Derivative of a Function of the Diag function












0












$begingroup$


Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:



$$
F(U)=traceleft(diag(U) A diag(U) right)
$$
where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.



$$
{partial F(U) over partial U } \
$$



I am more interested in the method used. Thanks in advance.






derivatives matrix-calculus trace







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$endgroup$

















    0












    $begingroup$


    Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:



    $$
    F(U)=traceleft(diag(U) A diag(U) right)
    $$
    where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.



    $$
    {partial F(U) over partial U } \
    $$



    I am more interested in the method used. Thanks in advance.






    derivatives matrix-calculus trace







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:



      $$
      F(U)=traceleft(diag(U) A diag(U) right)
      $$
      where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.



      $$
      {partial F(U) over partial U } \
      $$



      I am more interested in the method used. Thanks in advance.






      derivatives matrix-calculus trace







      share|cite|improve this question









      $endgroup$




      Suppose there is a vector $U in mathbb{R}^n$. How would you find the derivative of:



      $$
      F(U)=traceleft(diag(U) A diag(U) right)
      $$
      where $A in mathbb{R}^{n times n} succ 0 $ and where $diag(cdot)$ creates a diagonal matrix with $(cdot)$ on the leading diagonal. Where the derivative is taken with respect to the vector $U$, i.e.



      $$
      {partial F(U) over partial U } \
      $$



      I am more interested in the method used. Thanks in advance.






      derivatives matrix-calculus trace




      derivatives matrix-calculus trace






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 9:45









      p32fr4p32fr4

      377




      377






















          1 Answer
          1






          active

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          2












          $begingroup$

          Posting the solution I identified.



          Due to the trace operator evaluating the above is equivalent to evaluating:



          $$
          {partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
          left( begin{align}
          begin{array}
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
          vdots &\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
          end{array}end{align}right) \
          $$



          which becomes:



          $$
          {partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
          $$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
            $endgroup$
            – greg
            Dec 7 '18 at 23:39












          • $begingroup$
            That is a nice alternative thanks @greg
            $endgroup$
            – p32fr4
            Dec 21 '18 at 10:03











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Posting the solution I identified.



          Due to the trace operator evaluating the above is equivalent to evaluating:



          $$
          {partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
          left( begin{align}
          begin{array}
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
          vdots &\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
          end{array}end{align}right) \
          $$



          which becomes:



          $$
          {partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
          $$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
            $endgroup$
            – greg
            Dec 7 '18 at 23:39












          • $begingroup$
            That is a nice alternative thanks @greg
            $endgroup$
            – p32fr4
            Dec 21 '18 at 10:03
















          2












          $begingroup$

          Posting the solution I identified.



          Due to the trace operator evaluating the above is equivalent to evaluating:



          $$
          {partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
          left( begin{align}
          begin{array}
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
          vdots &\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
          end{array}end{align}right) \
          $$



          which becomes:



          $$
          {partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
          $$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
            $endgroup$
            – greg
            Dec 7 '18 at 23:39












          • $begingroup$
            That is a nice alternative thanks @greg
            $endgroup$
            – p32fr4
            Dec 21 '18 at 10:03














          2












          2








          2





          $begingroup$

          Posting the solution I identified.



          Due to the trace operator evaluating the above is equivalent to evaluating:



          $$
          {partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
          left( begin{align}
          begin{array}
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
          vdots &\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
          end{array}end{align}right) \
          $$



          which becomes:



          $$
          {partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
          $$






          share|cite|improve this answer









          $endgroup$



          Posting the solution I identified.



          Due to the trace operator evaluating the above is equivalent to evaluating:



          $$
          {partial left(sumlimits_{i=1}^{n} u_i A_{(i,i)}u_iright)over partial U }=
          left( begin{align}
          begin{array}
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_1}&\
          vdots &\
          {partial left(sumlimits_{i=1}^{n} u_iA_{(i,i)}u_iright)over partial u_n}&\
          end{array}end{align}right) \
          $$



          which becomes:



          $$
          {partial left( trace(diag(U) A diag(U)) right)over partial U} = 2 diag( A) U
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 10:47









          p32fr4p32fr4

          377




          377








          • 2




            $begingroup$
            You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
            $endgroup$
            – greg
            Dec 7 '18 at 23:39












          • $begingroup$
            That is a nice alternative thanks @greg
            $endgroup$
            – p32fr4
            Dec 21 '18 at 10:03














          • 2




            $begingroup$
            You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
            $endgroup$
            – greg
            Dec 7 '18 at 23:39












          • $begingroup$
            That is a nice alternative thanks @greg
            $endgroup$
            – p32fr4
            Dec 21 '18 at 10:03








          2




          2




          $begingroup$
          You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
          $endgroup$
          – greg
          Dec 7 '18 at 23:39






          $begingroup$
          You can also derive this using two identities involving the Hadamard product $${rm Diag}(x),A,{rm Diag}(y)=Aodot xy^T$$ $${rm Tr}(Aodot xy^T)=x^T(Iodot A)y=phi$$ So that $$dphi=y^T(Iodot A)dx+x^T(Iodot A)dy$$ Setting $y=x=u$ this becomes $$dphi=2u^T(Iodot A)du$$ $$frac{partialphi}{partial u}=2(Iodot A)u$$
          $endgroup$
          – greg
          Dec 7 '18 at 23:39














          $begingroup$
          That is a nice alternative thanks @greg
          $endgroup$
          – p32fr4
          Dec 21 '18 at 10:03




          $begingroup$
          That is a nice alternative thanks @greg
          $endgroup$
          – p32fr4
          Dec 21 '18 at 10:03


















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