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Jacobian of the inverse function



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Partial derivatives of a function which is constant on the diagonalShow that the given function is a diffeomorphismImplicit function theorem conclusion notation?Can we get a contradiction of the inverse function theorem?Existence of vector valued function with transpose jacobianWhat are the inverse operations of the “Partial derivative” and the “Total derivative”?The connection between the Jacobian, Hessian and the gradient?Using the chain rule to find the derivative of the inverse of a functionAn application of the inverse function theorem?Coordinate invariance of Jacobian?










0












$begingroup$


Let $F=$$y_i(x_1,x_2,..,x_n)$$_i=1^n$:$R^nrightarrow $$R^n$ be a map with continous nonvanishing partial derivatives then its jacobian at a point $p$ is defined by the matrix $M_i,j(F)=partial y_i/partial x_jlvert_p $. Now by the inverse function theorem $F$ has an inverse $F^-1=$$x_i(y_1,y_2,..,y_n)$ in a neigbourhood of $F(p)$ and so its jacobian must be defined by $M_i,j(F^-1)=partial x_i/partial y_jlvert_F(p) $. However from the univariate version of the inverse function theorem $partial x_i/partial y_jlvert_F(p)=(partial y_i/partial x_jlvert_p)^-1$



CONCLUSION:



Therefore one can conclude that jacobian of the inverse function $F^-1$ at the point $F(p)$ is the same as the matrix of the jacobian of $F$ at $p$ but with its terms reciprocated.



By the multivariable case of inverse function theorem my conclusion is wrong infact it follows from the chain rule that $M_i,j(F^-1)=M_i,j(F)^-1$.



Therefore my conclusion is wrong however I dont understand what is wrong behind my reasoning.



Thanks in advance










share|cite|improve this question











$endgroup$











  • $begingroup$
    "with its terms reciprocated" Utterly false. The Jacobian matrix of $F^-1$ is the inverse of the Jacobian matrix of $F$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 27 at 18:18






  • 1




    $begingroup$
    Try a simple example such as $F:(x,y)mapsto(x+y,y)$ to see how this fails.
    $endgroup$
    – amd
    Mar 27 at 18:22
















0












$begingroup$


Let $F=$$y_i(x_1,x_2,..,x_n)$$_i=1^n$:$R^nrightarrow $$R^n$ be a map with continous nonvanishing partial derivatives then its jacobian at a point $p$ is defined by the matrix $M_i,j(F)=partial y_i/partial x_jlvert_p $. Now by the inverse function theorem $F$ has an inverse $F^-1=$$x_i(y_1,y_2,..,y_n)$ in a neigbourhood of $F(p)$ and so its jacobian must be defined by $M_i,j(F^-1)=partial x_i/partial y_jlvert_F(p) $. However from the univariate version of the inverse function theorem $partial x_i/partial y_jlvert_F(p)=(partial y_i/partial x_jlvert_p)^-1$



CONCLUSION:



Therefore one can conclude that jacobian of the inverse function $F^-1$ at the point $F(p)$ is the same as the matrix of the jacobian of $F$ at $p$ but with its terms reciprocated.



By the multivariable case of inverse function theorem my conclusion is wrong infact it follows from the chain rule that $M_i,j(F^-1)=M_i,j(F)^-1$.



Therefore my conclusion is wrong however I dont understand what is wrong behind my reasoning.



