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Math Help - Linear Independence

  1. #1
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    Linear Independence

    This is like the other question I posted. I get so confused when they start talking about u + v and all these variables. Give me #'s and I'm good. But Variables throw me off for some reason. Anyway, I can't figure this out either

    Let u and v be distinct vectors in R^n. Prove that {u,v} is linearly independent if and only if {u+v, u-v} is linearly independant.

    TIA!!!!
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  2. #2
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    Quote Originally Posted by TreeMoney View Post
    This is like the other question I posted. I get so confused when they start talking about u + v and all these variables. Give me #'s and I'm good. But Variables throw me off for some reason. Anyway, I can't figure this out either

    Let u and v be distinct vectors in R^n. Prove that {u,v} is linearly independent if and only if {u+v, u-v} is linearly independant.

    TIA!!!!
    Suppose {u,v} are linearly independent, and that {u+v,u-v} are not,

    Then there exist a, b in R, at least one of which !=0, such that:

    a(u+v) + b(u-v)=0,

    so:

    (a+b)u + (a-b)v=0

    but now at least one of (a+b) and (a-b) !=0, so we conclude that
    u and v are linearly dependent - a contradiction.

    So we conclude that if {u.v} are linearly independent then so are {u+v,u-v}.

    The second half of the proof proceeds like the above but starting from
    the assumption that {u+v,u-v} are linearly independent and that [u,v}
    are not.

    RonL
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  3. #3
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    Quote Originally Posted by TreeMoney View Post
    Prove that {u,v} is linearly independent if and only if {u+v, u-v} is linearly independant.
    We now that if,
    Cu+Kv=0 then, C=K=0
    We need to show if,
    C(u+v)+K(u-v)=0 then, C=K=0
    That means, (scalar multiplication is commutatiove in vector space)
    Cu+Cv+Ku-Kv=0
    Thus,
    (C+K)u+(C-K)v=0
    By our initial assumption of linear independence if {u,v}
    C+K=0
    C-K=0
    The only solutions is 0.

    Now, for the converse.
    If, C(u+v)+K(u-v)=0
    Then, C=K=0
    Thus, we have,
    Cu+Kv=0 we need to show C=K=0
    Do the following trick, (skipping the distributative commutative stuff)
    (C+K)(u+v)+(C-K)(u-v)=0
    Thus,
    C+K=0
    C-K=0
    Thus,
    C=K=0
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