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Thread: 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 $\displaystyle u_{1}, u_{2}..., u_{k}$ are linearly independent vectors in $\displaystyle R^n$.

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

    Show that $\displaystyle u_{1}, u_{2}..., u_{k}, u_{k+1}$ 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 $\displaystyle u_{1}, u_{2}..., u_{k}$ are linearly independent vectors in $\displaystyle R^n$.

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

    Show that $\displaystyle u_{1}, u_{2}..., u_{k}, u_{k+1}$ 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 $\displaystyle a_1u_{1}+a_2u_{2}...+a_ku_{k}+a_{k+1}u_{k+1}=0$

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

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

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

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

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

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

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


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

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

    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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