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Math Help - Properties of Vector Spaces

  1. #1
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    Properties of Vector Spaces

    Hello. I have just started doing proofs involving Linear Algebra and I need some guidance and a little push in the right direction. Here is the result I am trying to prove:
    Let V be a vector space containing nonzero vectors u and v. Prove that if u \neq \alphav for each \alpha \in \mathbb{R}, then u \neq \beta(u + v) for each \beta \in \mathbb{R}.

    Now I am thinking I can prove this using the contrapositive. I believe the contrapositive is: If u = \beta(u+v) for some \beta \in \mathbb{R}, then u = \alphav for some \alpha \in \mathbb{R}.

    If this is the correct negation then I know that u = \beta(u+v) = \betau+ \betav so I need to find a way to make \betau+ \betav = \alphav. Im not quite sure how to go about doing this so I am hoping someone here can push me in the right direction. Thanks
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  2. #2
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    Re: Properties of Vector Spaces

    Quote Originally Posted by Twinsfan85 View Post
    Hello. I have just started doing proofs involving Linear Algebra and I need some guidance and a little push in the right direction. Here is the result I am trying to prove:
    Let V be a vector space containing nonzero vectors u and v. Prove that if u \neq \alphav for each \alpha \in \mathbb{R}, then u \neq \beta(u + v) for each \beta \in \mathbb{R}.

    Now I am thinking I can prove this using the contrapositive. I believe the contrapositive is: If u = \beta(u+v) for some \beta \in \mathbb{R}, then u = \alphav for some \alpha \in \mathbb{R}.

    If this is the correct negation then I know that u = \beta(u+v) = \betau+ \betav so I need to find a way to make \betau+ \betav = \alphav. Im not quite sure how to go about doing this so I am hoping someone here can push me in the right direction. Thanks
    I'lltake over here:

    u=\beta(u+v)~\implies~u=\beta u + \beta v ~\implies~u-\beta u = \beta v ~\implies~ \\ (1-\beta) u = \beta v ~\implies~ u = \underbrace{\frac{\beta}{1-\beta}}_{call \ this\ \alpha} \cdot v
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  3. #3
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    Re: Properties of Vector Spaces

    one caveat: clearly β = 1 would be a "bad choice". so we need to know u = β(u + v) never happens when β = 1. which isn't too hard:

    u = u + v implies v = 0, contradicting that v is, by stipulation, a non-zero vector.
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