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Inverse Function Thm

  1. Nov 3, 2006 #1
    Im not sure whether this is a "Homework Question", but it is a question regarding the proof of the Inverse Function Theorem. It starts like this:
    Let k be the linear transformation Df(a). Then k is non-singular, since det(f '(a)) != 0. Now D((k^-1(f(a))) = D(k^-1)(f(a)) (Df(a)) = k^-1 (Df(a)) is the identity linear transformation.

    Heres what i dont understand:
    If the theorem is true for k^-1 (f) then it is clearly true for f. Therefore we may assume at the outset the k is the identity.

    Can anyone explain this?
     
  2. jcsd
  3. Nov 3, 2006 #2

    mathwonk

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    you are trying to prove a certain function is a local homeomorphism. if it is them composing it with an invertible linear map will not chNGE THIS, AND ALSO IOF IT IS NOT COMPOSING WITH an invertible linear map will not changw that.

    so we may compose it with an invertible linear map before starting the proof.

    i.e. if we wnT to prove f is invertible, and if L is kown to be invertible, then if we prove fT is invertible, we may conclude that also fTT^I(-1) = f is invertible.


    the purpose of this reduction is to be able to simplify the derivative.
     
  4. Nov 5, 2006 #3
    Ok so f:Rn->Rn. and by the fact that k: Rn->Rn is a homeomorphism on
    an open set:
    If k^-1 (f) is a homeomorphism on an open set then f is a
    homeomorphism on an open set. Thus it suffices to prove that k^-1 (f)
    is a homeomorphism on an open set. (An open set containing the point a
    where d is continuously differentiable).

    But why can you assume that k is the identity map?
     
  5. Nov 5, 2006 #4

    matt grime

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    Becuase you've just shown that the result is true for arbitrary k (satisfying the hypotheses) if and only if it is true for the identity.


    This is perfectly normal. Any result in linear algebra about a vector v can often be translated to showing it for the zero vector only.

    The analytic version is to simply rescale so that Df, which just a matrix of derivatives, is the identity.
     
  6. Nov 5, 2006 #5
    ok so if the theorem is true for k = I then it is true for arbitrary k and from what I said before we can conclude that it is true for f?
     
  7. Nov 5, 2006 #6

    matt grime

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    It's just a change of basis argument - draw a picture in 2-d for the y=f(x) case to see what it's saying: if the slope is non-zero at a point we may assume that it is 1. I.e. if f'(0)=2, say, then f(x)/2 is a function whose derivative is 1 at x=0. (the general case is more complicated, it is not just dividing by a number, but the principle is the same).


    You can also assume that a=(0,0,..,0) as well, by similar arguments.
     
    Last edited: Nov 5, 2006
  8. Nov 5, 2006 #7
    Well actually if its true for k = I then k^-1 f reduces to f and therefore its true for f. It seems a little too simplified of an assumption. Is my logic correct?
     
  9. Nov 5, 2006 #8
    ok the one variable case makes sense. If its true for f(x)/2 then it is true for f(x).
     
  10. Nov 5, 2006 #9
    ok then i think my post number 7 is flawed because when were assuming that k = I were changing the nature of the function (like from f(x) to f(x)/2) so its actually what I said in post num 5 thats true right?
     
  11. Nov 5, 2006 #10

    matt grime

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    No, you're missing the point. The assumption is not that 'because k=I, we then have that k^-1f=f' at all. I mean, it's true, but not relevant.

    The assumption is that we may assume f satisfies Df(a)=I, because if it didn't the function k^-1f would satisfy Df(a)=I, and if k^-1f is invertible, so is f.

    We may always assume in these cases that a=(0,0,..,0), and Df(a)=I if it helps, and other things too just by a change of coordinates.
     
    Last edited: Nov 5, 2006
  12. Nov 5, 2006 #11
    Ok im still not sure. I understand that its true for arbitrary k if and only if it is true for k=I. But we are not sure that k = I. And by assuming k = I arent you changing the function?

    Im thinking we are supposed to somehow use the fact that D(k^-1 f)(a) = k^-1 Df(a) is the identity linear transformation.
     
  13. Nov 5, 2006 #12
    Ok so if Df(a) is not I, then you move on to k^-1 f which satisfies D(k^-1 f) = I. And if you can prove it for D(k^-1 f) then you proved it for f. But what makes you assume k = I?
     
  14. Nov 5, 2006 #13

    mathwonk

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    i thionk i understand your question. you are puzzled because they are changing notation. i.e. if the derivative k of f is not I then consider the derivative of k^-1f which is I. then call that new derivative k gain, to save letters. now the derivative of k^-1f , which is I, is still being called "k", although that is confusing.

    get it?

    i.e. instead of saying "we can assume k = I" they should more accurately have said "thus we only have to prove the result for functions whose derivative is I. so if k is the derivative, we may assume k =I".
     
  15. Nov 5, 2006 #14

    matt grime

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    I am really baffled by these questions. We are allowed to assume that k=I since we have shown that we may replace f by a function g(x)=Df(a)^-1f(x) that has Dg(a)=I, and that the inverse function theorem will be true for f if and only if it is true for g. Thus replacing f with g we can assume Dg=I from the beginning. What part of that don't you understand?
     
  16. Nov 5, 2006 #15

    matt grime

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    Yes, we are changing the function. But it doesn't matter the result is true for the original function if and only if it is true for the one we replace it by.
     
  17. Nov 5, 2006 #16
    ya so we now prove it for the function g = Df(a)^-1 f which has Dg(a) = I. Ok that makes sense. But in the next part of the proof it says:

    Whenever f(a+h) = f(a) we have
    |f(a+h)-f(a)-k(h)|/|h| = |h|/|h| = 1

    So is he still talking about the original f and the original k, or is he talking about the g and k = Dg(a). This is what confused me, but from what you guys are saying im assuming that hes talking about the g.
     
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