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Math Help - Proof: Addition of vector makes the set remain linearly independent

  1. #1
    owq
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    Proof: Addition of vector makes the set remain linearly independent

    Let's say u_{1}, u_{2}..., u_{k} are linearly independent vectors in R^n.

    Suppose u_{k+1} is a vector in R^n and u_{k+1} is not a linear combination of u_{1}, u_{2}..., u_{k}.

    Show that u_{1}, u_{2}..., u_{k}, u_{k+1} are linearly independent.

    I have some idea, but I can't think of a rigorous proof. Any ideas?
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  2. #2
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    Quote Originally Posted by owq View Post
    Let's say u_{1}, u_{2}..., u_{k} are linearly independent vectors in R^n.

    Suppose u_{k+1} is a vector in R^n and u_{k+1} is not a linear combination of u_{1}, u_{2}..., u_{k}.

    Show that u_{1}, u_{2}..., u_{k}, u_{k+1} are linearly independent.

    I have some idea, but I can't think of a rigorous proof. Any ideas?
    I will try to get you started.
    Let a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0

    Now we can have two cases:
    1. a_{k+1} not equal to 0
    Prove that this is not possible.
    So we are left with
    2. a_{k+1}=0. Under this what can you say about a_1,a_2,...,a_k?

    What can we conclude finally?
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  3. #3
    owq
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    Thanks.

    Can we say that since a_1u_{1}+a_2u_{2}...+a_ku_{k} not equals a_{k+1}u_{k+1}, the equation will never be equal to zero unless a_{k+1}=0?

    Does this work: since u_{k+1} and span( u_{1}, u_{2}..., u_{k}) are linearly independent, and u_{1}, u_{2}..., u_{k} are linearly independent, thus u_{1}, u_{2}..., u_{k}, u_{k+1} are linearly independent?
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  4. #4
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    Quote Originally Posted by owq View Post
    Thanks.

    Can we say that since a_1u_{1}+a_2u_{2}...+a_ku_{k} not equals a_{k+1}u_{k+1}, the equation will never be equal to zero unless a_{k+1}=0?
    I am also new to the subject. Though what you say is correct in essence a more 'rigorous' argument will be
    If a_{k+1} is not equal to 0, it has an inverse a_{k+1}^{-1}. If we multiply the entire equation a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0
    by a_{k+1}^{-1} we get <br />
b_1u_{1}+b_2u_{2}...+b_ku_{k}+u_{k+1}=0

    Which is not possible - hence a_{k+1} is not equal to 0.

    (Any comments from other senior members here will be helpful.)


    Does this work: since u_{k+1} and span( u_{1}, u_{2}..., u_{k}) are linearly independent, and u_{1}, u_{2}..., u_{k} are linearly independent, thus u_{1}, u_{2}..., u_{k}, u_{k+1} are linearly independent?
    Mu question would be - How do you define (and then prove) independence of a vector with a set span( u_{1}, u_{2}..., u_{k})?
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  5. #5
    owq
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    If a_{k+1} is not equal to 0, it has an inverse a_{k+1}^{-1}.
    If we multiply the entire equation a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0 by a_{k+1}^{-1} we get

    u_{k+1} = -a_{k+1}^{-1}a_1u_{1}+-a_{k+1}^{-1}a_2u_{2}...+-a_{k+1}^{-1}a_ku_{k} which is not possible.
    Hence a_{k+1} = 0.

    I see. Is there any other way to solve this?
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  6. #6
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    Not that I know of. I feel this one is pretty fine, as we have just relied on axioms of field and vector space. I guess, any other approach should use a similar argument(about the existence of inverse - as I think that is the key axiom used here). Thanks
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