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About the Jacobian determinant and the bijection

  1. Nov 21, 2008 #1

    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,

    Last edited: Nov 21, 2008
  2. jcsd
  3. Nov 21, 2008 #2
    F'(a) is a linear map and its being bijective is (by linear algebra) equivalent to det F'(a) not being equal to zero.
  4. Nov 21, 2008 #3
    Thanks and, could you tell me the name of this proposition?
  5. Nov 22, 2008 #4
    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" [Broken]
    Last edited by a moderator: May 3, 2017
  6. Nov 22, 2008 #5
    MERCI beaucoup~~ I will check it..
    Last edited by a moderator: May 3, 2017
  7. Nov 22, 2008 #6


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    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 any one of the variables as a function of the other two in some neighborhood of (x0, y0, z0).
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