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Math Help - Linear transformations - Proving that a very special set generates the target space

  1. #1
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    Linear transformations - Proving that a very special set generates the target space

    Hi! this is not actually a homework problem, just a problem from a linear algebra book in spanish ( Algebra lineal, Elon Lages Lima) that has been bothering me A LOT lately because the "hint" says "it's actually much easier than it looks" :S any help would be appreciated a lot


    1. The problem statement, all variables and given/known data

    Let E and F be vector spaces (of any dimension) and let X be a subset of F with the following property (which is only a conditional):


    IF a given linear transformation
    A: E ---> F satisfies that X is a subset of Im(A)

    THEN this linear transformation A is surjective. ... (*)

    Prove that X is a generating set for F.


    2. Relevant equations

    E1)Im(A) = set of images of elements of E under A.
    E2)A is surjective if and only if it transforms generating sets into generating sets.
    E3)A has (at least one) right inverse if it is surjective

    3. The attempt at a solution
    I assumed the antecedent of (*) since I think I can start the proof that way. Then I can assert that on one hand that A is surjective and using E2) I concluded that Im(A) is a generating set which is redundant.
    I also tried to take all y in F that are in Im(A) but not in X (because I assumed that X is inside Im(A) ) and get the inverse image of these elements so I can map these into X. But then I realized that I don't have the freedom to do that (I can at most assign arbitrary values to members of a basis).

    I tried to use the contrapositive of (*) and assume that there exists some w in F such that it is not the image of any v in E. then it follows that there exists some x in X such that it is not the image of any v in E. But this got me nowhere.

    Please help with this problem. Thanks
    Last edited by arkanoid; December 22nd 2010 at 05:07 PM. Reason: clearer
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  2. #2
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    So we are letting a competing forum be a linear transformation?
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  3. #3
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    LOl, I'm sorry, I must admit that I am a mathematical physicist and I was using the physics forum (to no avail) first and I copied and pasted the words... (resulting in the url address showing up there instead of the latex-ed formulas)... I am editing my post... sorry for the inconveniences
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by arkanoid View Post
    LOl, I'm sorry, I must admit that I am a mathematical physicist and I was using the physics forum (to no avail) first and I copied and pasted the words... (resulting in the url address showing up there instead of the latex-ed formulas)... I am editing my post... sorry for the inconveniences
    You have that in particular if A(E)=\text{span }X then A(E)=F...so?
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    Hi Drexel28, thank you for the idea but I still don't see how I can assume that there is a certain A such that Im(A) = Span(X). Working with the conditional: If Im(A) = Span(X) is FALSE then I can still assume that the consequent, that is, Im(A) = F (surjectivity) is true but then I cannot infer that Span(X) = F.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by arkanoid View Post
    Hi Drexel28, thank you for the idea but I still don't see how I can assume that there is a certain A such that Im(A) = Span(X). Working with the conditional: If Im(A) = Span(X) is FALSE then I can still assume that the consequent, that is, Im(A) = F (surjectivity) is true but then I cannot infer that Span(X) = F.
    Maybe I am misunderstanding the question. Let me phrase it in my own words and see if it's what you intended

    "Let E,F be vector spaces and X\subseteq F be such that X\subseteq A(E)\implies A(E)=F. Prove that \text{span }X=F"

    Right? If so, then what's wrong with my proof? Assuming that it's possible for A(E)=\text{span }X (e.g. that \dim E\geqslant \dim F) then we have that A(E)\supeteq X and so by our assumption it must be true that A(E)=F. But, putting these two together gives \text{span }X=A(E)=F or \text{span }X=F.

    What am I missing?
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  7. #7
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    Hi again!
    Indeed, the problem is like you said. well, I'm having a bit of skepticism because I don't know why I can assume that A(E) = Span(X). I know that the property is just a conditional but to be able to use both equalities : Span(X) = A(E) and A(E) = F I need to show that both equalities are true... I just assumed that the first one is true, the validity of the second one does not depend on the validity of the first one (because the conditional can have a true consequent with a false antecedent).
    Basically, this Proposition should also be true even when they have infinite dimension and even is E has a (possibly infinite but) smaller dimension than F. I know that in this case there is no way to have a surjective A but X would still have to be a generating set...
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