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Finding inverse of linear mapping

  1. Sep 13, 2011 #1
    so for a mapping f:X->Y

    where X,Y are Normed Vector Spaces

    if I have a function f(x) = y such that x in X and y in Y, how do I explicitly find f inverse?

    I sat down to do this and realize I've only been trained in the Reals where you switch the x,y and then solve for y. But this won't work for my function as my x,y are vectors (not scalars) and I don't have all the operations available like 1/x and ln(x).
     
  2. jcsd
  3. Sep 14, 2011 #2
    By a mapping do you mean a bounded linear map? And by inverse do you require your inverse to be bounded linear as well? Maybe you are only working with isometries?

    If you do not impose more conditions on f, X or Y than that f is bounded linear, then the problem is impossible in general. Just let X be the trivial normed vector space {0} and let Y be the normed vector space [itex]\mathbb{C}[/itex]. Then we may define [itex]f : X \to Y[/itex] by f(0) = 0. Clearly however f cannot have an inverse because X and Y are not of equal cardinality. The only function [itex]g : Y \to X[/itex] is g(y) = 0, but then f(g(1)) = f(0) = 0 so g is not a right inverse.
     
  4. Sep 14, 2011 #3
    I hope my function is bounded and continuous. My ultimate goal is to show a homeomorphism from a normed vector space to another. So the function I pick must be bicontinuous (and therefore bounded). So let's assume I was smart enough to pick a bicontinuous and bounded function.
     
  5. Sep 14, 2011 #4
    That is sufficient. Since f is a homeomorphism it has a continuous inverse g : Y -> X. You can show fairly easily that g is linear. g is linear and continuous and therefore bounded.

    Whether this is explicit enough or not I do not know, but I doubt you will find a much more explicit construction unless you are willing to severely restrict what X and Y are allowed to be.
     
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