Laplacian of $ nabla^2 f( A x)$ in terms of Laplacian of $ nabla^2_y f( y)$












1












$begingroup$


How to find the Laplacian
begin{align}
nabla^2_x f( A x)
end{align}
where $n times m$ matrix and we know know the laplacian of
begin{align}
nabla^2_y f( y)
end{align}



This question is about chain rule for the Laplacian.

I search but could not find a formula for it.



I think that the answer is
begin{align}
nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
end{align}






matrix-calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    How to find the Laplacian
    begin{align}
    nabla^2_x f( A x)
    end{align}
    where $n times m$ matrix and we know know the laplacian of
    begin{align}
    nabla^2_y f( y)
    end{align}



    This question is about chain rule for the Laplacian.

    I search but could not find a formula for it.



    I think that the answer is
    begin{align}
    nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
    end{align}






    matrix-calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      How to find the Laplacian
      begin{align}
      nabla^2_x f( A x)
      end{align}
      where $n times m$ matrix and we know know the laplacian of
      begin{align}
      nabla^2_y f( y)
      end{align}



      This question is about chain rule for the Laplacian.

      I search but could not find a formula for it.



      I think that the answer is
      begin{align}
      nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
      end{align}






      matrix-calculus







      share|cite|improve this question











      $endgroup$




      How to find the Laplacian
      begin{align}
      nabla^2_x f( A x)
      end{align}
      where $n times m$ matrix and we know know the laplacian of
      begin{align}
      nabla^2_y f( y)
      end{align}



      This question is about chain rule for the Laplacian.

      I search but could not find a formula for it.



      I think that the answer is
      begin{align}
      nabla^2_x f( A x)= {rm Tr}( AA^T ) nabla^2_y f( y) |_{y=Ax}
      end{align}






      matrix-calculus




      matrix-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 0:14







      Lisa

















      asked Dec 13 '18 at 23:23









      LisaLisa

      643313




      643313






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          The gradient of the function $f(y)$ is the vector $g$ which satisfies
          $$df = g^Tdy$$
          and the hessian is the matrix $H$ which satisfies
          $$dg = H,dy$$
          and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.



          Suppose a new independent vector $x$ is introduced with $,y=Ax$.



          It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.



          To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
          $$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
          Working these out is straightforward
          $$eqalign{
          df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
          g_* &= A^Tg crcr
          dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
          H_* &= A^THA cr
          L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
          }$$
          The last expression is the laplacian in terms of the new variable.

          Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
            $endgroup$
            – Lisa
            Dec 18 '18 at 15:32












          • $begingroup$
            @Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
            $endgroup$
            – greg
            Dec 18 '18 at 17:03










          • $begingroup$
            I see now. Thank you.
            $endgroup$
            – Lisa
            Dec 18 '18 at 17:41











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          The gradient of the function $f(y)$ is the vector $g$ which satisfies
          $$df = g^Tdy$$
          and the hessian is the matrix $H$ which satisfies
          $$dg = H,dy$$
          and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.



          Suppose a new independent vector $x$ is introduced with $,y=Ax$.



          It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.



          To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
          $$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
          Working these out is straightforward
          $$eqalign{
          df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
          g_* &= A^Tg crcr
          dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
          H_* &= A^THA cr
          L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
          }$$
          The last expression is the laplacian in terms of the new variable.

          Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
            $endgroup$
            – Lisa
            Dec 18 '18 at 15:32












          • $begingroup$
            @Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
            $endgroup$
            – greg
            Dec 18 '18 at 17:03










          • $begingroup$
            I see now. Thank you.
            $endgroup$
            – Lisa
            Dec 18 '18 at 17:41
















          3












          $begingroup$

          The gradient of the function $f(y)$ is the vector $g$ which satisfies
          $$df = g^Tdy$$
          and the hessian is the matrix $H$ which satisfies
          $$dg = H,dy$$
          and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.



          Suppose a new independent vector $x$ is introduced with $,y=Ax$.



          It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.



          To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
          $$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
          Working these out is straightforward
          $$eqalign{
          df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
          g_* &= A^Tg crcr
          dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
          H_* &= A^THA cr
          L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
          }$$
          The last expression is the laplacian in terms of the new variable.

          Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
            $endgroup$
            – Lisa
            Dec 18 '18 at 15:32












          • $begingroup$
            @Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
            $endgroup$
            – greg
            Dec 18 '18 at 17:03










          • $begingroup$
            I see now. Thank you.
            $endgroup$
            – Lisa
            Dec 18 '18 at 17:41














          3












          3








          3





          $begingroup$

          The gradient of the function $f(y)$ is the vector $g$ which satisfies
          $$df = g^Tdy$$
          and the hessian is the matrix $H$ which satisfies
          $$dg = H,dy$$
          and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.



          Suppose a new independent vector $x$ is introduced with $,y=Ax$.



          It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.



          To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
          $$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
          Working these out is straightforward
          $$eqalign{
          df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
          g_* &= A^Tg crcr
          dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
          H_* &= A^THA cr
          L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
          }$$
          The last expression is the laplacian in terms of the new variable.

          Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.






          share|cite|improve this answer











          $endgroup$



          The gradient of the function $f(y)$ is the vector $g$ which satisfies
          $$df = g^Tdy$$
          and the hessian is the matrix $H$ which satisfies
          $$dg = H,dy$$
          and the laplacian is the trace of the hessian $,,L={rm Tr}(H)$.



          Suppose a new independent vector $x$ is introduced with $,y=Ax$.



          It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.



          To find the gradient and hessian wrt $x$, one wants modifie${rm d^*}$ quantities which satisfy
          $$df=g_*^Tdx,,,,,,,dg_*=H_*,dx,,,,,,,L_*={rm Tr}(H_*)$$
          Working these out is straightforward
          $$eqalign{
          df &= g^Tdy = g^TA,dx = (A^Tg)^Tdx = g_*^Tdx cr
          g_* &= A^Tg crcr
          dg_* &= A^Tdg = A^T(H,(dy)) = A^TH(A,dx) = H_*,dx cr
          H_* &= A^THA cr
          L_* &= {rm Tr}(A^THA) = {rm Tr}(AA^TH) cr
          }$$
          The last expression is the laplacian in terms of the new variable.

          Unfortunately, knowing the value of ${rm Tr}(H)$ is of no help in evaluating ${rm Tr}(AA^TH)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 5:31

























          answered Dec 14 '18 at 2:45









          greggreg

          7,6201821




          7,6201821












          • $begingroup$
            Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
            $endgroup$
            – Lisa
            Dec 18 '18 at 15:32












          • $begingroup$
            @Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
            $endgroup$
            – greg
            Dec 18 '18 at 17:03










          • $begingroup$
            I see now. Thank you.
            $endgroup$
            – Lisa
            Dec 18 '18 at 17:41


















          • $begingroup$
            Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
            $endgroup$
            – Lisa
            Dec 18 '18 at 15:32












          • $begingroup$
            @Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
            $endgroup$
            – greg
            Dec 18 '18 at 17:03










          • $begingroup$
            I see now. Thank you.
            $endgroup$
            – Lisa
            Dec 18 '18 at 17:41
















          $begingroup$
          Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
          $endgroup$
          – Lisa
          Dec 18 '18 at 15:32






          $begingroup$
          Thank you so much. I wanted to an example. Let $f(x)=|x |^2$. I know that $nabla^2 f(x)=2 n$ where $n$ the dimension of $x$. Then $nabla^2 f(A x)= 2n Tr (A A^T )$, right?
          $endgroup$
          – Lisa
          Dec 18 '18 at 15:32














          $begingroup$
          @Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
          $endgroup$
          – greg
          Dec 18 '18 at 17:03




          $begingroup$
          @Lisa That's not right. Just substitute your sample function in the above formulas and you'll find $$eqalign{ H &=2I &implies L=2,{rm Tr}(I) &= 2n cr H_*&=A^T(2I)A &implies L_*=2,{rm Tr}(A^TA) &= 2|A|^2_F cr }$$
          $endgroup$
          – greg
          Dec 18 '18 at 17:03












          $begingroup$
          I see now. Thank you.
          $endgroup$
          – Lisa
          Dec 18 '18 at 17:41




          $begingroup$
          I see now. Thank you.
          $endgroup$
          – Lisa
          Dec 18 '18 at 17:41


















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