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Homework Help: Vector space

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the dimnesion and a basis of vector space V

    2. Relevant equations
    V is the set of all vectors (a,b,c) in R^3 with a+2b-4c=0

    3. The attempt at a solution
    (4c-2b,b,c) = b(-2,1,0) + c(4,0,1)
    so {(-2,1,0),(4,0,1)} is the basis of the SUBSPACE of V right?

    how do I get the 3rd linearly independent vector so that it forms the basis of V?
    Is it (1,0,0)? (0,1,0) seems possible... but the basis of a vector space isn't unique right?
  2. jcsd
  3. Jan 24, 2010 #2
    This is a good start, and the right idea needed to solve the problem.

    THE subspace? V has many subspaces and you're not asked to find bases of them. What you want is a basis of V. Remember that a basis of V is a set of vectors that span V and is linearly independent. Now consider:
    1. For any vector of the form (4c-2b,b,c) (i.e. a vector in V) can you express it as a linear combination of (-2,1,0) and (4,0,1). If the answer is yes, then (-2,1,0) and (4,0,1) span V and therefore we need not find more basis elements.
    2. Is {(-2,1,0),(4,0,1)} linearly independent?
    If you can answer both 1 and 2 in the affirmative, then {(-2,1,0),(4,0,1)} is a basis of V.

    Why do you believe you need a 3rd vector? You would only need a third vector if the dimension of V is 3, but it could be smaller (2 for instance).

    Neither (1,0,0) nor (0,1,0) is actually in V so they can't be used. No a basis isn't unique, but your set from the first part of your answer is a good guess for a basis.
  4. Jan 24, 2010 #3
    I think I've read somewhere that if there is m linearly independent vectors in R^n where m<n, these vectors form the subspace

    since we have R^3 now and found 2 linearly independent vectors, so I thought these 2 vectors form a subspace...

    does that mean basis of V is JUST {(-2,1,0) , (4,0,1)} in this case?
  5. Jan 24, 2010 #4
    I guess I was thrown of by your use of the. It's true that m linearly independent vectors span a subspace of R^n. In your case V is the subspace of R^n.

    They do, but they do form the subspace V or another?

    Yes, but you need to confirm it by proving that they span V and that they are linearly independent. Of course this implies that V has dimension 2. So in this case {(-2,1,0), (4,0,1)} span a subspace of R^3, and this subspace is V.
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