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Math Help - Orthogonal Projections Proof

  1. #1
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    Orthogonal Projections Proof

    Let x Rn and let p be the orthogonal projection of x onto W where W is a subspace of Rn. Provethat for all y W ,
    ||x(p+y)||2 =||xp||2 +||y||2.

    Now expanding out using the defn of norm (dot product) doesn't get me very far, any hints?
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  2. #2
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    Re: Orthogonal Projections Proof

    \|(x-p)+y)\|^2= <(x-p)+y,(x-p)+y>\text{  } =

    <x-p,x-p> -2<x-p,y>+<y,y> \text{  }=

    \|x-p\|^2+\|y\|^2-2<x-p,y>

    now y is in W, x-p is orthogonal to W so

    <x-p,y>=0
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  3. #3
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    Re: Orthogonal Projections Proof

    [deleted]
    Last edited by romsek; December 12th 2013 at 10:18 PM. Reason: being an idiot
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  4. #4
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    Re: Orthogonal Projections Proof

    Quote Originally Posted by romsek View Post
    \|(x-p)+y)\|^2= <(x-p)+y,(x-p)+y>\text{  } =

    <x-p,x-p> -2<x-p,y>+<y,y> \text{  }=

    \|x-p\|^2+\|y\|^2-2<x-p,y>

    now y is in W, x-p is orthogonal to W so

    <x-p,y>=0
    bah this is all wrong. It should be

    \|(x-p)-y)\|^2= <(x-p)-y,(x-p)-y>\text{  } =

    <x-p,x-p> -2<x-p,y>+<y,y> \text{  }=

    \|x-p\|^2+\|y\|^2-2<x-p,y>

    and the same result.
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