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Inner Product Space and Orthogonality proof question (is this the correct way?)

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Let V be the inner product space. Show that if w is orthogonal to each of the vectors u1, u2,...,ur, then it is orthogonal to every vector in the span{u1,u2,....,ur}.


    2. Relevant equations

    u.v=0 to be orthogonal
    If u and v are vectors in an inner product space, then ||u+v||^2 = ||u||^2 + ||v||^2

    3. The attempt at a solution
    So we know ||u+v||^2 = ||u||^2 + ||v||^2 for ortho vectors

    for u1: <u1,w>=0
    ||u1+w||^2 = <u1+w, u1+w>
    = <u1,u1> + 2<u1,w> + <w,w>
    = ||u1||^2 + ||w||^2

    then i do the same for u2 and ur.

    Is this the correct way to go about this?

    Thanks
     
  2. jcsd
  3. Nov 21, 2012 #2

    Dick

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    No, it isn't. What's the definition of span?
     
  4. Nov 21, 2012 #3
    So w has to be orthogonal to all of the vectors in the span{u1,u2,....,ur}

    k1u1+k2u2+......+krur=b to test for spanning where b is an arbitrary vector.
    would 'b' be w in this case?
     
  5. Nov 21, 2012 #4

    Dick

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    No, b would be an arbitrary vector in span{u1,u2,....,ur}. So you want to show b.w=0. What is (k1u1+k2u2+......+krur).w?
     
  6. Nov 21, 2012 #5
    i am sooo confused.

    b and w are both vectors.
    so b=(b1,b2,...br)
    w=(w1,w2,....wr) w vector is perpendicular to b vector (b lies in the span of u1,u2,...ur)

    so b1w1+b2w2+....+brwr=0

    *i noticed u edited what u asked me.

    What is (k1u1+k2u2+......+krur).w?

    k1u1w1+k2u2w2+.......+krurwr = 0 (im assuming it has to equal zero to be orthogonal)
     
  7. Nov 21, 2012 #6

    Dick

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    Look b=(k1u1+k2u2+......+krur). That's an arbitrary vector in the span. The k's are numbers and the u's are vectors. w is just w. Compute (k1u1+k2u2+......+krur).w by using the distributive property. You know (a+b).c=a.c+b.c where a,b,c are vectors, yes? You also know (xa).c=x(a.c) where a and c are vectors and x is a number, yes? Use those properties.
     
  8. Nov 21, 2012 #7
    Compute (k1u1+k2u2+......+krur).w
    =w.k1u1 + w.k2u2 + ...... + w.krur
    =k1(w.u1) + k2(w.u2) + ...... + kr(w.ur)

    sorry about bein really dumb with this, im horrible with proofs. but good with numbers....
     
  9. Nov 21, 2012 #8

    Dick

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    I'm sure you are good with numbers. 'Abstract' throws some people. Just pretend they are numbers. Now what are w.u1, w.u2, etc?
     
  10. Nov 21, 2012 #9
    w.u1=0
    w.u2=0
    ..
    w.ur=0

    they are all orthogonal to each other.

    =k1(w.u1) + k2(w.u2) + ...... + kr(w.ur)
    =k1(0) +k2(0) +.......+ kr(0)
    =0

    and btw man i really appreciate u walkin me through this proof. This helps me sooo much for this proof and just solving proofs in general.
     
  11. Nov 21, 2012 #10
    btw, u must know proofs well and understand them well.
    Do you have any tips for studying them/remembering proofs for tests/exams?
     
  12. Nov 21, 2012 #11

    Dick

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    I'll give you one that's pretty important. Look up the definition of any terms you are a little vague on. That's why my first hint was "What's the definition of span?". Your first attempt didn't even use that. That's a pretty good hint you are missing something. Other than that, practice makes perfect.
     
  13. Nov 21, 2012 #12
    ok, i will for sure do that while im studying.
    Thanks for all the help.
     
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