Inverse function theorem for partial derivatives of a vector function












0












$begingroup$


I have a simple vector function:



$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$



The inverse is obviously:



$$mathbf{y}^{-1} = {1over a}mathbf{x}$$



The inverse function theorem (at least for scalars) states that:



$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$



But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:



$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$



What is the relation between the above derivatives, then?






multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have a simple vector function:



    $$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$



    The inverse is obviously:



    $$mathbf{y}^{-1} = {1over a}mathbf{x}$$



    The inverse function theorem (at least for scalars) states that:



    $$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$



    But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:



    $$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$



    What is the relation between the above derivatives, then?






    multivariable-calculus partial-derivative







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have a simple vector function:



      $$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$



      The inverse is obviously:



      $$mathbf{y}^{-1} = {1over a}mathbf{x}$$



      The inverse function theorem (at least for scalars) states that:



      $$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$



      But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:



      $$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$



      What is the relation between the above derivatives, then?






      multivariable-calculus partial-derivative







      share|cite|improve this question











      $endgroup$




      I have a simple vector function:



      $$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$



      The inverse is obviously:



      $$mathbf{y}^{-1} = {1over a}mathbf{x}$$



      The inverse function theorem (at least for scalars) states that:



      $$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$



      But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:



      $$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$



      What is the relation between the above derivatives, then?






      multivariable-calculus partial-derivative




      multivariable-calculus partial-derivative






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 9 '18 at 17:22







      user99914

















      asked Feb 25 '15 at 2:19









      LiborLibor

      8781822




      8781822






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix



          $$J_F(x) = begin{pmatrix}
          dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
          vdots & ddots & vdots\
          dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$



          And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :



          $$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$



          You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
            $endgroup$
            – Libor
            Feb 25 '15 at 2:34












          • $begingroup$
            The Jacobian gives you the partial derivatives...
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:37










          • $begingroup$
            But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:40












          • $begingroup$
            Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:43












          • $begingroup$
            Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:48





















          0












          $begingroup$

          The derivative w.r.t. $a$ is



          $${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$



          and the derivative of inverse is the reciproval of above, i.e.



          $${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$



          The notation caused confusion about what is the inverse function. The inverse is function of $a$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What is $1/mathbf x$ when $mathbf x$ is a vector?
            $endgroup$
            – M. Winter
            Jan 10 '18 at 13:17











          Your Answer





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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix



          $$J_F(x) = begin{pmatrix}
          dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
          vdots & ddots & vdots\
          dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$



          And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :



          $$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$



          You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
            $endgroup$
            – Libor
            Feb 25 '15 at 2:34












          • $begingroup$
            The Jacobian gives you the partial derivatives...
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:37










          • $begingroup$
            But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:40












          • $begingroup$
            Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:43












          • $begingroup$
            Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:48


















          0












          $begingroup$

          The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix



          $$J_F(x) = begin{pmatrix}
          dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
          vdots & ddots & vdots\
          dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$



          And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :



          $$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$



          You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
            $endgroup$
            – Libor
            Feb 25 '15 at 2:34












          • $begingroup$
            The Jacobian gives you the partial derivatives...
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:37










          • $begingroup$
            But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:40












          • $begingroup$
            Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:43












          • $begingroup$
            Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:48
















          0












          0








          0





          $begingroup$

          The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix



          $$J_F(x) = begin{pmatrix}
          dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
          vdots & ddots & vdots\
          dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$



          And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :



          $$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$



          You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem






          share|cite|improve this answer











          $endgroup$



          The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix



          $$J_F(x) = begin{pmatrix}
          dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
          vdots & ddots & vdots\
          dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$



          And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :



          $$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$



          You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 25 '15 at 16:03









          Libor

          8781822




          8781822










          answered Feb 25 '15 at 2:29









          TryssTryss

          13k1229




          13k1229












          • $begingroup$
            I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
            $endgroup$
            – Libor
            Feb 25 '15 at 2:34












