Trace of product of gradient












2












$begingroup$


Consider vector field $v$. We know, that $Tr(nabla v)=div ~ v$. Is is true that $$Tr(nabla vnabla v)=(div ~ v)^2$$
Thanks for any hint.






analysis vector-analysis trace







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  • $begingroup$
    @mathcounterexamples.net $nabla v$ is matrix, so $nabla v nabla v$ is product of two matricies
    $endgroup$
    – lojdmoj
    Jan 2 at 14:33










  • $begingroup$
    @mathcounterexamples.net $url {math.stackexchange.com/questions/156880/…}$
    $endgroup$
    – lojdmoj
    Jan 2 at 14:47
















2












$begingroup$


Consider vector field $v$. We know, that $Tr(nabla v)=div ~ v$. Is is true that $$Tr(nabla vnabla v)=(div ~ v)^2$$
Thanks for any hint.






analysis vector-analysis trace







share|cite|improve this question









$endgroup$












  • $begingroup$
    @mathcounterexamples.net $nabla v$ is matrix, so $nabla v nabla v$ is product of two matricies
    $endgroup$
    – lojdmoj
    Jan 2 at 14:33










  • $begingroup$
    @mathcounterexamples.net $url {math.stackexchange.com/questions/156880/…}$
    $endgroup$
    – lojdmoj
    Jan 2 at 14:47














2












2








2





$begingroup$


Consider vector field $v$. We know, that $Tr(nabla v)=div ~ v$. Is is true that $$Tr(nabla vnabla v)=(div ~ v)^2$$
Thanks for any hint.






analysis vector-analysis trace







share|cite|improve this question









$endgroup$




Consider vector field $v$. We know, that $Tr(nabla v)=div ~ v$. Is is true that $$Tr(nabla vnabla v)=(div ~ v)^2$$
Thanks for any hint.






analysis vector-analysis trace




analysis vector-analysis trace






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 2 at 14:07









lojdmojlojdmoj

877




877












  • $begingroup$
    @mathcounterexamples.net $nabla v$ is matrix, so $nabla v nabla v$ is product of two matricies
    $endgroup$
    – lojdmoj
    Jan 2 at 14:33










  • $begingroup$
    @mathcounterexamples.net $url {math.stackexchange.com/questions/156880/…}$
    $endgroup$
    – lojdmoj
    Jan 2 at 14:47


















  • $begingroup$
    @mathcounterexamples.net $nabla v$ is matrix, so $nabla v nabla v$ is product of two matricies
    $endgroup$
    – lojdmoj
    Jan 2 at 14:33










  • $begingroup$
    @mathcounterexamples.net $url {math.stackexchange.com/questions/156880/…}$
    $endgroup$
    – lojdmoj
    Jan 2 at 14:47
















$begingroup$
@mathcounterexamples.net $nabla v$ is matrix, so $nabla v nabla v$ is product of two matricies
$endgroup$
– lojdmoj
Jan 2 at 14:33




$begingroup$
@mathcounterexamples.net $nabla v$ is matrix, so $nabla v nabla v$ is product of two matricies
$endgroup$
– lojdmoj
Jan 2 at 14:33












$begingroup$
@mathcounterexamples.net $url {math.stackexchange.com/questions/156880/…}$
$endgroup$
– lojdmoj
Jan 2 at 14:47




$begingroup$
@mathcounterexamples.net $url {math.stackexchange.com/questions/156880/…}$
$endgroup$
– lojdmoj
Jan 2 at 14:47










2 Answers
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$begingroup$

By considering $nabla v$ to be the Jacobian of the vector field $v$. One can show that:



$$nabla v nabla v = begin{bmatrix}
nabla v_1^T\
dots\
nabla v_n^T
end{bmatrix} begin{bmatrix}
partial_{x_1}v & dots & partial_{x_n}v
end{bmatrix}
= begin{bmatrix}
left< nabla v_1, partial_{x_1}vright>& dots & dots\
dots & dots & dots \
dots & dots & left< nabla v_n, partial_{x_n}vright>
end{bmatrix}$$



Hence,



$$Tr(nabla v nabla v) = sum_{i=1}^{n} left< nabla v_i, partial_{x_i} vright > ;;(1)$$



And $$left< nabla v_i, partial_{x_i} vright > = sum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i};;(2)$$



Therefore, from (1) and (2):



$$Tr(nabla v nabla v) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i} ;;(3)$$



Now $$text{div}(v) = sum_{i=1}^n frac{partial v_i}{partial x_i}$$



$$left ( text{div}(v) right )^2 = left (sum_{i=1}^n frac{partial v_i}{partial x_i} right )left (sum_{k=1}^n frac{partial v_k}{partial x_k} right ) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_i} frac{partial v_k}{partial x_k};;(4)$$



From (3) and (4) we deduce that $Tr(nabla v nabla v) neq left ( text{div}(v) right )^2$.



