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Math Help - Linear combination proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Linear combination proof

    Q: provided that S_{0}=\{\vec{u_{1}},...,\vec{u_{n}}\} is a linearly independent set of vectors and that the set S_{1}=\{\vec{u_{1}},...,\vec{u_{n}},\vec{v}\} is linearly dependent, prove that \vec{v} is a linear combination of the \vec{u_{i}}'s.

    A: Since S_{0} is linearly independent the vector equation,

    c_{1}\vec{u_1}+c_{2}\vec{u_2}+. . .+c_{n}\vec{u_n}=\vec{0}

    has only the trival solution, c_{1}=0, c_{2}=0,. . .,c_{k}=0. Conversely, the set S_{1} is linearly dependent; thus, there exists a vector in S_{1} that can be written as a linear combination. Since S_{1}=\{\{S_{0}\},\vec{v}\} and S_{0} is known to be linearly independent we can narrow our search to just one vector, the vector \vec{v}. So, we have a new vector equation,

    c_{1}\vec{u_1}+c_{2}\vec{u_2}+. . .+c_{n}\vec{u_n}+k\vec{v}=\vec{0}

    where k\in{\mathbb{R}}

    Solving for \vec{v}

    \vec{v}=-\frac{c_{1}}{k}\vec{u_1}-. . .-\frac{c_{n}}{k}\vec{u_n}

    \therefore \vec{v} can be written as linear comination of \vec{u_{i}}'s.

    I dunno, I feel a chunck is missing. Is that a sufficient proof?

    Thanks



    Last edited by Danneedshelp; July 13th 2009 at 03:47 PM.
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  2. #2
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    You need to explain why we know that t\not=0.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    You need to explain why we know that t\not=0.
    are you talking about the constant k?
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  4. #4
    Senior Member Danneedshelp's Avatar
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    Well, k\neq{0}, because the set S_{1} is linear dependent and as I tried explaining, every vector in the set is linearly independent except for the vector \vec{v}. This implies, when we write our new set as a linear combination of the zero vector, k cannot equal zero, because if k=0 we would have only the trivail solution and in turn S_{1} would be linearly independent. So, S_{1} is dependent \forall{k\neq{0}}\in{\mathbb{R}}.

    Is that the right train of thought?
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  5. #5
    Super Member Gamma's Avatar
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    right track

    I think things look good, you just gotta be more careful with how you say things and set up the proof. You are given that \{u_1,...,u_n\} is L.I. and that when you add that vector v to it \{u_1,...,u_n, v\}, it makes it L.D.

    So you say since \{u_1,...,u_n, v\} is LD, there exist a nontrivial linear combination such that

    c_1u_1+...+c_nu_n + kv=0

    Suppose for a moment that k=0. Then at least one of those c_i\not = 0. But then you have just found a nontrivial linear combination of \{u_1,...,u_n\} which sums to 0, in particular c_1u_1+...+c_nu_n=0 which contradicts the linear independence of \{u_1,...,u_n\}. Thus, you know k \not = 0 and you can proceed using your argument.

    Might be worth noting why \frac{1}{k} exists. I don't know whether you are in algebra or linear algebra, but you can have modules which are basically vector spaces but the coefficients come from a ring where multiplicative inverses need not exist. You are in a vector space which is over a field, so it makes sense to divide by k since you have shown it to be a nonzero element of the field which means it is a unit (is invertible).
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