          • $begingroup$
            The Jacobian gives you the partial derivatives...
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:37










          • $begingroup$
            But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:40












          • $begingroup$
            Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:43












          • $begingroup$
            Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:48




















          • $begingroup$
            I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
            $endgroup$
            – Libor
            Feb 25 '15 at 2:34












          • $begingroup$
            The Jacobian gives you the partial derivatives...
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:37










          • $begingroup$
            But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:40












          • $begingroup$
            Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
            $endgroup$
            – Tryss
            Feb 25 '15 at 2:43












          • $begingroup$
            Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
            $endgroup$
            – Libor
            Feb 25 '15 at 2:48


















          $begingroup$
          I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
          $endgroup$
          – Libor
          Feb 25 '15 at 2:34






          $begingroup$
          I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
          $endgroup$
          – Libor
          Feb 25 '15 at 2:34














          $begingroup$
          The Jacobian gives you the partial derivatives...
          $endgroup$
          – Tryss
          Feb 25 '15 at 2:37




          $begingroup$
          The Jacobian gives you the partial derivatives...
          $endgroup$
          – Tryss
          Feb 25 '15 at 2:37












          $begingroup$
          But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
          $endgroup$
          – Libor
          Feb 25 '15 at 2:40






          $begingroup$
          But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
          $endgroup$
          – Libor
          Feb 25 '15 at 2:40














          $begingroup$
          Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
          $endgroup$
          – Tryss
          Feb 25 '15 at 2:43






          $begingroup$
          Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
          $endgroup$
          – Tryss
          Feb 25 '15 at 2:43














          $begingroup$
          Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
          $endgroup$
          – Libor
          Feb 25 '15 at 2:48






          $begingroup$
          Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
          $endgroup$
          – Libor
          Feb 25 '15 at 2:48













          0












          $begingroup$

          The derivative w.r.t. $a$ is



          $${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$



          and the derivative of inverse is the reciproval of above, i.e.



          $${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$



          The notation caused confusion about what is the inverse function. The inverse is function of $a$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What is $1/mathbf x$ when $mathbf x$ is a vector?
            $endgroup$
            – M. Winter
            Jan 10 '18 at 13:17
















          0












          $begingroup$

          The derivative w.r.t. $a$ is



          $${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$



          and the derivative of inverse is the reciproval of above, i.e.



          $${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$



          The notation caused confusion about what is the inverse function. The inverse is function of $a$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What is $1/mathbf x$ when $mathbf x$ is a vector?
            $endgroup$
            – M. Winter
            Jan 10 '18 at 13:17














          0












          0








          0





          $begingroup$

          The derivative w.r.t. $a$ is



          $${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$



          and the derivative of inverse is the reciproval of above, i.e.



          $${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$



          The notation caused confusion about what is the inverse function. The inverse is function of $a$.






          share|cite|improve this answer











          $endgroup$



          The derivative w.r.t. $a$ is



          $${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$



          and the derivative of inverse is the reciproval of above, i.e.



          $${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$



          The notation caused confusion about what is the inverse function. The inverse is function of $a$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 25 '15 at 18:29

























          answered Feb 25 '15 at 17:55









          LiborLibor

          8781822




          8781822












          • $begingroup$
            What is $1/mathbf x$ when $mathbf x$ is a vector?
            $endgroup$
            – M. Winter
            Jan 10 '18 at 13:17


















          • $begingroup$
            What is $1/mathbf x$ when $mathbf x$ is a vector?
            $endgroup$
            – M. Winter
            Jan 10 '18 at 13:17
















          $begingroup$
          What is $1/mathbf x$ when $mathbf x$ is a vector?
          $endgroup$
          – M. Winter
          Jan 10 '18 at 13:17




          $begingroup$
          What is $1/mathbf x$ when $mathbf x$ is a vector?
          $endgroup$
          – M. Winter
          Jan 10 '18 at 13:17


















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