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Homework Help: Basis of a subspace?

  1. Feb 15, 2009 #1
    Hi, I had a basic linear algebra question

    Question #1

    1. The problem statement, all variables and given/known data

    Find a basis for the subspace of R3 for which the components in all of the vectors sum to zero.

    2. Relevant equations

    If u and v are in w and w is a subspace, then a*u + b*v is in w.

    3. The attempt at a solution

    w = {v in R3 : v1 + v2 + v3 = 0}

    Okay, so let's say you have Ax = b, where the column space of A is the basis B, and b is a vector which is in w.

    I really don't know how to work with this problem beyond that. I can imagine a basis looking something like:

    [1, 0, 0], [0, -1/2, 0], [0, 0, 1/2]

    Because if you add those vectors together, all of the components sum to 0. And those are indeed linearly independent. But I don't know if those are the right basis vectors.


  2. jcsd
  3. Feb 15, 2009 #2
    Call the subspace described in the problem W.

    If v = (x, y, z) is in W, then x+y+z=0. One equation, three unknowns => 2 parameters, so let y=s, z=t. Then we have v = (-y - z, y, z) = (-1, 1, 0)s + (-1, 0, 1)t.

    (Note that the condition that x+y+z=0 for each v=(x,y,z) in W is equivalent to saying that W is the perpendicular subspace of span(1, 1, 1).)
  4. Feb 15, 2009 #3


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    Or, slightly different approach, since v1+ v2+ v3= 0, v3= -v1- v2. Let v1= 1, v2= 0 so v3= -1. We have (1, 0, -1). Let v1= 0, v2= 1 so v3= -1. We have (0, 1, -1). Those are basis vectors. That's not the same two vectors as Unco got but there are an infinite number of different bases for this subspace.
  5. Feb 15, 2009 #4
    Oh, so when he says "Find a basis", he doesn't mean find all of the bases, he just means find a single vector in the basis?

    So if I had another question "Find a basis for a subspace of R3 in which all vectors satisfy:

    (1 1 0) v = 0

    Then I could just give a vector like:

    (-1, 1, 0) and then say that I found a basis?
  6. Feb 15, 2009 #5


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    It means find all of the vectors in a single basis
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