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Thread: Linear Transformations and Compositions

  1. #1
    Kyo is offline
    Junior Member
    Sep 2012

    Angry Linear Transformations and Compositions

    Hi guys, I'd like some help with this question. It's been incredibly frustrating trying to solve this for hours, but getting nowhere.

    Suppose U -> V -> W, with transformations T (U to V) and S (V to W).

    1) Suppose ST is one-to-one and T is onto, show that S is one-to-one.

    My answer 1)
    Suppose (ST)(u) is one-to-one, and suppose T is onto:
    (ST)(u) = S(T(0)) = 0 = S(T(u)) = S(v)
    Therefore v = 0.

    2) Suppose ST is onto and S is one-to-one, show that T is onto.

    This one has me beat and I'm not sure how to approach it at all.

    Can someone please (hopefully) verify that I did 1) correctly, and help with 2)?

    Thank you!
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Linear Transformations and Compositions

    your proof of one is hard to follow.

    we want to prove that S is 1-1, which means proving S(v) = 0 implies v = 0.

    suppose S(v) = 0. since v is in V, and T is onto v, v = T(u) for some u in U.

    therefore: S(v) = S(T(u)) = ST(u).

    since S(v) = 0, ST(u) = 0.

    since ST is 1-1, u = 0.

    thus v = T(u) = T(0) = 0, which is what we desired to prove.

    2) here, we must find for any given v in V, some u in U with T(u) = v.

    now, we know that given any w in W, we have u in U, with ST(u) = w.

    so consider w = S(v), for our given v above.

    we have w = S(v) = ST(u), for some u in U (since ST is onto).

    since S is 1-1, and S(v) = ST(u) = S(T(u)), v = T(u), and we are done.
    Thanks from Kyo
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