what happens when we multiply matrices together.



                  So yes, to prove distributivity you will need something more. If you consider the summation representation the proof is quite straightforward though.







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 29 at 14:19


























                  answered Jul 29 at 14:14









                  Daniel Mroz

                  851314




                  851314






















                       

                      draft saved


                      draft discarded


























                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866109%2fdistributive-law-for-matrices%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What is the equation of a 3D cone with generalised tilt?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                      Read more

                      Color the edges and diagonals of a regular polygon

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                      Read more

                      Relationship between determinant of matrix and determinant of adjoint?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                      Read more