Going from matrix form to summation for calculating the derivative












1












$begingroup$


I have troubles calculating the derivatives of:
$frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$



I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?






summation matrix-calculus







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    1












    $begingroup$


    I have troubles calculating the derivatives of:
    $frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$



    I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?






    summation matrix-calculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have troubles calculating the derivatives of:
      $frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$



      I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?






      summation matrix-calculus







      share|cite|improve this question









      $endgroup$




      I have troubles calculating the derivatives of:
      $frac{dy}{dx}$ if $y= vec{x}^{T}matrix{W}vec{x}$



      I think it would be easier if I go to summation form and then take it from there. However, i'm not sure on how to do that? Same thing when I have a summation to go back to matrix form. Is there an easy step by step guide for this?






      summation matrix-calculus




      summation matrix-calculus






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      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 31 at 8:53









      Jesper.LindbergJesper.Lindberg

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          $begingroup$

          Hint:



          Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$



          Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$



          Then, we have
          begin{align}
          (mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
          end{align}






          share|cite|improve this answer









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            0












            $begingroup$

            Hint:



            Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$



            Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$



            Then, we have
            begin{align}
            (mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
            end{align}






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Hint:



              Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$



              Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$



              Then, we have
              begin{align}
              (mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
              end{align}






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint:



                Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$



                Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$



                Then, we have
                begin{align}
                (mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
                end{align}






                share|cite|improve this answer









                $endgroup$



                Hint:



                Let $mathbf{A}$ be a matrix of size $n times m$ with elements $A_{ik}$ for $i = 1, 2 dots, n$ and $k = 1, 2, dots, m$



                Let $mathbf{B}$ be a matrix of size $m times p$ with elements $B_{kj}$ for $k = 1, 2 dots, m$ and $j = 1, 2, dots, p$



                Then, we have
                begin{align}
                (mathbf{A} , mathbf{B})_{ij} ; = ; sum_{k=1}^{m}A_{ik} , B_{kj}
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 31 at 9:01









                K_inverseK_inverse

                288213




                288213






























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