Results 1 to 3 of 3

Math Help - Orthogonal projection

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Orthogonal projection

    I'm unable to do the following problem : Let V be a vector space over \mathbb{K} whose dimension is n and with an inner product <,>. Let W be a subspace of V with dimension m and \{ w_1,...,w_m \} be an orthonormal basis of W.
    We define the orthogonal projection over W as P_W(v)=\sum_{i=1}^{m} <v,w_i>w_i.
    Demonstrate that ||v-P_W(v) || \leq || v-w||, \forall w \in W.
    My attempt : I suppose it's true and want to see that it's really true ( don't want to fall over any contradiction).
    So ||v-P_W(v) || \leq || v-w|| \Leftrightarrow ||v-P_W(v) ||^2 \leq || v-w||^2 \Leftrightarrow <v,-P_W(v)> \leq <v,-w>. And I didn't reach anything.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by arbolis View Post
    I'm unable to do the following problem : Let V be a vector space over \mathbb{K} whose dimension is n and with an inner product <,>. Let W be a subspace of V with dimension m and \{ w_1,...,w_m \} be an orthonormal basis of W.
    We define the orthogonal projection over W as P_W(v)=\sum_{i=1}^{m} <v,w_i>w_i.
    Demonstrate that ||v-P_W(v) || \leq || v-w||, \forall w \in W.
    My attempt : I suppose it's true and want to see that it's really true ( don't want to fall over any contradiction).
    So ||v-P_W(v) || \leq || v-w|| \Leftrightarrow ||v-P_W(v) ||^2 \leq || v-w||^2 \Leftrightarrow <v,-P_W(v)> \leq <v,-w>. And I didn't reach anything.
    Okay, here's how I would do it:

    Let \{w_1,...,w_m \} \subset \{v_1,...,v_n \} where w_i=v_i for 1 \leq i \leq m and \{v_i \} is an orthonormal basis for V, and so we have v= \sum_{i=1}^ n \ {<v,v_i>v_i} and so if u=P_W (v)= \sum_{i=1}^ m \ {<v,v_i>v_i} :

    \Vert v-u \Vert ^2 = \Vert  \sum_{i=1}^ n \ {<v,v_i>v_i} - \sum_{i=1}^m \ {<v,v_i>v_i} \Vert ^2 = \Vert \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2

    Now, pick w \in W then w= \sum_{i=1}^ m \ {<w,v_i>v_i} and so:

    \Vert v-w \Vert ^2 = \Vert \sum_{i=1}^ m \ {<v-w,v_i>v_i} + \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2 = \Vert \sum_{i=1}^ m \ {<v-w,v_i>v_i} \Vert ^2 + \Vert \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2 \geq \Vert \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2 = \Vert v-u \Vert ^2

    Another thing, in your post you have \Vert a-b \Vert ^2 = <a,-b> but it's actually <a,a> - <a,b> - <b,a> + <b,b>.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thank you very much Jose. I'll think about it tomorrow. To tell you the truth I was so tired when I posted my thread that I actually didn't think a lot but rather used my short term memory to post what I did today.
    I need a rest. My exam is on Monday so I'm pushing hard!
    See you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Orthogonal Projection
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 7th 2011, 05:30 PM
  2. Orthogonal projection
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 18th 2010, 06:58 AM
  3. Orthogonal Projection
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 17th 2010, 07:06 AM
  4. Replies: 1
    Last Post: December 2nd 2009, 11:14 PM
  5. orthogonal projection
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 6th 2008, 08:00 PM

Search Tags


/mathhelpforum @mathhelpforum