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Dimension of subspace of V^n with orthogonal vectors

  1. Jan 17, 2010 #1
    in a space V^n, prove that the set of all vectors {v1,v2,..}, orthogonal to any v≠0, form a subspace V^(n-1).

    i know that a subspace of V^n must be at least one dimension less and the set of vector v1,v2,... build a orthogonal basis, but how can one show with this preconditions that the subspace has to be of dimension (n-1)?
     
  2. jcsd
  3. Jan 17, 2010 #2
    For a fixed non-zero vector v extend (v) to a basis [itex](v,b_2,b_3,\ldots,b_n)[/itex] of V^n. Now what is the dimension of [itex]\text{Span}(b_2,\ldots,b_n)[/itex] and how this space relate to all vectors orthogonal to v?
     
  4. Jan 17, 2010 #3
    so would v1+ span(b2,...,bn) be a subspace of v^n ?
    but why should (v,b2,b3,...,bn) be a basis, when v isnt normalized to a unit vector?

    thanks rasmhop for your reply
     
  5. Jan 17, 2010 #4
    I'm not talking about an orthonormal basis or anything like that, just a basis. In general in a vector space V of dimension n, for m linearly independent vectors [itex]b_1,\ldots,b_m[/itex] we can find vectors [itex]b_{m+1},\ldots,b_n[/itex] such that [itex]b_1,\ldots,b_n[/itex] is a basis for V.

    The idea of my first reply was that we let [itex]W = \text{Span}(b_2,\ldots,b_n)[/itex] and let W' be the vector space of vectors orthogonal to v. Now it's possible to show W=W' by showing that if a vector is in W', then it must be in W; and if a vector is in W, then it is in W'.
     
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