About the Jacobian determinant and the bijection

[simpleeyelid]

Hello!

I am having problems with the inverse function theorem.

In some books it says to be locally inversible: first C1, 2nd Jacobian determinant different from 0

And I saw some books say to be locally inversible, it suffices to change the 2NDto "F'(a) is bijective"..

How could these two be equivalent.

Thank you for your kindness in advance,

Sincerely

 

[Pere Callahan]

F'(a) is a linear map and its being bijective is (by linear algebra) equivalent to det F'(a) not being equal to zero.

 

[simpleeyelid]

F'(a) is a linear map and its being bijective is (by linear algebra) equivalent to det F'(a) not being equal to zero.

Thanks and, could you tell me the name of this proposition?

 

[Pere Callahan]

Thanks and, could you tell me the name of this proposition?

I'm not sure if this proposition has a special name. You may look up in wikipedia the equivalent conditions for a matrix to be invertible (which means that the associated linear map is an vector space isomorphism, i.e. bijective.)

http://en.wikipedia.org/wiki/Matrix_inversion"

 

[simpleeyelid]

I'm not sure if this proposition has a special name. You may look up in wikipedia the equivalent conditions for a matrix to be invertible (which means that the associated linear map is an vector space isomorphism, i.e. bijective.)

http://en.wikipedia.org/wiki/Matrix_inversion"

MERCI beaucoup~~ I will check it..

 

[HallsofIvy]

Thanks and, could you tell me the name of this proposition?

The general theorem is the "implicit function theorem" which basically says if the Jacobian of f(x,y,z) is not 0 at (x₀, y₀, z₀) then we can solve for anyone of the variables as a function of the other two in some neighborhood of (x₀, y₀, z₀).