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Math Help - Linear Algebra proof

  1. #1
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    Linear Algebra proof

    I have shown that if a set of vectors (v1,...,vn) is linearly independant, then every vector v that is an element of <v1,...,vn> has a unique representation as a linear combination of vectors v1,...,vn.

    Now I need to show that the converse statement is true, that is if a vector v has a unique representation as a linear combination of vectors v1,...,vn then the set of vectors (v1,...,vn) are linearly independant. Does anyone have suggestions on you show that?
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  2. #2
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    Suppose we can write a vector x as a_1v_1+...+a_nv_n and as b_1v_1+...+b_nv_n. We want to show a_1=b_1,...,a_n=b_n for uniqueness. So that means, a_1v_1+...+a_nv_n = b_1v_1+...+b_nv_n. Subtract them, (a_1-b_1)v_1+...+(a_n-b_n)v_n = 0. Since these vector are linearly independent it means there is only the trivial representation. And so that means, a_1 - b_1 = 0 ... a_n - b_n =0 so a_1 = b_1 ... a_n=b_n. Q.E.D.
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  3. #3
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    That is how I showed the first part - that there is a unique representation. I don't think that works to show the converse that if there is a unique representation, then the set of vectors (v1,...,vn) are independant.
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  4. #4
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    Correct not everything can be expressed in linearly independent vectors.
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  5. #5
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    A nice google search helped me find the answer to my question:

    Elements of Operator Theory - Google Book Search

    Thanks for your help, though!
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  6. #6
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    What is so hard? Consider \{ (0,1,0),(1,0,0) \} this is linearly independent but (0,0,1) cannot be obtained as a linear combination of those two elements.
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  7. #7
    MHF Contributor kalagota's Avatar
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    just try to understand the proof..

    Suppose v has a unique representation of linear combinations of vectors v_1, v_2, ... ,v_n, i.e.
    v = \sum_{i=1}^n a_iv_i = a_1v_1 + a_2v_2 + ... + a_nv_n

    and suppose the S:= \{ v_1, v_2, ..., v_n \} is linearly dependent.

    then, \exists v_j \in S which is a linear combination of the other vectors in S (assuming you have proven this already) i.e.

    v_j = - (b_j^{-1}b_1v_1 + b_j^{-1}b_2v_2 + ... + b_j^{-1}b_{j-1}v_{j-1} + b_j^{-1}b_{j+1}v_{j+1} + ... + b_j^{-1}b_nv_n ---> (*)

    from
    v = \sum_{i=1}^n a_iv_i = a_1v_1 + a_2v_2 + ... + a_jv_j + ... + a_nv_n, we replace v_j by (*), and since V is an abelian group, we can rearrange and combine like terms, so that,

    v = \sum_{i=1}^n a_iv_i = (a_1 - a_jb_j^{-1}b_1)v_1 + (a_2 - a_jb_j^{-1}b_2)v_2 + ... + (a_{j-1} - a_jb_j^{-1}b_{j-1})v_{j-1} +
    (a_{j+1} - a_jb_j^{-1}b_{j+1})v_{j+1} + ... + (a_n - a_jb_j^{-1}b_n)v_n

    and clearly, a_k - a_jb_j^{-1}b_k = a_k \Leftrightarrow a_jb_j^{-1}b_k = 0, for k=1,2,...,n
    but we have to note that S must not contain the zero vector since v has a unique representation of linear comb of vectors in S. hence, a_k - a_jb_j^{-1}b_k \neq a_k for all k.
    which contradicts now that v has a unique representation of linear comb of the vectors in S.

    therefore, S must be lin. indep set. QED
    Last edited by kalagota; November 15th 2007 at 02:41 PM.
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