Is there an easy way to compute the jacobian of a normalized vector?











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If $x in mathbb{R}^3$ I want to compute the jacobian of the following function



$$
f(x) = frac{x}{lVert x rVert }
$$



If I proceed I get a matrix whose elements are



$$
a_{ij} = begin{cases}
frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
-frac{x_i x_j}{lVert x rVert^3} &i neq j
end{cases}
$$



Is this the most compact form?
The derivation is based on the product rule componentwise.






multivariable-calculus jacobian







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    If $x in mathbb{R}^3$ I want to compute the jacobian of the following function



    $$
    f(x) = frac{x}{lVert x rVert }
    $$



    If I proceed I get a matrix whose elements are



    $$
    a_{ij} = begin{cases}
    frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
    -frac{x_i x_j}{lVert x rVert^3} &i neq j
    end{cases}
    $$



    Is this the most compact form?
    The derivation is based on the product rule componentwise.






    multivariable-calculus jacobian







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      If $x in mathbb{R}^3$ I want to compute the jacobian of the following function



      $$
      f(x) = frac{x}{lVert x rVert }
      $$



      If I proceed I get a matrix whose elements are



      $$
      a_{ij} = begin{cases}
      frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
      -frac{x_i x_j}{lVert x rVert^3} &i neq j
      end{cases}
      $$



      Is this the most compact form?
      The derivation is based on the product rule componentwise.






      multivariable-calculus jacobian







      share|cite|improve this question















      If $x in mathbb{R}^3$ I want to compute the jacobian of the following function



      $$
      f(x) = frac{x}{lVert x rVert }
      $$



      If I proceed I get a matrix whose elements are



      $$
      a_{ij} = begin{cases}
      frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
      -frac{x_i x_j}{lVert x rVert^3} &i neq j
      end{cases}
      $$



      Is this the most compact form?
      The derivation is based on the product rule componentwise.






      multivariable-calculus jacobian




      multivariable-calculus jacobian






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 at 9:08

























      asked Aug 14 at 18:09









      user8469759

      1,3361616




      1,3361616






















          1 Answer
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          I get the same result:
          $$
          begin{align}
          J_{ij}
          &= partial_j left( frac{x}{lVert x rVert} right)_i \
          &= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
          &= frac{delta_{ij}}{lVert x rVert} +
          x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
          &= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
          end{align}
          $$






          share|cite|improve this answer





















          • Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
            – user8469759
            Aug 14 at 18:55













          Your Answer





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          up vote
          0
          down vote













          I get the same result:
          $$
          begin{align}
          J_{ij}
          &= partial_j left( frac{x}{lVert x rVert} right)_i \
          &= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
          &= frac{delta_{ij}}{lVert x rVert} +
          x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
          &= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
          end{align}
          $$






          share|cite|improve this answer





















          • Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
            – user8469759
            Aug 14 at 18:55

















          up vote
          0
          down vote













          I get the same result:
          $$
          begin{align}
          J_{ij}
          &= partial_j left( frac{x}{lVert x rVert} right)_i \
          &= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
          &= frac{delta_{ij}}{lVert x rVert} +
          x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
          &= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
          end{align}
          $$






          share|cite|improve this answer





















          • Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
            – user8469759
            Aug 14 at 18:55















          up vote
          0
          down vote










          up vote
          0
          down vote









          I get the same result:
          $$
          begin{align}
          J_{ij}
          &= partial_j left( frac{x}{lVert x rVert} right)_i \
          &= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
          &= frac{delta_{ij}}{lVert x rVert} +
          x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
          &= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
          end{align}
          $$






          share|cite|improve this answer












          I get the same result:
          $$
          begin{align}
          J_{ij}
          &= partial_j left( frac{x}{lVert x rVert} right)_i \
          &= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
          &= frac{delta_{ij}}{lVert x rVert} +
          x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
          &= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
          end{align}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 14 at 18:20









          mvw

          31.3k22252




          31.3k22252












          • Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
            – user8469759
            Aug 14 at 18:55




















          • Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
            – user8469759
            Aug 14 at 18:55


















          Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
          – user8469759
          Aug 14 at 18:55






          Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
          – user8469759
          Aug 14 at 18:55




















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