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Nice little proof, can any one do it?

  1. Sep 5, 2009 #1
    A linear transformation F is said to be one-to-one if it satisfies the following condition: if F(u) = F(v) then u = v. Prove that F is one-to-one if and only if Ker(F) = {0}.
     
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  3. Sep 5, 2009 #2

    arildno

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    Well, what are your thoughts about this?
     
  4. Sep 5, 2009 #3

    arildno

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    To help you along a bit:

    a) IF one-to one..
    You are to show that this implies the kernel of F contains just the 0 element.

    Now, you can prove that this must be true, by way of contradiction:

    ASSUME that the i) linear transformation F is both ii) one-to-one AND has iii) non-zero vectors in its kernel.

    Show that i)+iii) implies that F is NOT one-to-one!
     
  5. Sep 5, 2009 #4
    hmmm yeah my friend posed this proof to me and asked me to try and find a good solution to it if i could, to be honest i have idea! its possible that i just lack the algebra to do so
     
  6. Sep 5, 2009 #5
    Um, this proof doesn't really require any algebra, just the basic properties that define a linear transformation. For the reverse direction, suppose F(u) = F(v) so that F(u) - F(v) = F(u-v).

    For the forward direction, you could also prove that F(0) = 0, which means that {0} is contained in Ker(F), so you just have to prove that Ker(F) is contained in {0}. Let v be in Ker(F), find out what this means and remember that you should have proved 0 = F(0).
     
  7. Sep 6, 2009 #6
    ah thanks snipez90, pretty sure ive got it out!
     
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