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Not Satisfied with Linear Algebra Theorem

  1. Feb 15, 2013 #1

    I am just not satisfied with the following theorem (I don't know it's name):

    Let T:R^n -> R^m be a linear transformation. Then T is one-to-one if and only if the equation T(x) = 0 has only the trivial solution.

    The "proof" involves saying that if T is not one-to-one, then there are two different vectors U and V such that T(U)=T(V)= some vector B. And since T is linear it follows that T(U-V) = T(U)-T(V) = B - B = 0. It then concludes by saying "hence there are nontrivial solutions to T(X)= 0. So, either the two conditions in the theorem are both true or they are both false."

    I just don't see how that proved the theorem in any way, perhaps because I don't fully understand which two conditions it is talking about.

    Could anyone help me here? Thank you.
  2. jcsd
  3. Feb 15, 2013 #2


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    Hey Vorde.

    A 1-1 would imply that T(U-V) = T(U) - T(V) != 0 for U != V. If however this were false then it would imply that for some U != V that T(U) = T(V) proving that the mapping is not 1-1.

    But then you have to take into account the trivial solution (i.e. the zero vector) as a special case where T(U) = 0 for U = 0.

    The formal proof for 1-1-ness is to show that for U != V then T(U) != T(V) for all U and V in the domain of the mapping.
  4. Feb 15, 2013 #3
    Hey Chiro,

    Right, so I understand that part. What I don't understand is how it is enough to show that there aren't any non-zero vectors that map to the zero-vector to know that the mapping is one-to-one everywhere.

    Why couldn't the zero vector be the only vector that maps to the zero vector but still have non 1-1-ness elsewhere?
  5. Feb 15, 2013 #4


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    We know that the zero vector always maps to the zero vector, but we also know that if everything is 1-1, then it means that only the zero vector maps to the zero vector and everything else maps to some other vector (that isn't the zero vector).
  6. Feb 15, 2013 #5


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    Take two vectors, u and v.
    Suppose T(u) = T(v).
    Then T(u) - T(v) = 0.
    But be linearity, T(u) - T(v) = T(u-v).
    Then only vector that maps to 0 is the zero vector.
    Hence, u - v = 0, so u = v, proving T was 1-to-1.
  7. Feb 15, 2013 #6
    Okay, but that only works if you posit that the transformation is 1-1, and the theorem doesn't start with the assumption that the transformation is 1-1. I understand that the zero vector will always map to itself, just not why that says anything about the rest of the transformation.

    Ah, wait, I think I might see it now.
  8. Feb 15, 2013 #7


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    Let [itex]T[/itex] be injective (sorry I absolutely hate the term 1 - 1 god knows why it is even used). Because T is linear, [itex]T(0) = 0[/itex]. This immediately implies this is the only vector for which this is true because if [itex]\exists v\in V :T(v) = 0[/itex], then the injectivity of [itex]T[/itex] implies that [itex]v = 0[/itex]. Now let [itex]T(v) = 0[/itex] be true only for the zero element then if [itex]T(v) = T(w)[/itex] we have that [itex]T(v - w) = 0\Rightarrow v = w[/itex] thus [itex]T[/itex] is injective. This is essentially what the proof you quoted is saying but the quoted proof is more concise.

    EDIT: Seems like people responded while I was typing this up but I guess I'll leave it here anyways =D.
  9. Feb 15, 2013 #8
    Okay, thank you to all who have been helping.

    What was really bothering me was that I didn't see why the trivial solution had to be the only solution for the zero vector. But I just went back and thought about it and now I can see that this must be the case.

    That assumed, I can follow the rest of the theorem.

    Once again, thank you all.
  10. Feb 15, 2013 #9


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    T is one to one if
    we must have

    now suppose
    only when

    by linearity
    then we must have
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