Derivative of matrix expression $(Y − Abeta)^TW(Y − Abeta)$ wrt $beta$.












1












$begingroup$


$Y$ and $beta$ are $1 times n$ matrices and $W$ is a diagonal $n times n$ matrix.



What is the best way to think about how to simplify this expression and its derivative to get the expression below? What are the simple rules I should remember to get this?



$2A^TWAbeta − 2A^TW^TY$






matrices matrix-equations matrix-calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    $Y$ and $beta$ are $1 times n$ matrices and $W$ is a diagonal $n times n$ matrix.



    What is the best way to think about how to simplify this expression and its derivative to get the expression below? What are the simple rules I should remember to get this?



    $2A^TWAbeta − 2A^TW^TY$






    matrices matrix-equations matrix-calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      $Y$ and $beta$ are $1 times n$ matrices and $W$ is a diagonal $n times n$ matrix.



      What is the best way to think about how to simplify this expression and its derivative to get the expression below? What are the simple rules I should remember to get this?



      $2A^TWAbeta − 2A^TW^TY$






      matrices matrix-equations matrix-calculus







      share|cite|improve this question











      $endgroup$




      $Y$ and $beta$ are $1 times n$ matrices and $W$ is a diagonal $n times n$ matrix.



      What is the best way to think about how to simplify this expression and its derivative to get the expression below? What are the simple rules I should remember to get this?



      $2A^TWAbeta − 2A^TW^TY$






      matrices matrix-equations matrix-calculus




      matrices matrix-equations matrix-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 16:28







      user3408780

















      asked Dec 8 '18 at 22:41









      user3408780user3408780

      62




      62






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          Define the vectors
          $$eqalign{
          g &= (Ab-y) &implies dg=A,db cr
          h &= (W!Ab-Wy) &implies dh=W!A,db cr
          }$$
          Write the function in terms of these new variables and find its differential and gradient.
          $$eqalign{
          f &= g^Th cr
          df &= h^Tdg + g^Tdh cr
          &= (h^TA+g^TWA),db cr
          &= (A^Th+A^TW^Tg)^T,db cr
          frac{partial f}{partial b}
          &= A^Th+A^TW^Tg cr
          &= A^T(WAb-Wy)+A^TW^T(Ab-y) cr
          &= A^T(W+W^T)Ab - A^T(W+W^T)y cr
          }$$
          If $W=W^T$ this can be simplied to
          $$eqalign{
          frac{partial f}{partial b}
          &= 2A^TWAb - 2A^TWy cr
          }$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes forgot to mention W is a diagonal matrix
            $endgroup$
            – user3408780
            Dec 9 '18 at 15:43



















          0












          $begingroup$

          Throughout I implicitly sum over repeated indices. You're differentiating a scalar $Z_i W_{ij}Z_j,,Z_i:=Y_i-A_{ik}beta_k$ with respect to a vector $beta$, giving a vector whose $k$th component is obtained by differentiating with respect to $beta_k$. Since $partial_k:=tfrac{partial}{partialbeta_k}impliespartial_k Z_i=-A_{ik}$, the product rule obtains $partial_k(Z_i W_{ij}Z_j)=-A_{ik}W_{ij}Z_j-Z_i W_{ij}A_{jk}$. Now you can use the rules of matrix multiplication to rewrite this neatly as $-[Z^T(W+W^T)A]_k$, making the derivative $-Z^T(W+W^T)A=(Abeta-Y)^T(W+W^T)A$.



          Now for a sanity check, which is worthwhile with any calculation this complex. If $W$ were a scalar instead, we'd have $partial_k Z^TWZ=Wpartial_k (Z^TZ)=-2Z^TWA$. But when we reinstate $W$'s matrix status, we note its antisymmetric part doesn't even contribute to the scalar we're differentiating, so without loss of generality $W$ should be replaced throughout with its symmetric part $(W+W^T)/2$. That gives $-Z^T(W+W^T)A$ instead, as we've found.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Define the vectors
            $$eqalign{
            g &= (Ab-y) &implies dg=A,db cr
            h &= (W!Ab-Wy) &implies dh=W!A,db cr
            }$$
            Write the function in terms of these new variables and find its differential and gradient.
            $$eqalign{
            f &= g^Th cr
            df &= h^Tdg + g^Tdh cr
            &= (h^TA+g^TWA),db cr
            &= (A^Th+A^TW^Tg)^T,db cr
            frac{partial f}{partial b}
            &= A^Th+A^TW^Tg cr
            &= A^T(WAb-Wy)+A^TW^T(Ab-y) cr
            &= A^T(W+W^T)Ab - A^T(W+W^T)y cr
            }$$
            If $W=W^T$ this can be simplied to
            $$eqalign{
            frac{partial f}{partial b}
            &= 2A^TWAb - 2A^TWy cr
            }$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yes forgot to mention W is a diagonal matrix
              $endgroup$
              – user3408780
              Dec 9 '18 at 15:43
















