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Question for the spanning set

  1. Feb 15, 2009 #1
    support that p is in M and M=span{v,w},that means p=av+bw, right?
    then, could either a or b be zero?

    regarded to this following problem:
    Is is possible that {[1 2 0]^T,[2 0 3]^T} can span the subspace U={[r s 0]^T / r and s in R}?


    I thought if s=2r and the scalar for [2 0 3]^T could be zero, it might be possible?
     
  2. jcsd
  3. Feb 15, 2009 #2

    HallsofIvy

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    U is a two dimensional vector space so [r 2r 0]^T is only a one dimemsional subspace of it. [1 2 0]^T spans that subspace of U but not U.

    The question is, "suppose [r s 0]^T is a general vector in U. Do there exist numbers x, y such that [r s 0]^T= x[1 2 0]^T+ y[2 0 3]^T?" If so then we would have [r s 0]^T= [x 2x 0]^T+ [2y 0 3y]= [x+ 2y 2x 3y]. The means we must have r= x+ 2y, s= 2x, 0= 3y. From the last equation, obviously y= 0. But then we r= x, s= 2x. Since r and s can be any two real numbers, there does not exist a number x satifying both equations.
     
  4. Feb 16, 2009 #3
    I see, thx.
    how about my first question, I would still like to know the answer to that:
    support that p is in M and M=span{v,w},that means p=av+bw, right?
    then, could either a or b be zero?


    Also, I have just seen a problem that asks for a spanning set for the zero subspace {0} of R^n.
    Then, I let [0 0 0 0]^T be the zero subspace of R^4, so is this a spanning set for that zero subspace:
    span{[0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0]}
     
    Last edited: Feb 16, 2009
  5. Feb 16, 2009 #4

    HallsofIvy

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  6. Feb 16, 2009 #5
    thx.
    I am sorry that I have got a more problem right away. Could you help again?

    Do you know how to determine if a set of some given vectors in R^n can span R^n?
    regarded to this question:

    Determine if the following given vectors span R^4.
    {[1 3 -5 0] [-2 1 0 0] [0 2 1 -1] [1 -4 5 0]}.


    I thought they could because I seem to be able to reduce the matrix with these vectors as columns into an identity matrix. But the answer in the book says they can't. I was getting confused.
     
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