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Thread: Need help with the 2nd part of a two part question on Linear combinations

  1. #1
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    Need help with the 2nd part of a two part question on Linear combinations

    (a) Show that w is a linear combination of u and v.
    (b) Use your previous answer to show that u is a linear combination of v and w.

    I did (a), but I'm not sure how I'm supposed to use it to to (b). What is the method here?
    Last edited by MrJank; Aug 31st 2018 at 09:05 AM.
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  2. #2
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    Re: Need help with the 2nd part of a two part question on Linear combinations

    Quote Originally Posted by MrJank View Post
    (a) Show that w is a linear combination of u and w
    (b) Use your previous answer to show that u is a linear combination of v and w.
    I did (a), but I'm not sure how I'm supposed to use it to to (b). What is the method here?
    Are you very sure that you have written part (a) correctly?
    $w=0\cdot u+ 1\cdot w$ that is trivial, but does help with part (b).
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    Re: Need help with the 2nd part of a two part question on Linear combinations

    Sorry, I fixed it. Show w is a linear comb of u and v
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    Re: Need help with the 2nd part of a two part question on Linear combinations

    Quote Originally Posted by MrJank View Post
    Sorry, I fixed it. Show w is a linear comb of u and v
    Good then, this must be part of a larger problem. What are $u~\&~v~?$
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    Re: Need help with the 2nd part of a two part question on Linear combinations

    u v w
    [1] [2] [-1]
    [1] [1] [1]
    [-2] [-3] [0]

    Like i said, I've got part (a), I'm just unsure how to use it to get b.
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  6. #6
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    Re: Need help with the 2nd part of a two part question on Linear combinations

    it's established that $w$ is a linear combination of $u$ and $v$, i.e.

    $w = \alpha u + \beta v,~\forall \alpha,~\beta \neq 0$

    so

    $u = \dfrac 1 \alpha \left(w - \beta v\right) = \dfrac 1 \alpha w - \dfrac \beta \alpha v$

    and thus $u$ is a linear combination of $w$ and $v$
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  7. #7
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    Re: Need help with the 2nd part of a two part question on Linear combinations

    That really makes sense.

    Ty.
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