            1












            $begingroup$

            Define the vectors
            $$eqalign{
            g &= (Ab-y) &implies dg=A,db cr
            h &= (W!Ab-Wy) &implies dh=W!A,db cr
            }$$
            Write the function in terms of these new variables and find its differential and gradient.
            $$eqalign{
            f &= g^Th cr
            df &= h^Tdg + g^Tdh cr
            &= (h^TA+g^TWA),db cr
            &= (A^Th+A^TW^Tg)^T,db cr
            frac{partial f}{partial b}
            &= A^Th+A^TW^Tg cr
            &= A^T(WAb-Wy)+A^TW^T(Ab-y) cr
            &= A^T(W+W^T)Ab - A^T(W+W^T)y cr
            }$$
            If $W=W^T$ this can be simplied to
            $$eqalign{
            frac{partial f}{partial b}
            &= 2A^TWAb - 2A^TWy cr
            }$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yes forgot to mention W is a diagonal matrix
              $endgroup$
              – user3408780
              Dec 9 '18 at 15:43














            1












            1








            1





            $begingroup$

            Define the vectors
            $$eqalign{
            g &= (Ab-y) &implies dg=A,db cr
            h &= (W!Ab-Wy) &implies dh=W!A,db cr
            }$$
            Write the function in terms of these new variables and find its differential and gradient.
            $$eqalign{
            f &= g^Th cr
            df &= h^Tdg + g^Tdh cr
            &= (h^TA+g^TWA),db cr
            &= (A^Th+A^TW^Tg)^T,db cr
            frac{partial f}{partial b}
            &= A^Th+A^TW^Tg cr
            &= A^T(WAb-Wy)+A^TW^T(Ab-y) cr
            &= A^T(W+W^T)Ab - A^T(W+W^T)y cr
            }$$
            If $W=W^T$ this can be simplied to
            $$eqalign{
            frac{partial f}{partial b}
            &= 2A^TWAb - 2A^TWy cr
            }$$






            share|cite|improve this answer









            $endgroup$



            Define the vectors
            $$eqalign{
            g &= (Ab-y) &implies dg=A,db cr
            h &= (W!Ab-Wy) &implies dh=W!A,db cr
            }$$
            Write the function in terms of these new variables and find its differential and gradient.
            $$eqalign{
            f &= g^Th cr
            df &= h^Tdg + g^Tdh cr
            &= (h^TA+g^TWA),db cr
            &= (A^Th+A^TW^Tg)^T,db cr
            frac{partial f}{partial b}
            &= A^Th+A^TW^Tg cr
            &= A^T(WAb-Wy)+A^TW^T(Ab-y) cr
            &= A^T(W+W^T)Ab - A^T(W+W^T)y cr
            }$$
            If $W=W^T$ this can be simplied to
            $$eqalign{
            frac{partial f}{partial b}
            &= 2A^TWAb - 2A^TWy cr
            }$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 2:40









            greggreg

            8,2951823




            8,2951823












            • $begingroup$
              Yes forgot to mention W is a diagonal matrix
              $endgroup$
              – user3408780
              Dec 9 '18 at 15:43


















            • $begingroup$
              Yes forgot to mention W is a diagonal matrix
              $endgroup$
              – user3408780
              Dec 9 '18 at 15:43
















            $begingroup$
            Yes forgot to mention W is a diagonal matrix
            $endgroup$
            – user3408780
            Dec 9 '18 at 15:43




            $begingroup$
            Yes forgot to mention W is a diagonal matrix
            $endgroup$
            – user3408780
            Dec 9 '18 at 15:43











            0












            $begingroup$

            Throughout I implicitly sum over repeated indices. You're differentiating a scalar $Z_i W_{ij}Z_j,,Z_i:=Y_i-A_{ik}beta_k$ with respect to a vector $beta$, giving a vector whose $k$th component is obtained by differentiating with respect to $beta_k$. Since $partial_k:=tfrac{partial}{partialbeta_k}impliespartial_k Z_i=-A_{ik}$, the product rule obtains $partial_k(Z_i W_{ij}Z_j)=-A_{ik}W_{ij}Z_j-Z_i W_{ij}A_{jk}$. Now you can use the rules of matrix multiplication to rewrite this neatly as $-[Z^T(W+W^T)A]_k$, making the derivative $-Z^T(W+W^T)A=(Abeta-Y)^T(W+W^T)A$.