Sorry for the computational "proof". I hope I find a more geometric one in the future.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    A simple counterexample in $mathbb R^n$.



    Take the identity vector field $J: x mapsto x$. Then the Jacobian $nabla J$ is the identity matrix $I_n$. So



    $$Tr(nabla J cdot nabla J)= Tr(I_n)= n neq n^2= (div I)^2$$






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      0












      $begingroup$

      By considering $nabla v$ to be the Jacobian of the vector field $v$. One can show that:



      $$nabla v nabla v = begin{bmatrix}
      nabla v_1^T\
      dots\
      nabla v_n^T
      end{bmatrix} begin{bmatrix}
      partial_{x_1}v & dots & partial_{x_n}v
      end{bmatrix}
      = begin{bmatrix}
      left< nabla v_1, partial_{x_1}vright>& dots & dots\
      dots & dots & dots \
      dots & dots & left< nabla v_n, partial_{x_n}vright>
      end{bmatrix}$$



      Hence,



      $$Tr(nabla v nabla v) = sum_{i=1}^{n} left< nabla v_i, partial_{x_i} vright > ;;(1)$$



      And $$left< nabla v_i, partial_{x_i} vright > = sum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i};;(2)$$



      Therefore, from (1) and (2):



      $$Tr(nabla v nabla v) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i} ;;(3)$$



      Now $$text{div}(v) = sum_{i=1}^n frac{partial v_i}{partial x_i}$$



      $$left ( text{div}(v) right )^2 = left (sum_{i=1}^n frac{partial v_i}{partial x_i} right )left (sum_{k=1}^n frac{partial v_k}{partial x_k} right ) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_i} frac{partial v_k}{partial x_k};;(4)$$



      From (3) and (4) we deduce that $Tr(nabla v nabla v) neq left ( text{div}(v) right )^2$.



      Sorry for the computational "proof". I hope I find a more geometric one in the future.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        By considering $nabla v$ to be the Jacobian of the vector field $v$. One can show that:



        $$nabla v nabla v = begin{bmatrix}
        nabla v_1^T\
        dots\
        nabla v_n^T
        end{bmatrix} begin{bmatrix}
        partial_{x_1}v & dots & partial_{x_n}v
        end{bmatrix}
        = begin{bmatrix}
        left< nabla v_1, partial_{x_1}vright>& dots & dots\
        dots & dots & dots \
        dots & dots & left< nabla v_n, partial_{x_n}vright>
        end{bmatrix}$$



        Hence,



        $$Tr(nabla v nabla v) = sum_{i=1}^{n} left< nabla v_i, partial_{x_i} vright > ;;(1)$$



        And $$left< nabla v_i, partial_{x_i} vright > = sum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i};;(2)$$



        Therefore, from (1) and (2):



        $$Tr(nabla v nabla v) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i} ;;(3)$$



        Now $$text{div}(v) = sum_{i=1}^n frac{partial v_i}{partial x_i}$$



        $$left ( text{div}(v) right )^2 = left (sum_{i=1}^n frac{partial v_i}{partial x_i} right )left (sum_{k=1}^n frac{partial v_k}{partial x_k} right ) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_i} frac{partial v_k}{partial x_k};;(4)$$



        From (3) and (4) we deduce that $Tr(nabla v nabla v) neq left ( text{div}(v) right )^2$.



