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Math Help - Span and linearly independence question

  1. #1
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    Span and linearly independence question

    Suppose that {u; v} is a linearly independent set of vectors and w is not in the span of {u; v}. Show that the set {u, v,w} is linearly independent.

    So we no that a linearly independent set:
    x1v1+x2v2+.......xpvp=0 where x has only the trivial soltuion
    So does that mean i have to plug in u and w into that equation to show it will equal 0?
    Im confused on how to show it when w is not in the span. Any Help would be appriciated
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  2. #2
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    Quote Originally Posted by kensington View Post
    Suppose that {u; v} is a linearly independent set of vectors and w is not in the span of {u; v}. Show that the set {u, v,w} is linearly independent.

    So we no that a linearly independent set:
    x1v1+x2v2+.......xpvp=0 where x has only the trivial soltuion
    So does that mean i have to plug in u and w into that equation to show it will equal 0?
    Im confused on how to show it when w is not in the span. Any Help would be appriciated


    Supose au+bv+cw=0\,,\,\,a,b,c\mbox{ scalars }\Longrightarrow cw = -au-bw . If c\neq 0 then

    dividing by c we'd get that w is a lin. comb. of u,v, so it must be c=0\Longrightarrow au+bv=0\Longrightarrow \mbox{ also } a=b=0\Longrightarrow we're done.

    Tonio
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  3. #3
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    in a different way,
    Let a,b,c be scalars such that au+bv+cw=0
    if we had c\neq  0 we would have w=-\frac{1}{c}\left ( au+bv \right ) which implies w\in \texttt{Span}(v,u) and that's not true by assumption.Therefore c=0.As a result, we go back to au+bv=0.And.. we're done.
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