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Math Help - linear independence proof

  1. #1
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    linear independence proof

    Can someone show me how to do this question, ?


    suppose { u,v,w} is linearly dependant than , w can be written as a linear combination of u and v, for some scalars c1, ,c2,

    c1u + c2v .....kw = 0




    w = (-c1/k)u - (c2/k)v

    i am not sure where to go from here.
    Attached Thumbnails Attached Thumbnails linear independence proof-screen-shot-2014-06-11-11.06.04.png  
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  2. #2
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    Re: linear independence proof

    If the dimension of the space is 2, that's the max number of linearly independent vectors you can have by definition of dim.

    EDIT: a more direct way of looking at it:
    By def of span, all vectors in S are linear combinations of u and v,including w
    Last edited by Hartlw; June 11th 2014 at 07:10 AM.
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  3. #3
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    Re: linear independence proof

    Quote Originally Posted by Tweety View Post
    Can someone show me how to do this question, ?


    suppose { u,v,w} is linearly dependant than , w can be written as a linear combination of u and v, for some scalars c1, ,c2, c1u + c2v .....kw = 0
    You can't prove this- it isn't true. Take u= <1, 0, 0>, v= <0, 0, 0> and w= <0, 1, 0> then w cannot be written as a linear combination of u and v. Another counter example would be u= <1, 0, 0>, v= <3, 0,0>, and w= <0, 1, 0>.

    But that is NOT WHAT you want to prove. You want to include "if {u, v} is linearly independent".
    If u, v, w are dependent there exist scalars, a, b, c, not all 0, such that au+ bv+ cw= 0. IF c= 0, we would have
    au+ bv= 0 with either a or b or both non-zero which contradicts the fact that {a, b} is linearly independent. Therefore c is non-zero and then we can solve for w: w= -(au+ bv)/c= -(a/c)u- (b/c)v.


    w = (-c1/k)u - (c2/k)v

    i am not sure where to go from here.
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  4. #4
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    Re: linear independence proof

    I assumed you were trying to prove thumbnail, which I thought said, if u,v span S, and w belongs to S, u,v,w are linearly dependent, which is true.
    But it said, linearly INdepent, which makes the proposition false, ie, canít be proved.
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  5. #5
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    Re: linear independence proof

    Given:

    $\{u,v\}$ is linearly independent.

    $w \not\in \text{span}(\{u,v\})$.

    To prove:

    $\{u,v,w\}$ is linearly independent.

    Suppose that $\alpha u + \beta v + \gamma w = 0$.

    If $\gamma \neq 0$, we have:

    $w = \dfrac{-\alpha}{\gamma}u + \dfrac{-\beta}{\gamma}v$, contradicting that $w$ is not in the span of $u$ and $v$.

    Hence $\gamma = 0$, and thus $\alpha u + \beta v = 0$, whence by the linear independence of $\{u,v\}$, we have $\alpha = \beta = 0$, which shows that $\{u,v,w\}$ is linearly independent.

    It is always a good idea in these sorts of problems to prove what you are told to prove, and not something else which you think is the same.
    Thanks from Tweety
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  6. #6
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    Re: linear independence proof

    Step 1): Identify the question. Assume itís the thumbnail (I canít remember them, so):

    ďu, v, and w are elements of a vector space V.
    {u,v} is a linearly independent set, and w ɇ span {u,v}.
    Prove that {u,v,w} is an independent set.Ē

    Step 2): The solution.
    Assume {u,v,w} is dependent. Implies w ϵ span {u,v}. Contradiction.
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  7. #7
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    Re: linear independence proof

    Quote Originally Posted by Hartlw View Post
    Step 1): Identify the question. Assume it’s the thumbnail (I can’t remember them, so):

    “u, v, and w are elements of a vector space V.
    {u,v} is a linearly independent set, and w ɇ span {u,v}.
    Prove that {u,v,w} is an independent set.”

    Step 2): The solution.
    Assume {u,v,w} is dependent. Implies w ϵ span {u,v}. Contradiction.
    Step 3): read post #3 and reflect a bit.
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