Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By Deveno

Math Help - Subspaces and orthogonality

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    10

    Subspaces and orthogonality

    Hi everyone

    Stuck on this problem, hoping someone here could point me in the right direction.

    We have two subspaces of an inner product space. The dimension of one is larger than the other (and both finite). The question is if there is at least one nonzero vector in the larger subspace that is orthogonal to all vectors of the smaller one.

    Clearly this seems to be true for R2 and R3 but I don't know how to generalize this and in general I just suck at proving things
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,369
    Thanks
    736

    Re: Subspaces and orthogonality

    in an inner product space, for any subspace W of a vector space V, you can define W = {v in V: <v,w> = 0, for all w in W}, the orthogonal complement of W.

    moreover, V = W⊕W.

    if we have another subspace U, with dim(U) > dim(W), then there must exist some vector u in U, that is not in W (otherwise U would be a subspace of W, and thus dim(U) ≤ dim(W)).

    since u is not in W, it cannot be 0 (because 0 is in every subspace), and therefore it is a non-zero element of W. hence <u,w> = 0 for all w in W.

    *****

    here is an example from R4, with the standard dot product:

    let W = span({(1,0,0,1),(1,1,0,0)}), and let U = span({1,0,0,0),(0,1,0,0),(0,0,1,0)}).

    an element of W is of the form: (a+b,b,0,a). first let's find a basis for W:

    writing a typical element of R4 as (x,y,z,w) we have:

    (a+b,b,0,a).(x,y,z,w) = 0, so: ax+by+aw = 0. letting b = 0, a ≠ 0, we get:

    ax+aw = 0, so w = -x.

    letting a = 0, b ≠ 0, we get: by = 0, so y = 0. so every element of W is of the form:

    (x,0,z,-x) = x(1,0,0,-1) + z(0,0,1,0), so {(1,0,0,-1),(0,0,1,0)} is a basis for W.

    now we just need to find an element of U in W, which is easy: (0,0,1,0) clearly fits the bill. as a check:

    (a+b,b,0,a).(0,0,1,0) = (a+b)(0) + b(0) + (0)(1) + a(0) = 0 + 0 + 0 + 0 = 0, and clearly (0,0,1,0) is NOT in W, for every element of W has a 0 in the 3rd coordinate.
    Last edited by Deveno; August 14th 2012 at 06:07 PM.
    Thanks from Robinator
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Orthogonality and Subspaces
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 12th 2010, 04:14 AM
  2. Orthogonality
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 2nd 2009, 03:41 PM
  3. Orthogonality
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 30th 2009, 07:48 PM
  4. Replies: 0
    Last Post: October 13th 2009, 03:48 PM
  5. Orthogonality
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 28th 2007, 06:15 PM

Search Tags


/mathhelpforum @mathhelpforum