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Linear Algebra Proof

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that a vector k is in 'W perp' if and only if k is orthogonal to every vector in the spanning set of W, where W is a subspace of Rn

    3. The attempt at a solution
    It's so obviously true that I don't know how to prove it! :S

    Here's what I did:
    Let {w1, w2SUB], .... wm} be a spanning for W.
    Let w be a vector in W where
    w = c1w1 + c2w2 + ... + cmwm
    and all the weights c1, c2,.....,cm are nonzero
    Suppose k is in 'W perp' but is not orthogonal to every vector in the spanning set.
    Then k.w
    = k. (c1w1 + c2w2 + ... + cmwm)
    = c1k.w1 + c2k.w2 +... + cmk.wm
    = a, where a is a non-zero number
    But to be in 'W perp', the dot product of k and any vector in W must be zero
    Therefore, k has to be orthogonal to every vector in the spanning set of W to be in 'W perp"

    Is this right?? I can't think of any other way to do it :S
     
    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 11, 2008 #2

    HallsofIvy

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    Your definition of 'Wperp' is the space of all vector orthogonal to every member of W, right? And you are given that a vector is orthogonal to every member of a spanning set?

    Do a direct solution, not indirect. If k.wi= 0, what is k.w?
     
  4. Oct 11, 2008 #3
    Yep 'W perp' (W perpendicular) is the set of all vectors orthogonal to W.
    Oops I realised I made some typo errors. The 'x's in the last two lines were meant to be 'k's. I've edited it now.
    What do you mean by a direct solution?
    And what's 'wi'? :S
     
  5. Oct 11, 2008 #4

    gabbagabbahey

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    He means that your basis set is [itex]\{ \vec{w}_i \}= \{\vec{w}_1,\vec{w}_2, \ldots \vec{w}_n \}[/itex] and if [itex]\vec{k}[/itex] is perpendicular to each of your spanning vectors, [itex]\vec{k} \cdot \vec{w}_i=0 \quad \forall i \in [1,n][/itex]. What does that make [itex]\vec{k} \cdot \vec{w}[/itex]?

    And if [itex]\vec{k}[/itex] is perpendicular to every vector in W, then [itex]\vec{k} \cdot \vec{w}=0[/itex]. What does that make [itex]\vec{k} \cdot \vec{w}_i=0[/itex] for each i?
     
    Last edited: Oct 11, 2008
  6. Oct 11, 2008 #5
    If k is perpendicular to each of the spanning vectors, then k.w = 0
    and if k is perpendicular to every vector in W, then k.wi = 0

    Is that all I have to write? :S
     
  7. Oct 11, 2008 #6

    gabbagabbahey

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    Well, you should explicitly show that each of those claims is true; but other than that, yes; since any vector for which k.w is zero, is in W_perp.
     
  8. Oct 11, 2008 #7
    Oh ok cool :). Thanks guys.
     
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