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Proof: Orthagonal Complement

  1. Apr 30, 2007 #1
    Prove that the orthogonal complement of a subspace of (Rn) is itself a subspace of (Rn)

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    Let V be the orthogonal complement of S, S a subspace of (Rn).

    Let the set of vectors that span (Rn) be written as the columns of matrix A.

    consider the homogenous equation

    A(transpose)u=0

    The solution space of the vectors u will all dot with any row vector from A transpose equaling zero.

    So the null space of A transpose is the subspace V.

    By (Fundamental Subspace Theroem) Two subspaces, Column space of a matricies transpose and the nullspace of that same matrix form a direct sum of (Rn).

    Thus V is also a subspace of (Rn)

    Does this make sense?
    Am I trying way too hard here becuase this seems like it should be an easy one.
     
    Last edited: Apr 30, 2007
  2. jcsd
  3. Apr 30, 2007 #2

    StatusX

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    Yea, that seems about right, although I think you want the columns to span S, not R^n.

    It would probably be a good exercise to try to prove this directly from the definition of the orthogonal complement of a subspace, rather than appealing to that theorem.
     
  4. Apr 30, 2007 #3

    HallsofIvy

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    I wouldn't do anything that complicated. In particular, there is no reason to appeal to "matrices". A subset of a vector space is a subspace if and only if au+ bv is a membere of subset whenever u and v are members of the subset and a and b scalars. If u and v are members of v and w a member of S, what is the innerproduct of (au+bv) with w?
     
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