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Linear algebra subspace proof

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that if S is a subspace of R1, then either S={0} or S=R1.

    Trying to come up with a proof I dissected each statement, I know that in order for S to be

    a subspace the zero vector must lie within the subset. So I know S={0} is true. I then

    checked an arbitary vector x1 which lies on R1 to make sure it

    was closed under scalar multiplication, and addition, and that checked out as well.

    Not sure if I am on the right track.

    Thanks


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2013 #2

    jbunniii

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    You need to specify what field of scalars you are using. The statement is true if the field is ##\mathbb{R}##, false if it is, for example, ##\mathbb{Q}##.

    Assuming you are using ##\mathbb{R}## for the field of scalars, use the fact that a subspace must be closed under scalar multiplication.
     
  4. Feb 10, 2013 #3
    Your ideas are right, but it's not really a proof unless you do everything step-by-step.
    For example, for the first possible set: ##S = \{0\}##, show that all three properties of a subspace are satisfied.
     
  5. Feb 10, 2013 #4

    jbunniii

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    Why is that necessary? The problem statement tells you that ##S## is a subspace. All you need to do is show that if it's not ##\{0\}## then it must be all of ##\mathbb{R}^1##.

    So: if ##S## is not ##\{0\}##, then ##S## contains a nonzero element, say ##s \neq 0##. Since ##S## is a subspace, it is closed under scalar multiplication. Thus it must contain ##\alpha s## for every ##\alpha \in F## where ##F## is the scalar field (presumably ##\mathbb{R}##). Therefore...?
     
  6. Feb 11, 2013 #5
    s=r1?
     
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