Gradient of $||Ax - y||^2$ with respect to $A$












1














How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.



Attempt so far



$$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$



$$ nabla_A(x^TAy) = xy^T$$



Where I am stuck



I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.






vector-analysis matrix-calculus







share|cite|improve this question





























    1














    How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.



    Attempt so far



    $$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$



    $$ nabla_A(x^TAy) = xy^T$$



    Where I am stuck



    I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.






    vector-analysis matrix-calculus







    share|cite|improve this question



























      1












      1








      1







      How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.



      Attempt so far



      $$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$



      $$ nabla_A(x^TAy) = xy^T$$



      Where I am stuck



      I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.






      vector-analysis matrix-calculus







      share|cite|improve this question















      How do I proceed to find $nabla_A||Ax - y||^2$ where $A in mathbb{R}^{ntimes n}$ and $x,y in mathbb{R}^n$ and the norm is the Euclidean norm.



      Attempt so far



      $$||Ax - y||^2 = (Ax-y)^T(Ax-y) = x^TA^TAx - 2x^TAy + y^Ty $$



      $$ nabla_A(x^TAy) = xy^T$$



      Where I am stuck



      I don't know how to tackle the $x^TA^TAx$ term since if I try to apply chain rule, I will have to differentiate a matrix with respect to a matrix.






      vector-analysis matrix-calculus




      vector-analysis matrix-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 9:31









      user550103

      6711315




      6711315










      asked Nov 22 at 6:38









      Anant Joshi

      451211




      451211






















          2 Answers
          2






          active

          oldest

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          1














          Before we start deriving the gradient, some facts and notations for brevity:




          • Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$

          • Cyclic properties of Trace/Frobenius product
            begin{align}
            A : B C
            &= BC : A \
            &= A C^T : B \
            &= {text{etc.}} cr
            end{align}


          Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.



          Now, we can obtain the differential first, and then the gradient.
          begin{align}
          df
          &= dleft( Ax-y:Ax-y right) \
          &= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
          &= 2 left(Ax - yright) : dA x \
          &= 2left( Ax-yright)x^T : dA\
          end{align}



          Thus, the gradient is
          begin{align}
          frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
          end{align}






          share|cite|improve this answer























          • What is the difference between differential and gradient?
            – Anant Joshi
            Nov 22 at 9:19










          • see this math.stackexchange.com/questions/289923/…
            – user550103
            Nov 22 at 9:20



















          0














          For matrices, the most easy way is often to get back to definition of differentiability, i.e. :



          $$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$



          With $L_A$ a linear map.



          We begin with :
          $$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$



          Then we have :



          $$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$



          To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$



          Thus :



          $$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$



          With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.






          share|cite|improve this answer























          • Why is the final answer a number and not a matrix?
            – Anant Joshi
            Nov 22 at 9:13










          • Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
            – nicomezi
            Nov 22 at 9:21











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          Before we start deriving the gradient, some facts and notations for brevity:




          • Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$

          • Cyclic properties of Trace/Frobenius product
            begin{align}
            A : B C
            &= BC : A \
            &= A C^T : B \
            &= {text{etc.}} cr
            end{align}


          Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.



          Now, we can obtain the differential first, and then the gradient.
          begin{align}
          df
          &= dleft( Ax-y:Ax-y right) \
          &= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
          &= 2 left(Ax - yright) : dA x \
          &= 2left( Ax-yright)x^T : dA\
          end{align}



          Thus, the gradient is
          begin{align}
          frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
          end{align}






          share|cite|improve this answer























          • What is the difference between differential and gradient?
            – Anant Joshi
            Nov 22 at 9:19










          • see this math.stackexchange.com/questions/289923/…
            – user550103
            Nov 22 at 9:20
















          1














          Before we start deriving the gradient, some facts and notations for brevity:




          • Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$

          • Cyclic properties of Trace/Frobenius product
            begin{align}
            A : B C
            &= BC : A \
            &= A C^T : B \
            &= {text{etc.}} cr
            end{align}


          Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.



          Now, we can obtain the differential first, and then the gradient.
          begin{align}
          df
          &= dleft( Ax-y:Ax-y right) \
          &= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
          &= 2 left(Ax - yright) : dA x \
          &= 2left( Ax-yright)x^T : dA\
          end{align}



          Thus, the gradient is
          begin{align}
          frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
          end{align}






          share|cite|improve this answer























          • What is the difference between differential and gradient?
            – Anant Joshi
            Nov 22 at 9:19










          • see this math.stackexchange.com/questions/289923/…
            – user550103
            Nov 22 at 9:20














          1












          1








          1






          Before we start deriving the gradient, some facts and notations for brevity:




          • Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$

          • Cyclic properties of Trace/Frobenius product
            begin{align}
            A : B C
            &= BC : A \
            &= A C^T : B \
            &= {text{etc.}} cr
            end{align}


          Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.