            Now for a sanity check, which is worthwhile with any calculation this complex. If $W$ were a scalar instead, we'd have $partial_k Z^TWZ=Wpartial_k (Z^TZ)=-2Z^TWA$. But when we reinstate $W$'s matrix status, we note its antisymmetric part doesn't even contribute to the scalar we're differentiating, so without loss of generality $W$ should be replaced throughout with its symmetric part $(W+W^T)/2$. That gives $-Z^T(W+W^T)A$ instead, as we've found.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Throughout I implicitly sum over repeated indices. You're differentiating a scalar $Z_i W_{ij}Z_j,,Z_i:=Y_i-A_{ik}beta_k$ with respect to a vector $beta$, giving a vector whose $k$th component is obtained by differentiating with respect to $beta_k$. Since $partial_k:=tfrac{partial}{partialbeta_k}impliespartial_k Z_i=-A_{ik}$, the product rule obtains $partial_k(Z_i W_{ij}Z_j)=-A_{ik}W_{ij}Z_j-Z_i W_{ij}A_{jk}$. Now you can use the rules of matrix multiplication to rewrite this neatly as $-[Z^T(W+W^T)A]_k$, making the derivative $-Z^T(W+W^T)A=(Abeta-Y)^T(W+W^T)A$.



              Now for a sanity check, which is worthwhile with any calculation this complex. If $W$ were a scalar instead, we'd have $partial_k Z^TWZ=Wpartial_k (Z^TZ)=-2Z^TWA$. But when we reinstate $W$'s matrix status, we note its antisymmetric part doesn't even contribute to the scalar we're differentiating, so without loss of generality $W$ should be replaced throughout with its symmetric part $(W+W^T)/2$. That gives $-Z^T(W+W^T)A$ instead, as we've found.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Throughout I implicitly sum over repeated indices. You're differentiating a scalar $Z_i W_{ij}Z_j,,Z_i:=Y_i-A_{ik}beta_k$ with respect to a vector $beta$, giving a vector whose $k$th component is obtained by differentiating with respect to $beta_k$. Since $partial_k:=tfrac{partial}{partialbeta_k}impliespartial_k Z_i=-A_{ik}$, the product rule obtains $partial_k(Z_i W_{ij}Z_j)=-A_{ik}W_{ij}Z_j-Z_i W_{ij}A_{jk}$. Now you can use the rules of matrix multiplication to rewrite this neatly as $-[Z^T(W+W^T)A]_k$, making the derivative $-Z^T(W+W^T)A=(Abeta-Y)^T(W+W^T)A$.



                Now for a sanity check, which is worthwhile with any calculation this complex. If $W$ were a scalar instead, we'd have $partial_k Z^TWZ=Wpartial_k (Z^TZ)=-2Z^TWA$. But when we reinstate $W$'s matrix status, we note its antisymmetric part doesn't even contribute to the scalar we're differentiating, so without loss of generality $W$ should be replaced throughout with its symmetric part $(W+W^T)/2$. That gives $-Z^T(W+W^T)A$ instead, as we've found.






                share|cite|improve this answer









                $endgroup$



                Throughout I implicitly sum over repeated indices. You're differentiating a scalar $Z_i W_{ij}Z_j,,Z_i:=Y_i-A_{ik}beta_k$ with respect to a vector $beta$, giving a vector whose $k$th component is obtained by differentiating with respect to $beta_k$. Since $partial_k:=tfrac{partial}{partialbeta_k}impliespartial_k Z_i=-A_{ik}$, the product rule obtains $partial_k(Z_i W_{ij}Z_j)=-A_{ik}W_{ij}Z_j-Z_i W_{ij}A_{jk}$. Now you can use the rules of matrix multiplication to rewrite this neatly as $-[Z^T(W+W^T)A]_k$, making the derivative $-Z^T(W+W^T)A=(Abeta-Y)^T(W+W^T)A$.



                Now for a sanity check, which is worthwhile with any calculation this complex. If $W$ were a scalar instead, we'd have $partial_k Z^TWZ=Wpartial_k (Z^TZ)=-2Z^TWA$. But when we reinstate $W$'s matrix status, we note its antisymmetric part doesn't even contribute to the scalar we're differentiating, so without loss of generality $W$ should be replaced throughout with its symmetric part $(W+W^T)/2$. That gives $-Z^T(W+W^T)A$ instead, as we've found.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 8 '18 at 22:59









                J.G.J.G.

                26.7k22742




                26.7k22742






























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