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Orthogonal Matrices and span

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data

    If w is orthogonal to u and v, then show that w is also orthogonal to span ( u , v )

    2. Relevant equations

    two orthogonal vectors have a dot product equalling zero

    3. The attempt at a solution

    I can see this geometrically in my mind, and I know that w . u = 0 and w . v = 0
    but I don't know or understand how I can show this for its span in writing.
     
  2. jcsd
  3. Jan 17, 2009 #2
    a vector in span(u,v) is of the form au+bv. So w . (au+bv) = 0, using the distributivity of the dot product.
     
  4. Jan 17, 2009 #3
    you said that span(u,v) is in this form au+bv

    so
    w . (au+bv) = 0
    w . au + w . bv = 0

    where a and b are any scalar numbers
    and that's all? There's no more to it?

    [edit]
    thanks grief. That one little bit helped a lot!
     
    Last edited: Jan 17, 2009
  5. Jan 17, 2009 #4
    No, Grief said a vector in the subspace span(u,v) is of the form au+bv for scalars a and b. To say that a vector is orthogonal to a subspace means that the vector is orthogonal to each vector in that subspace.

    You need to show that given a vector x in span(u,v), we have w.x=0 . From above, x is of the form au+bv, so you want to show that w.(au+bv)=0. This means beginning with w.(au+bv) and showing it equals 0. As Grief already said, to do so just requires distributivity and recognising that w being orthogonal to u and to v means that w.u=0 and w.v=0.
     
  6. Jan 18, 2009 #5
    To make sure I got this right one more time:
    a vector in span(u,v) is in this form au+bv

    making (au+bv) dot w = 0 shows it is orthogonal, meaning any vector in that span(u,v) is orthogonal to w
    then it'll become w . au + w . bv = 0
    then w . au = a(w.u) = 0 since w.u = 0 (orthogonal)
    w . bv = 0 since w . v = 0 (orthogonal)
    so then 0 = 0

    right?
     
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