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Math Help - Induction of linear independent vectors

  1. #1
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    Induction of linear independent vectors

    This is perhaps an easy question- more of a question about mathematical induction than linear algebra, perhaps:

    Let V be a vector-space over a field \mathbb{F}; v_1,...,v_n are linear independent vectors in V. Prove by induction that the vectors w_i = \sum_{k=1} ^{i} v_k, 1 \leq i \leq n, are linear independent.

    So, the question is asking to prove that the sum of linearly independent vectors is a vector which is linearly independent.

    The condition for linear independence is:
    \sum_{k=1} ^{i} a_k v_k = 0, 1 \leq i \leq n, when and only when a_i = 0 for all 1 \leq i \leq k

    So v_1 = a_1v_1 = 0 = a_1w_1, but how do you make an induction step out of this? For w_2, the proposition is that v_1 + v_2 is also linearly independent:

    w_2 = v_1 + v_2 = a_1v_1 + a_2v_2 ; a_1=a_2=0, so you can bracket the scalars out... hmm? But I still can't see how you can make an induction step. I'm sure it's kind of easy, but that's only making it more frustrating.
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  2. #2
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    Could you use the unit-vectors to prove this, or perhaps the dimension of the vector space? dim(V) = n?
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  3. #3
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    Quote Originally Posted by bmp05 View Post
    This is perhaps an easy question- more of a question about mathematical induction than linear algebra, perhaps:

    Let V be a vector-space over a field \mathbb{F}; v_1,...,v_n are linear independent vectors in V. Prove by induction that the vectors w_i = \sum_{k=1} ^{i} v_k, 1 \leq i \leq n, are linear independent.

    So, the question is asking to prove that the sum of linearly independent vectors is a vector which is linearly independent.

    The condition for linear independence is:
    \sum_{k=1} ^{i} a_k v_k = 0, 1 \leq i \leq n, when and only when a_i = 0 for all 1 \leq i \leq k

    So v_1 = a_1v_1 = 0 = a_1w_1, but how do you make an induction step out of this? For w_2, the proposition is that v_1 + v_2 is also linearly independent:

    w_2 = v_1 + v_2 = a_1v_1 + a_2v_2 ; a_1=a_2=0, so you can bracket the scalars out... hmm? But I still can't see how you can make an induction step. I'm sure it's kind of easy, but that's only making it more frustrating.
    the induction is over n, the number of vectors. for n = 1, there's nothing to prove. suppose the claim is true for n (induction hypothesis) and take any n + 1 linearly independent vectors

    v_1, \cdots , v_{n+1}. suppose c_1v_1 + c_2(v_1+v_2) + \cdots + c_n(v_1 + \cdots + v_n) + c_{n+1}(v_1 + \cdots + v_{n+1})=0. if we show that c_1=c_2 = \cdots = c_{n+1}=0, we're done. first note that if c_{n+1}=0, then

    c_1=c_2 = \cdots = c_n = 0 by induction hypothesis and so we're done. if c_{n+1} \neq 0, then we'll have: v_{n+1} = -c_{n+1}^{-1}c_1v_1 - c_{n+1}^{-1}c_2(v_1 + v_2) - \cdots - c_{n+1}^{-1}(c_n + c_{n+1})(v_1 + \cdots + v_n), which is not

    possible because v_1, \cdots , v_{n+1} are linearly independent. this completes our induction.
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  4. #4
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    So, you're proving the proposition for all of n- per definition- and then you prove it for n+1? But once again, this relies on the definition of linear independence. It's a strange question, I think. But thanks for your help NCA- btw. can you recommend a good text book for a first course in linear algebra?
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