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Math Help - Proof!

  1. #1
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    Lightbulb Proof!

    1. Prove that if u is orthogonal to both v and w, then u is orthogonal to both v + w.
    2. Prove that if u is orthogonal to both v and w, then u is orthogonal to sv + tw for all scalar s and t.


    i think...
    u * (vw) = 0 -> u*v + u*w = 0
    u * (v + w) = 0 -> u*v + u*w = 0
    .....
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by aeubz View Post
    1. Prove that if u is orthogonal to both v and w, then u is orthogonal to both v + w.
    2. Prove that if u is orthogonal to both v and w, then u is orthogonal to sv + tw for all scalar s and t.


    i think...
    u * (vw) = 0 -> u*v + u*w = 0
    u * (v + w) = 0 -> u*v + u*w = 0
    .....
    I don't understand why vw in what you wrote ??

    If u is orthogonal to v, it means that u.v=0, where . denotes the dot product.
    We have :
    u.v=0
    u.w=0

    What is the dot product of u with v+w ?
    u.(v+w)=u.v+u.w, by a property (*) of the dot product.
    And this equals 0 by the hypothesis. Hence u is orthogonal to v+w.


    Same reasoning for the other one (you'll have to use properties (*) and (**))

    ------------------------------------------------------------------
    (*) \vec{x} \cdot (\vec{y}+\vec{z})=\vec{x} \cdot \vec{y}+\vec{x} \cdot \vec{z}

    (**) \forall \lambda \in \mathbb{R} ~,~ \vec{x} \cdot (\lambda \vec{y})=\lambda (\vec{x} \cdot \vec{y})
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  3. #3
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    Another question !!

    Thx Moo!

    Prove that (u + v) * (u - v ) = ||u||^2 - ||v||^2 for all vectors u and v in R^n.

    u.u - u.v + v.u - v.v
    u.u - v.v
    ||u||^2 - ||v||^2

    ??

    And is ||u + v||^2 same thing as (u+v).(u+v)
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  4. #4
    Moo
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    Quote Originally Posted by aeubz View Post
    Another question !!

    Thx Moo!

    Prove that (u + v) * (u - v ) = ||u||^2 - ||v||^2 for all vectors u and v in R^n.

    u.u - u.v + v.u - v.v
    u.u - v.v
    ||u||^2 - ||v||^2

    ??

    And is ||u + v||^2 same thing as (u+v).(u+v)
    Hmmm basically, we have this :

    ||\vec{u}||^2=\vec{u} \cdot \vec{u}

    For any vector u. So it is yes to all your questions ^^


    Note that in order to say that -u.v+v.u=0, you have to use the commutative property of the dot product
    Last edited by Moo; September 25th 2008 at 10:27 PM. Reason: typo
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  5. #5
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    Thx again Moo!

    does anyone maybe have a link to prove the properties of the transpose?
    Transpose - Wikipedia, the free encyclopedia

    My teacher said we should know how to prove these...
    In the website link, # 1 , 2 , 3 , 4 , and (A^r)^T = (A^T)^r for all nonnegative integers.
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  6. #6
    Moo
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    Quote Originally Posted by aeubz View Post
    Thx again Moo!

    does anyone maybe have a link to prove the properties of the transpose?
    Transpose - Wikipedia, the free encyclopedia

    My teacher said we should know how to prove these...
    In the website link, # 1 , 2 , 3 , 4 , and (A^r)^T = (A^T)^r for all nonnegative integers.
    You're welcome !

    But you'd better post new questions in new threads
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  7. #7
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    Quote Originally Posted by Moo View Post
    ||\vec{u}||=\vec{u} \cdot \vec{u}
    For any vector u.
    It should be: ||\vec{u}||^{\color{red}2}=\vec{u} \cdot \vec{u}
    For any vector u.
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