        Sorry for the computational "proof". I hope I find a more geometric one in the future.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          By considering $nabla v$ to be the Jacobian of the vector field $v$. One can show that:



          $$nabla v nabla v = begin{bmatrix}
          nabla v_1^T\
          dots\
          nabla v_n^T
          end{bmatrix} begin{bmatrix}
          partial_{x_1}v & dots & partial_{x_n}v
          end{bmatrix}
          = begin{bmatrix}
          left< nabla v_1, partial_{x_1}vright>& dots & dots\
          dots & dots & dots \
          dots & dots & left< nabla v_n, partial_{x_n}vright>
          end{bmatrix}$$



          Hence,



          $$Tr(nabla v nabla v) = sum_{i=1}^{n} left< nabla v_i, partial_{x_i} vright > ;;(1)$$



          And $$left< nabla v_i, partial_{x_i} vright > = sum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i};;(2)$$



          Therefore, from (1) and (2):



          $$Tr(nabla v nabla v) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i} ;;(3)$$



          Now $$text{div}(v) = sum_{i=1}^n frac{partial v_i}{partial x_i}$$



          $$left ( text{div}(v) right )^2 = left (sum_{i=1}^n frac{partial v_i}{partial x_i} right )left (sum_{k=1}^n frac{partial v_k}{partial x_k} right ) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_i} frac{partial v_k}{partial x_k};;(4)$$



          From (3) and (4) we deduce that $Tr(nabla v nabla v) neq left ( text{div}(v) right )^2$.



          Sorry for the computational "proof". I hope I find a more geometric one in the future.






          share|cite|improve this answer









          $endgroup$



          By considering $nabla v$ to be the Jacobian of the vector field $v$. One can show that:



          $$nabla v nabla v = begin{bmatrix}
          nabla v_1^T\
          dots\
          nabla v_n^T
          end{bmatrix} begin{bmatrix}
          partial_{x_1}v & dots & partial_{x_n}v
          end{bmatrix}
          = begin{bmatrix}
          left< nabla v_1, partial_{x_1}vright>& dots & dots\
          dots & dots & dots \
          dots & dots & left< nabla v_n, partial_{x_n}vright>
          end{bmatrix}$$



          Hence,



          $$Tr(nabla v nabla v) = sum_{i=1}^{n} left< nabla v_i, partial_{x_i} vright > ;;(1)$$



          And $$left< nabla v_i, partial_{x_i} vright > = sum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i};;(2)$$



          Therefore, from (1) and (2):



          $$Tr(nabla v nabla v) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_k} frac{partial v_k}{partial x_i} ;;(3)$$



          Now $$text{div}(v) = sum_{i=1}^n frac{partial v_i}{partial x_i}$$



          $$left ( text{div}(v) right )^2 = left (sum_{i=1}^n frac{partial v_i}{partial x_i} right )left (sum_{k=1}^n frac{partial v_k}{partial x_k} right ) = sum_{i=1}^nsum_{k=1}^n frac{partial v_i}{partial x_i} frac{partial v_k}{partial x_k};;(4)$$



          From (3) and (4) we deduce that $Tr(nabla v nabla v) neq left ( text{div}(v) right )^2$.



          Sorry for the computational "proof". I hope I find a more geometric one in the future.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 14:50









          pedrothpedroth

          765




          765























              1












              $begingroup$

              A simple counterexample in $mathbb R^n$.



              Take the identity vector field $J: x mapsto x$. Then the Jacobian $nabla J$ is the identity matrix $I_n$. So



              $$Tr(nabla J cdot nabla J)= Tr(I_n)= n neq n^2= (div I)^2$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                A simple counterexample in $mathbb R^n$.



                Take the identity vector field $J: x mapsto x$. Then the Jacobian $nabla J$ is the identity matrix $I_n$. So



                $$Tr(nabla J cdot nabla J)= Tr(I_n)= n neq n^2= (div I)^2$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A simple counterexample in $mathbb R^n$.



                  Take the identity vector field $J: x mapsto x$. Then the Jacobian $nabla J$ is the identity matrix $I_n$. So



                  $$Tr(nabla J cdot nabla J)= Tr(I_n)= n neq n^2= (div I)^2$$






                  share|cite|improve this answer











                  $endgroup$



                  A simple counterexample in $mathbb R^n$.



                  Take the identity vector field $J: x mapsto x$. Then the Jacobian $nabla J$ is the identity matrix $I_n$. So



                  $$Tr(nabla J cdot nabla J)= Tr(I_n)= n neq n^2= (div I)^2$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 2 at 15:08

























                  answered Jan 2 at 15:02









                  mathcounterexamples.netmathcounterexamples.net

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