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Implicit function theorem intuition behind non-zero jacobian determinant

Prove this system of equations defines an implicit functionImplicit Function Theorem in Higher DimensionsApplication of Implicit Function TheoremQuestion on Inductive Proof of Implicit Function TheoremImplicit function theorem conclusion notation?Jacobian determinantRelaxing continuous differentiability in the implicit function theoremInterpreting Jacobian as instance of implicit function theorem testmultivariable implicit function theoremImplicit function theorem: The result about equivalence of partial derivatives

0

$begingroup$

Implicit function Theorem: In the general implicit function theorem for $m$ variables and $m$ implicit equations in the form
$$beginalign mathbf F(x_1,x_2,ldots,x_n, u_1, u_2, ldots, u_m) = 0 endalign$$
where $mathbf F=langle F_1, F_2,...,F_m rangle$

I have been introduced to the requirement that the square jacobian matrix for $mathbf F(u_1,...,u_n)$ must be invertible which means the determinant should be non zero. This is apparently analogous to requiring $fracpartial fpartial yne0$ for the $2D$ case $F(x,y)=0$.

Can someone please explain any intuition behind this requirement?

real-analysis multivariable-calculus implicit-function-theorem

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asked Mar 22 at 11:06

user523384user523384

227

$endgroup$

add a comment | 

0

$begingroup$

Implicit function Theorem: In the general implicit function theorem for $m$ variables and $m$ implicit equations in the form
$$beginalign mathbf F(x_1,x_2,ldots,x_n, u_1, u_2, ldots, u_m) = 0 endalign$$
where $mathbf F=langle F_1, F_2,...,F_m rangle$

I have been introduced to the requirement that the square jacobian matrix for $mathbf F(u_1,...,u_n)$ must be invertible which means the determinant should be non zero. This is apparently analogous to requiring $fracpartial fpartial yne0$ for the $2D$ case $F(x,y)=0$.

Can someone please explain any intuition behind this requirement?

real-analysis multivariable-calculus implicit-function-theorem

share|cite|improve this question

asked Mar 22 at 11:06

user523384user523384

227

$endgroup$

add a comment | 

$begingroup$

Implicit function Theorem: In the general implicit function theorem for $m$ variables and $m$ implicit equations in the form
$$beginalign mathbf F(x_1,x_2,ldots,x_n, u_1, u_2, ldots, u_m) = 0 endalign$$
where $mathbf F=langle F_1, F_2,...,F_m rangle$

I have been introduced to the requirement that the square jacobian matrix for $mathbf F(u_1,...,u_n)$ must be invertible which means the determinant should be non zero. This is apparently analogous to requiring $fracpartial fpartial yne0$ for the $2D$ case $F(x,y)=0$.

Can someone please explain any intuition behind this requirement?

real-analysis multivariable-calculus implicit-function-theorem

share|cite|improve this question

asked Mar 22 at 11:06

user523384user523384

227

$endgroup$

share|cite|improve this question

asked Mar 22 at 11:06

user523384user523384

227

share|cite|improve this question

asked Mar 22 at 11:06

user523384user523384

227

asked Mar 22 at 11:06

user523384user523384

227

asked Mar 22 at 11:06

user523384user523384

227

1 Answer
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1

$begingroup$

Consider the warm-up exercise, of formulating the implicit function theorem for linear functions. That is, where $mathbf F(x,u)=0$ can be given by a matrix multiplication formula like $$ Ax+Cu=b$$ where $x$ in an $n$-vector and $u$ an $m$-vector, with matrices $A$ and $C$ and fixed vector $b$. Linear algebra tells us this equation is always uniquely solveable for $u$ given $x$ precisely when the $mtimes m$ matrix $C$ is non-singular, that is, has non-vanishing determinant.

In this case $C$ is the Jacobian.

Now in the non-linear but differentiable case. A differentiable function is, intuitively, one which is well approximated by a linear one. One might expect that what holds in the exactly linear case carries over to the differentiable case. The technical content of the implicit function theorem is that this is so.

share|cite|improve this answer

edited Mar 22 at 20:23

answered Mar 22 at 14:02

kimchi loverkimchi lover

11.7k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this! It makes sense.
  $endgroup$
  – user523384
  Mar 23 at 9:57
  

  
  
  
  

add a comment | 

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1

$begingroup$

Consider the warm-up exercise, of formulating the implicit function theorem for linear functions. That is, where $mathbf F(x,u)=0$ can be given by a matrix multiplication formula like $$ Ax+Cu=b$$ where $x$ in an $n$-vector and $u$ an $m$-vector, with matrices $A$ and $C$ and fixed vector $b$. Linear algebra tells us this equation is always uniquely solveable for $u$ given $x$ precisely when the $mtimes m$ matrix $C$ is non-singular, that is, has non-vanishing determinant.

In this case $C$ is the Jacobian.

Now in the non-linear but differentiable case. A differentiable function is, intuitively, one which is well approximated by a linear one. One might expect that what holds in the exactly linear case carries over to the differentiable case. The technical content of the implicit function theorem is that this is so.

share|cite|improve this answer

edited Mar 22 at 20:23

answered Mar 22 at 14:02

kimchi loverkimchi lover

11.7k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this! It makes sense.
  $endgroup$
  – user523384
  Mar 23 at 9:57
  

  
  
  
  

add a comment | 

1

$begingroup$

Consider the warm-up exercise, of formulating the implicit function theorem for linear functions. That is, where $mathbf F(x,u)=0$ can be given by a matrix multiplication formula like $$ Ax+Cu=b$$ where $x$ in an $n$-vector and $u$ an $m$-vector, with matrices $A$ and $C$ and fixed vector $b$. Linear algebra tells us this equation is always uniquely solveable for $u$ given $x$ precisely when the $mtimes m$ matrix $C$ is non-singular, that is, has non-vanishing determinant.

In this case $C$ is the Jacobian.

Now in the non-linear but differentiable case. A differentiable function is, intuitively, one which is well approximated by a linear one. One might expect that what holds in the exactly linear case carries over to the differentiable case. The technical content of the implicit function theorem is that this is so.

share|cite|improve this answer

edited Mar 22 at 20:23

answered Mar 22 at 14:02

kimchi loverkimchi lover

11.7k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this! It makes sense.
  $endgroup$
  – user523384
  Mar 23 at 9:57
  

  
  
  
  

add a comment | 

$begingroup$

Consider the warm-up exercise, of formulating the implicit function theorem for linear functions. That is, where $mathbf F(x,u)=0$ can be given by a matrix multiplication formula like $$ Ax+Cu=b$$ where $x$ in an $n$-vector and $u$ an $m$-vector, with matrices $A$ and $C$ and fixed vector $b$. Linear algebra tells us this equation is always uniquely solveable for $u$ given $x$ precisely when the $mtimes m$ matrix $C$ is non-singular, that is, has non-vanishing determinant.

In this case $C$ is the Jacobian.

Now in the non-linear but differentiable case. A differentiable function is, intuitively, one which is well approximated by a linear one. One might expect that what holds in the exactly linear case carries over to the differentiable case. The technical content of the implicit function theorem is that this is so.

share|cite|improve this answer

edited Mar 22 at 20:23

answered Mar 22 at 14:02

kimchi loverkimchi lover

11.7k31229

$endgroup$

share|cite|improve this answer

edited Mar 22 at 20:23

answered Mar 22 at 14:02

kimchi loverkimchi lover

11.7k31229

answered Mar 22 at 14:02

kimchi loverkimchi lover

11.7k31229

answered Mar 22 at 14:02

kimchi loverkimchi lover

11.7k31229