Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By Deveno

Math Help - Linear Transformations and Compositions

  1. #1
    Kyo
    Kyo is offline
    Junior Member
    Joined
    Sep 2012
    From
    Canada
    Posts
    30

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,312
    Thanks
    693

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Transformations and the General Linear Group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 26th 2011, 10:50 AM
  2. Basic Linear Algebra - Linear Transformations Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 7th 2010, 03:59 PM
  3. Replies: 3
    Last Post: October 17th 2010, 05:13 PM
  4. Linear Transformations and Linear Independence
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 6th 2008, 07:36 PM
  5. Replies: 3
    Last Post: June 2nd 2007, 10:08 AM

Search Tags


/mathhelpforum @mathhelpforum