Thanks in advance










share|cite|improve this question











$endgroup$











  • $begingroup$
    "with its terms reciprocated" Utterly false. The Jacobian matrix of $F^-1$ is the inverse of the Jacobian matrix of $F$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 27 at 18:18






  • 1




    $begingroup$
    Try a simple example such as $F:(x,y)mapsto(x+y,y)$ to see how this fails.
    $endgroup$
    – amd
    Mar 27 at 18:22














0












0








0





$begingroup$


Let $F=$$y_i(x_1,x_2,..,x_n)$$_i=1^n$:$R^nrightarrow $$R^n$ be a map with continous nonvanishing partial derivatives then its jacobian at a point $p$ is defined by the matrix $M_i,j(F)=partial y_i/partial x_jlvert_p $. Now by the inverse function theorem $F$ has an inverse $F^-1=$$x_i(y_1,y_2,..,y_n)$ in a neigbourhood of $F(p)$ and so its jacobian must be defined by $M_i,j(F^-1)=partial x_i/partial y_jlvert_F(p) $. However from the univariate version of the inverse function theorem $partial x_i/partial y_jlvert_F(p)=(partial y_i/partial x_jlvert_p)^-1$



CONCLUSION:



Therefore one can conclude that jacobian of the inverse function $F^-1$ at the point $F(p)$ is the same as the matrix of the jacobian of $F$ at $p$ but with its terms reciprocated.



By the multivariable case of inverse function theorem my conclusion is wrong infact it follows from the chain rule that $M_i,j(F^-1)=M_i,j(F)^-1$.



Therefore my conclusion is wrong however I dont understand what is wrong behind my reasoning.



Thanks in advance










share|cite|improve this question











$endgroup$




Let $F=$$y_i(x_1,x_2,..,x_n)$$_i=1^n$:$R^nrightarrow $$R^n$ be a map with continous nonvanishing partial derivatives then its jacobian at a point $p$ is defined by the matrix $M_i,j(F)=partial y_i/partial x_jlvert_p $. Now by the inverse function theorem $F$ has an inverse $F^-1=$$x_i(y_1,y_2,..,y_n)$ in a neigbourhood of $F(p)$ and so its jacobian must be defined by $M_i,j(F^-1)=partial x_i/partial y_jlvert_F(p) $. However from the univariate version of the inverse function theorem $partial x_i/partial y_jlvert_F(p)=(partial y_i/partial x_jlvert_p)^-1$



CONCLUSION:



Therefore one can conclude that jacobian of the inverse function $F^-1$ at the point $F(p)$ is the same as the matrix of the jacobian of $F$ at $p$ but with its terms reciprocated.



By the multivariable case of inverse function theorem my conclusion is wrong infact it follows from the chain rule that $M_i,j(F^-1)=M_i,j(F)^-1$.



Therefore my conclusion is wrong however I dont understand what is wrong behind my reasoning.



Thanks in advance







multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 27 at 18:37







TheGeometer

















asked Mar 27 at 18:00









TheGeometerTheGeometer

942619




942619











  • $begingroup$
    "with its terms reciprocated" Utterly false. The Jacobian matrix of $F^-1$ is the inverse of the Jacobian matrix of $F$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 27 at 18:18






  • 1




    $begingroup$
    Try a simple example such as $F:(x,y)mapsto(x+y,y)$ to see how this fails.
    $endgroup$
    – amd
    Mar 27 at 18:22

















  • $begingroup$
    "with its terms reciprocated" Utterly false. The Jacobian matrix of $F^-1$ is the inverse of the Jacobian matrix of $F$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 27 at 18:18






  • 1




    $begingroup$
    Try a simple example such as $F:(x,y)mapsto(x+y,y)$ to see how this fails.
    $endgroup$
    – amd
    Mar 27 at 18:22
















$begingroup$
"with its terms reciprocated" Utterly false. The Jacobian matrix of $F^-1$ is the inverse of the Jacobian matrix of $F$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 27 at 18:18




$begingroup$
"with its terms reciprocated" Utterly false. The Jacobian matrix of $F^-1$ is the inverse of the Jacobian matrix of $F$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 27 at 18:18




1




1




$begingroup$
Try a simple example such as $F:(x,y)mapsto(x+y,y)$ to see how this fails.
$endgroup$
– amd
Mar 27 at 18:22





$begingroup$
Try a simple example such as $F:(x,y)mapsto(x+y,y)$ to see how this fails.
$endgroup$
– amd
Mar 27 at 18:22











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