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

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

    b) Define $\displaystyle v := I - \pi$. Show that $\displaystyle v$ is the projection map onto $\displaystyle W'$ i.e. show that $\displaystyle v$ is a linear map with range $\displaystyle W'$ and that $\displaystyle v^2 = v$.
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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 $\displaystyle V$ be a vector space equipped with a positive definite inner product. Let $\displaystyle W$ be a subspace of $\displaystyle V$. Let $\displaystyle W'$ be its orthogonal complement. Define a map $\displaystyle \pi$: $\displaystyle V$ -> $\displaystyle V$ as $\displaystyle \pi(v) = v_1$ where $\displaystyle v = v_1 + v_2$ with $\displaystyle v_1$ in $\displaystyle W$ and $\displaystyle v_2$ in $\displaystyle W'$. Show that:

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

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

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

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

    also $\displaystyle \pi^2(v_1)=\pi(\pi(v_1))=\pi(w_1)=w_1=\pi(v_1)$ and so $\displaystyle \pi^2=\pi.$ the rest of the problem is as easy as this.
    Last edited by NonCommAlg; Apr 3rd 2010 at 08:16 PM.
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