          Now, we can obtain the differential first, and then the gradient.
          begin{align}
          df
          &= dleft( Ax-y:Ax-y right) \
          &= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
          &= 2 left(Ax - yright) : dA x \
          &= 2left( Ax-yright)x^T : dA\
          end{align}



          Thus, the gradient is
          begin{align}
          frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
          end{align}






          share|cite|improve this answer














          Before we start deriving the gradient, some facts and notations for brevity:




          • Trace and Frobenius product relation $$leftlangle A, B Crightrangle={rm tr}(A^TBC) := A : B C$$

          • Cyclic properties of Trace/Frobenius product
            begin{align}
            A : B C
            &= BC : A \
            &= A C^T : B \
            &= {text{etc.}} cr
            end{align}


          Let $f := left|Ax-y right|^2 = Ax-y:Ax-y$.



          Now, we can obtain the differential first, and then the gradient.
          begin{align}
          df
          &= dleft( Ax-y:Ax-y right) \
          &= left(dA x : Ax-yright) + left(Ax-y : dA xright) \
          &= 2 left(Ax - yright) : dA x \
          &= 2left( Ax-yright)x^T : dA\
          end{align}



          Thus, the gradient is
          begin{align}
          frac{partial}{partial A} left( left|Ax-y right|^2 right)= 2left( Ax-yright)x^T.
          end{align}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 9:17

























          answered Nov 22 at 9:08









          user550103

          6711315




          6711315












          • What is the difference between differential and gradient?
            – Anant Joshi
            Nov 22 at 9:19










          • see this math.stackexchange.com/questions/289923/…
            – user550103
            Nov 22 at 9:20


















          • What is the difference between differential and gradient?
            – Anant Joshi
            Nov 22 at 9:19










          • see this math.stackexchange.com/questions/289923/…
            – user550103
            Nov 22 at 9:20
















          What is the difference between differential and gradient?
          – Anant Joshi
          Nov 22 at 9:19




          What is the difference between differential and gradient?
          – Anant Joshi
          Nov 22 at 9:19












          see this math.stackexchange.com/questions/289923/…
          – user550103
          Nov 22 at 9:20




          see this math.stackexchange.com/questions/289923/…
          – user550103
          Nov 22 at 9:20











          0














          For matrices, the most easy way is often to get back to definition of differentiability, i.e. :



          $$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$



          With $L_A$ a linear map.



          We begin with :
          $$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$



          Then we have :



          $$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$



          To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$



          Thus :



          $$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$



          With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.






          share|cite|improve this answer























          • Why is the final answer a number and not a matrix?
            – Anant Joshi
            Nov 22 at 9:13










          • Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
            – nicomezi
            Nov 22 at 9:21
















          0














          For matrices, the most easy way is often to get back to definition of differentiability, i.e. :



          $$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$



          With $L_A$ a linear map.



          We begin with :
          $$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$



          Then we have :



          $$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$



          To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$



          Thus :



          $$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$



          With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.






          share|cite|improve this answer























          • Why is the final answer a number and not a matrix?
            – Anant Joshi
            Nov 22 at 9:13










          • Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
            – nicomezi
            Nov 22 at 9:21














          0












          0








          0






          For matrices, the most easy way is often to get back to definition of differentiability, i.e. :



          $$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$



          With $L_A$ a linear map.



          We begin with :
          $$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$



          Then we have :



          $$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$



          To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$



          Thus :



          $$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$



          With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.






          share|cite|improve this answer














          For matrices, the most easy way is often to get back to definition of differentiability, i.e. :



          $$||(A+H)x - y||^2-||Ax - y||^2=L_A(H) + o(|H|)$$



          With $L_A$ a linear map.



          We begin with :
          $$||(A+H)x - y||^2=langle Ax+Hx-y , Ax+Hx-y rangle.$$



          Then we have :



          $$||(A+H)x - y||^2= langle Ax-y,Ax-yrangle + 2langle Hx,Ax -yrangle + langle Hx,Hxrangle.$$



          To conclude : $$||(A+H)x - y||^2-||Ax - y||^2=2langle Hx,Ax -yrangle + o (|H|)$$



          Thus :



          $$left(nabla_A||Ax - y||^2right)_{i,j}=2langle E_{i,j}x,Ax -yrangle$$



          With $E_{i,j}$ the matrix with a $1$ at row $i$ and column $j$ and $0$ otherwise.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 9:20

























          answered Nov 22 at 9:11









          nicomezi

          4,0741820




          4,0741820












          • Why is the final answer a number and not a matrix?
            – Anant Joshi
            Nov 22 at 9:13










          • Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
            – nicomezi
            Nov 22 at 9:21


















          • Why is the final answer a number and not a matrix?
            – Anant Joshi
            Nov 22 at 9:13










          • Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
            – nicomezi
            Nov 22 at 9:21
















          Why is the final answer a number and not a matrix?
          – Anant Joshi
          Nov 22 at 9:13




          Why is the final answer a number and not a matrix?
          – Anant Joshi
          Nov 22 at 9:13












          Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
          – nicomezi
          Nov 22 at 9:21




          Because I misunderstood the notation and went a bit fast at the end. Is it better now ?
          – nicomezi
          Nov 22 at 9:21


















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