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Math Help - Prove ||u-v|| = 2^(1/2)

  1. #1
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    Prove ||u-v|| = 2^(1/2)

    I need to prove that ||u-v|| = 2^(1/2) if {u,v} is an orthonormal set in the vector space V.

    Here is what I have so far, but do not know if I am even on the right page:

    ||u-v|| = <u-v, u-v>^(1/2), which is the definition of distance between vectors

    But where do I go from here?

    Thanks for any help.
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  2. #2
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    Re: Prove ||u-v|| = 2^(1/2)

    Quote Originally Posted by page929 View Post
    I need to prove that ||u-v|| = 2^(1/2) if {u,v} is an orthonormal set in the vector space V.
    \|u-v\|^2=(u-v)\cdot (u-v)=u\cdot u-2u\cdot v+v\cdot v=~?
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  3. #3
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    Re: Prove ||u-v|| = 2^(1/2)

    Quote Originally Posted by Plato View Post
    \|u-v\|^2=(u-v)\cdot (u-v)=u\cdot u-2u\cdot v+v\cdot v=~?
    u^2 - 2uv + v^2
    ||u||^2 - 2uv + ||v||^2
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  4. #4
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    Re: Prove ||u-v|| = 2^(1/2)

    the fact that {u,v} is an orthonormal set tells you you have "special values" for <u,u>, <u,v> and <v,v>. what are they?
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  5. #5
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    Re: Prove ||u-v|| = 2^(1/2)

    Quote Originally Posted by Deveno View Post
    the fact that {u,v} is an orthonormal set tells you you have "special values" for <u,u>, <u,v> and <v,v>. what are they?
    <u,u> = 1, <u,v> = 0, <v,v> = 1
    So, 1+0+1 = 2
    but I need ||u-v|| not ||u-v||^2, so I take the square root and my answer is 2^(1/2)
    Last edited by page929; October 2nd 2011 at 10:01 AM. Reason: SOLVED
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