Representing a function in vector form.












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$begingroup$


As far as I understand, according to linear algebra, linear functions, both single and multivariable, can be represented in vector form.



For instance,



$$z = aw + bx + cy + d$$



can be rewritten as



$$z = begin{bmatrix}a & b&c &|& dend{bmatrix} cdot begin{bmatrix}w \ x \ y \ 1end{bmatrix}$$



My questions are:




  1. Am I correct with the above?

  2. Is it possible to rewrite the function $bbox[lightgray] {z = w^2 + x^2 + y^2 + 8}$ in vector form?










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    0












    $begingroup$


    As far as I understand, according to linear algebra, linear functions, both single and multivariable, can be represented in vector form.



    For instance,



    $$z = aw + bx + cy + d$$



    can be rewritten as



    $$z = begin{bmatrix}a & b&c &|& dend{bmatrix} cdot begin{bmatrix}w \ x \ y \ 1end{bmatrix}$$



    My questions are:




    1. Am I correct with the above?

    2. Is it possible to rewrite the function $bbox[lightgray] {z = w^2 + x^2 + y^2 + 8}$ in vector form?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      As far as I understand, according to linear algebra, linear functions, both single and multivariable, can be represented in vector form.



      For instance,



      $$z = aw + bx + cy + d$$



      can be rewritten as



      $$z = begin{bmatrix}a & b&c &|& dend{bmatrix} cdot begin{bmatrix}w \ x \ y \ 1end{bmatrix}$$



      My questions are:




      1. Am I correct with the above?

      2. Is it possible to rewrite the function $bbox[lightgray] {z = w^2 + x^2 + y^2 + 8}$ in vector form?










      share|cite|improve this question









      $endgroup$




      As far as I understand, according to linear algebra, linear functions, both single and multivariable, can be represented in vector form.



      For instance,



      $$z = aw + bx + cy + d$$



      can be rewritten as



      $$z = begin{bmatrix}a & b&c &|& dend{bmatrix} cdot begin{bmatrix}w \ x \ y \ 1end{bmatrix}$$



      My questions are:




      1. Am I correct with the above?

      2. Is it possible to rewrite the function $bbox[lightgray] {z = w^2 + x^2 + y^2 + 8}$ in vector form?







      functions vectors multivariate-polynomial






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      asked Dec 29 '18 at 17:40









      user366312user366312

      648520




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          2 Answers
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          $begingroup$

          In the same vein as your example, one can write $$w^2+x^2+y^2+8=begin{bmatrix}w&x&y&1end{bmatrix}begin{bmatrix}1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&8end{bmatrix}begin{bmatrix}w\x\y\1end{bmatrix}.$$






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            0












            $begingroup$

            If you let $$ u=<x,y,w>$$
            Then $$z=u.u+8=||u||^2 +8$$






            share|cite|improve this answer









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              2 Answers
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              $begingroup$

              In the same vein as your example, one can write $$w^2+x^2+y^2+8=begin{bmatrix}w&x&y&1end{bmatrix}begin{bmatrix}1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&8end{bmatrix}begin{bmatrix}w\x\y\1end{bmatrix}.$$






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                1












                $begingroup$

                In the same vein as your example, one can write $$w^2+x^2+y^2+8=begin{bmatrix}w&x&y&1end{bmatrix}begin{bmatrix}1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&8end{bmatrix}begin{bmatrix}w\x\y\1end{bmatrix}.$$






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                  1












                  1








                  1





                  $begingroup$

                  In the same vein as your example, one can write $$w^2+x^2+y^2+8=begin{bmatrix}w&x&y&1end{bmatrix}begin{bmatrix}1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&8end{bmatrix}begin{bmatrix}w\x\y\1end{bmatrix}.$$






                  share|cite|improve this answer









                  $endgroup$



                  In the same vein as your example, one can write $$w^2+x^2+y^2+8=begin{bmatrix}w&x&y&1end{bmatrix}begin{bmatrix}1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&8end{bmatrix}begin{bmatrix}w\x\y\1end{bmatrix}.$$







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                  answered Dec 30 '18 at 0:38









                  amdamd

                  32k21053




                  32k21053























                      0












                      $begingroup$

                      If you let $$ u=<x,y,w>$$
                      Then $$z=u.u+8=||u||^2 +8$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        If you let $$ u=<x,y,w>$$
                        Then $$z=u.u+8=||u||^2 +8$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          If you let $$ u=<x,y,w>$$
                          Then $$z=u.u+8=||u||^2 +8$$






                          share|cite|improve this answer









                          $endgroup$



                          If you let $$ u=<x,y,w>$$
                          Then $$z=u.u+8=||u||^2 +8$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 29 '18 at 17:49









                          Mohammad Riazi-KermaniMohammad Riazi-Kermani

                          42.2k42061




                          42.2k42061






























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