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Math Help - [SOLVED] Hard Orthogonal Complement Proof

  1. #1
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    [SOLVED] Hard Orthogonal Complement Proof

    At least, I think it is hard. I don't know how to start this at all.

    Let V be a vector space equipped with a positive definite inner product. Let W be a subspace of V. Let W' be its orthogonal complement. Define a map \pi: V -> V as \pi(v) = v_1 where v = v_1 + v_2 with v_1 in W and v_2 in W'. Show that:

    a) The map \pi is welldefined, linear with range W and that \pi^2 = \pi. The map \pi is called the projectiion map onto the subspace W.

    b) Define v := I - \pi. Show that v is the projection map onto W' i.e. show that v is a linear map with range W' and that v^2 = v.
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  2. #2
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    Quote Originally Posted by Zalren View Post
    At least, I think it is hard. I don't know how to start this at all.

    Let V be a vector space equipped with a positive definite inner product. Let W be a subspace of V. Let W' be its orthogonal complement. Define a map \pi: V -> V as \pi(v) = v_1 where v = v_1 + v_2 with v_1 in W and v_2 in W'. Show that:

    a) The map \pi is welldefined, linear with range W and that \pi^2 = \pi. The map \pi is called the projectiion map onto the subspace W.

    b) Define v := I - \pi. Show that v is the projection map onto W' i.e. show that v is a linear map with range W' and that v^2 = v.
    very staightforward. note that V=W \oplus W'. let v_1=w_1+w_1', \ v_2 =w_2 + w_2' be in V, where w_j \in W, \ w_j' \in W'.

    to show \pi is well-define suppose that v_1=v_2. then w_1=w_2 and so \pi(v_1)=w_1=w_2=\pi(v_2). so \pi is well-defined.

    to show \pi is linear let c be a scalar. then \pi (cv_1 +v_2)=\pi (cw_1+w_2 + cw_1' + w_2')=cw_1+w_2=c \pi(v_1) + \pi(v_2).

    also \pi^2(v_1)=\pi(\pi(v_1))=\pi(w_1)=w_1=\pi(v_1) and so \pi^2=\pi. the rest of the problem is as easy as this.
    Last edited by NonCommAlg; April 3rd 2010 at 09:16 PM.
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