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Linear Algebra - Generators Proof

  1. Feb 12, 2006 #1
    Just curious if my proof is sufficient, again :smile:
    Let [itex]A_{1},... ,A_{r}[/itex] be generators of a subspace V of [itex]R^n[/itex]. Let W be the set of all elements of [itex]R^n[/itex] which are perpendicular to [itex]A_{1},... ,A_{r}[/itex]. Show that the vectors of W are perpendicular to every element of V.

    We must show that [itex]v \cdot w = 0[/itex], where [itex]v \in V[/itex], and [itex]w \in W[/itex]
    but since [itex]A_{1},... ,A_{r}[/itex] generate V, then v can be written as:
    [tex]v = x_{1}A_{1} + ... + x_{r}A_{r}[/tex]

    so, [tex]v \cdot w = (x_{1}A_{1} + ... + x_{r}A_{r}) \cdot w[/tex]
    [tex]= (x_{1}A_{1}) \cdot w + ... + (x_{r}A_{r}) \cdot w[/tex]
    [tex]= x_{1}(A_{1} \cdot w) + ... + x_{r}(A_{r} \cdot w)[/tex]
    [tex]= x_{1}(0) + ... + x_{r}(0)[/tex]
    [tex]= 0[/tex]

    So, [tex]v \cdot w = 0[/tex], and every vector of W is perpendicular to every element of V.
    Last edited: Feb 12, 2006
  2. jcsd
  3. Feb 12, 2006 #2


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    Use [ itex ] instead of [ tex ] inside of paragraphs.

    What sort of thing is


  4. Feb 12, 2006 #3
    Thanks, I copied what I wrote originally, and forgot to change the commas to plusses :redface:
  5. Feb 12, 2006 #4
    20 minutes later, and I finally got the latex correct :grumpy:
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