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Jacobian of the inverse function

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

multivariable-calculus

share|cite|improve this question

edited Mar 27 at 18:37

TheGeometer

asked Mar 27 at 18:00

TheGeometerTheGeometer

942619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "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
  
  

  
  
  
  

add a comment | 

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

multivariable-calculus

share|cite|improve this question

edited Mar 27 at 18:37

TheGeometer

asked Mar 27 at 18:00

TheGeometerTheGeometer

942619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "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
  
  

  
  
  
  

add a comment | 

$begingroup$

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

multivariable-calculus

share|cite|improve this question

edited Mar 27 at 18:37

TheGeometer

asked Mar 27 at 18:00

TheGeometerTheGeometer

942619

$endgroup$

share|cite|improve this question

edited Mar 27 at 18:37

TheGeometer

asked Mar 27 at 18:00

TheGeometerTheGeometer

942619

share|cite|improve this question

edited Mar 27 at 18:37

TheGeometer

asked Mar 27 at 18:00

TheGeometerTheGeometer

942619

asked Mar 27 at 18:00

TheGeometerTheGeometer

942619

asked Mar 27 at 18:00

TheGeometerTheGeometer

942619