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Math Help - Linear transformation between function spaces

  1. #1
    tun
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    Linear transformation between function spaces

    Hi all.

    I am a beginner to real maths and am looking at a problem from a book by Loomis and Sternberg.

    Let T be the linear transformation from \mathbb{R}^B to \mathbb{R}^A such that T(f) = f \circ \varphi .

    I need to show that T is an isomorphism if \varphi is a bijection by showing that:

    (a) \varphi injective -> T surjective;

    (b) \varphi surjective -> T injective.

    I'm not really sure where to start, and I can't even see the link between the properties of the 2 mappings; any hints would be greatly appreciated.

    Thanks
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  2. #2
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    Is \mathbb{R}^A=\{f\text{ such that }f: A \rightarrow \mathbb{R}\} (I hate this notation), and in what sense do you mean isomorphism?

    If I am right on the notation, I suppose it must be that \phi: A \rightarrow B. Suppose \varphi is injective. Let g:A\rightarrow \mathbb{R} be any function.
    We need to find a function f: B\rightarrow \mathbb{R} so that g=f\circ \varphi.
    To do this, first note that since \varphi is injective, it is actually a bijection A\rightarrow \text{Range}(\varphi),
    so there exists \gamma:\text{Range}(\varphi)\rightarrow A such that \gamma \circ \varphi(a)=a for all a\in A.
    Then define f(b)=0 if b\notin \text{Range}(\varphi) and f(b)=g(\gamma(b)) otherwise. Then f \circ \phi (a)=f(\phi(a))=g(\gamma(\phi(a)))=g(a). Since we have found a preimage for an arbitrary element, the function is onto.

    See if you can do (b) on your own now.
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  3. #3
    tun
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    Thanks for that, I was on completely the wrong lines - I was thinking it was something to do with the linearity. You were right about the notation. Out of interest, what do you use to represent spaces like \mathbb{R}^A?

    As for (b), if I'm right then it's easier than (a):

    Let f and g be functions from B to \mathbb{R}. We need to show that T(f) = T(g) \Rightarrow f = g, then T is injective.

    T(f) = T(g) means that f \circ \varphi = g \circ \varphi, so that f and g are equal for all elements in B that are in the range of \varphi. If \varphi is surjective, then the range of \varphi is all of B and so f and g are equal everywhere in their domain.

    Does this seem right? Thanks for your help!
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  4. #4
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    That looks right to me!

    As for the notation, I just define a set to be all such functions, i.e. let \mathcal{F}=\{ f:A\rightarrow B\text{ such that } f\text{ satisfies...}\}.
    It may be less "slick" but it sure prevents any and all confusion about what you are saying.
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