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Math Help - Linear Transformations and Linear Independence

  1. #1
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    Linear Transformations and Linear Independence

    Let L: V -> W be a linear transformation and let S = {v1, ... vn} be a set of vectors in V. Prove that if T = {L(v1), ... L(vn)} is linearly independent, then so is S. What can we say about the converse?

    So I gave the problem a try. Not sure if I got it right or if I did it backwards! Having trouble with the converse though, so feedback is greatly appreciated!

    Let's say L is one to one.

    Let a1L(v1) + ... + anL(vn) = 0w where a1, ... an are real numbers

    L(a1v1 + ... + anvn) = 0w = L(0v)

    Since L is one to one we conclude that a1v1 + ... + anvn = 0v

    This would imply that S is linearly independent because a1, ... an = 0.

    I'm not sure how I'd start with the converse. That would be that S is linearly independent so then T would be linearly independent. I got stuck there. Thanks guys!
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  2. #2
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    Quote Originally Posted by Brokescholar View Post
    Let L: V -> W be a linear transformation and let S = {v1, ... vn} be a set of vectors in V. Prove that if T = {L(v1), ... L(vn)} is linearly independent, then so is S. What can we say about the converse?

    Let's say L is one to one.
    No need to make this assumption. Say that \{\bold{v}_1,...,\bold{v}_n\} is not linear independent then there are a_i not all zero so that \sum_i a_i \bold{v}_i = \bold{0} but then L\left( \sum_i a_i \bold{v}_i \right) = L(\bold{0}) = \bold{0}. But L is linear and so \sum_i a_i L(\bold{v}_i) = \bold{0}. Therefore, \{ L(\bold{v}_1),...,L(\bold{v}_n)\} is not linearly independet. Now take the contrapositive of this statement to complete the proof.

    The converse is not true unless we know that L is a one-to-one linear transformation.
    Can you come up with a conterexample?
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