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Math Help - Showing the closest point in a set to a point outside the set is orthogonal to

  1. #1
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    Showing the closest point in a set to a point outside the set is orthogonal to

    Hi everyone,

    I have a bit of an algebra/convexity type problem. I've been asked to show that if V is a linear subspace in R^n and y is not in V, then show that x* in V is the closest point in V to y if and only if y-x* is orthogonal to V. I'm a bit stuck on this. I know any subspace must be a hyperplane, so intuitively, this makes sense, but I'm having trouble bridging the gap to make it rigorous. Any ideas?

    Thanks for any help!

    Best,

    Alex
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  2. #2
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    Quote Originally Posted by cutmeout View Post
    Hi everyone,

    I have a bit of an algebra/convexity type problem. I've been asked to show that if V is a linear subspace in R^n and y is not in V, then show that x* in V is the closest point in V to y if and only if y-x* is orthogonal to V. I'm a bit stuck on this. I know any subspace must be a hyperplane, so intuitively, this makes sense, but I'm having trouble bridging the gap to make it rigorous. Any ideas?

    Thanks for any help!

    Best,

    Alex
    let V^{\perp}=\{z \in \mathbb{R}^n: \ <z,v>=0, \ \forall v \in V \} be the orthogonal complement of V in \mathbb{R}^n. we know that \mathbb{R}^n=V \oplus V^{\perp}. so: y=v_0+w_0, for some unique v_0 \in V, \ w_0 \in V^{\perp}. now let v be any element
    of V. then: ||y-v||^2=||v_0 - v +w_0||^2=<v_0-v+w_0, \ v_0-v+w_0>=||v_0-v||^2 + ||w_0||^2. (note that <v_0-v, \ w_0>=<w,\ v_0-v>=0, because v_0-v \in V and w_0 \in V^{\perp}.)

    so we showed that for all v \in V: \ ||y-v||^2=||v_0-v||^2+||w_0||^2. now v_0,w_0 are constants. thus ||y-v|| is minimum if and only if ||v_0-v|| is minimum, i.e. v=v_0. but then y-v=w_0 \in V^{\perp}.

    conversely if y - v_1 =w_1 \in V^{\perp}, for some v_1 \in V, then y=v_1+w_1=v_0+w_0. thus v_1=v_0, \ w_1=w_0. hence ||y-v_1||=||y-v_0||, and as we showed: ||y-v_0||=\min_{v \in V} ||y-v||. \ \ \ \Box
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  3. #3
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    It makes so much more sense now. Thank you very much!
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