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myininaya Group Title

True/False?: Let V and W be vector spaces. Let L:V->W be a linear map. If kerL=0, then L is a bijection.

  • 2 years ago
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  1. myininaya Group Title
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    I think it is false but I need a counterexample to show me that. lol

    • 2 years ago
  2. myininaya Group Title
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    So kerL=0 is the elements v in V such that L(v)=0

    • 2 years ago
  3. myininaya Group Title
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    So we need to show that for some mapping L that L is not a bijection Remember a bijection has the following properties: A )x,y in V, L(x)=L(y) implies x=y (injective) B )if for every w in W there exists at least one v in V such that L(v)=w (surjective)

    • 2 years ago
  4. myininaya Group Title
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    So we know that L is injective. So L fails the surjective part of the definition of bijective. Assume that ker(L) = 0. If L(v1) = L(v2), then linearity of L tells that L(v1 - v2) = 0. Then ker(L) = 0 implies v1-v2 = 0, which shows that v1=v2 as desired.

    • 2 years ago
  5. myininaya Group Title
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    @jamesj @Zarkon I hope I don't bother you guys too much...I can't figure out such a function for my counterexample. You don't have to answer if you don't want to but of course you already knew that. lol

    • 2 years ago
  6. JamesJ Group Title
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    If the kernel is non-zero the the dimensionality of the image of V, L(V) is equal to V. This tells us intuitively that the map is 1:1. To formalize it, we need two things, as you observe a) surjectivity b) injectivity a) is trivial, and notice that L(V) need NOT equal W Now using your argument above, you've shown the L is injective, i.e., 1:1, because if L(v1) = L(v2), then v1 = v2. Hence indeed L is a bijection.

    • 2 years ago
  7. myininaya Group Title
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    So that is why I couldn't think of a function that fails the surjective part becacuse the statement was true. lol

    • 2 years ago
  8. myininaya Group Title
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    Thanks JamesJ.

    • 2 years ago
  9. JamesJ Group Title
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    sure

    • 2 years ago
  10. Zarkon Group Title
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    Wouldn't we need to know the dimensions of V and W in order to determine if it was indeed surjective?

    • 2 years ago
  11. JamesJ Group Title
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    I'm just saying it surjective onto its image and the sense of a function being a bijection being just one to one and onto. If the question is: is the map L a bijection *between V and W*, yes, then clearly we need that dim(V) = dim(W) in the finite case.

    • 2 years ago
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