Results 1 to 3 of 3

Math Help - Transform one-to-one

  1. #1
    Junior Member
    Joined
    Jun 2011
    Posts
    45

    Transform one-to-one

    The problem assumes V and W are vector spaces, and that T: V --> W is linear. We must prove that T is one to one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W;

    My attempt:
    I first assume T one to one. Then N(T) = {0}. Assume V is linearly independent, but its image in W was not. Then:
     \sum_{i=1}^{n} a_i w_i = 0. But  \sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i T(v_i) = T \sum_{i=1}^{n} a_i v_i. Since T is one to one, an inverse exists, so we get :
     T^{-1} T \sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i v_i = T^{-1} (0) = 0. But then we have a linear combination of vectors in v that sum to the zero vector, contradicting our assumption that V was linearly independent.

    I'm not sure if this part is right, and I'm also not sure how to prove going the other way (that if it carries linear independence, it's one to one).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Transform one-to-one

    Quote Originally Posted by jsndacruz View Post
    The problem assumes V and W are vector spaces, and that T: V --> W is linear. We must prove that T is one to one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W;

    My attempt:
    I first assume T one to one. Then N(T) = {0}. Assume V is linearly independent, but its image in W was not. Then:
     \sum_{i=1}^{n} a_i w_i = 0. But  \sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i T(v_i) = T \sum_{i=1}^{n} a_i v_i. Since T is one to one, an inverse exists, so we get :
     T^{-1} T \sum_{i=1}^{n} a_i w_i = \sum_{i=1}^{n} a_i v_i = T^{-1} (0) = 0. But then we have a linear combination of vectors in v that sum to the zero vector, contradicting our assumption that V was linearly independent.

    I'm not sure if this part is right, and I'm also not sure how to prove going the other way (that if it carries linear independence, it's one to one).
    You omitted a lot of steps...and words...but yeah, that's the right idea. To do the converse, why don't you try thinking about what would happen if \displaystyle \sum_i a_i v_i (where, presumably as you mean, \{v_1,\cdots,v_n\} is a basis for [tex]V[tex]) mapped to zero? This would tell us that \displaystyle \sum_i a_i T(v_i)=0, but what did we just assume about T, that tells us about \{T(v_1),\cdots,T(v_n)\}=T(\{v_1,\cdots,v_n\}). In turn, what does this tell us about \displaystyle \sum_i a_i v_i, and moreover what does it tells us about \ker T.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2011
    Posts
    45

    Re: Transform one-to-one

    Sorry for the sloppy derivation - was just trying to sketch out the proof first. And you were right in thinking I meant {v_1..v_n} was a basis. Thanks a lot!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: March 22nd 2014, 04:53 PM
  2. Laplace transform and Fourier transform what is the different?
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: December 29th 2010, 11:51 PM
  3. Z transform help
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: June 20th 2010, 11:57 PM
  4. Help for Z transform of t^2
    Posted in the Advanced Applied Math Forum
    Replies: 6
    Last Post: June 20th 2010, 04:57 AM
  5. Replies: 0
    Last Post: April 23rd 2009, 06:44 AM

/mathhelpforum @mathhelpforum