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Homework Help: Subspace spanned(Is this right?)

  1. Jan 25, 2010 #1
    so, if I wanna calculate the subspace spanned by A in:

    [tex]A = {(1,0,1) , (0,1,0)} in R^{3}[/tex]

    [tex] c_{1}(1,0,1)+c_{2}(0,1,0) = (x,y,z)[/tex]

    i can make a system:

    [tex]c_{1} = x[/tex]

    [tex]c_{2} = y[/tex]

    [tex]c_{1} = z[/tex]

    from which I can conclude that x = z, and so, the subspace spanned will be the plane given by x = z.

    Is this right?
     
  2. jcsd
  3. Jan 25, 2010 #2
    This is half a proof. You have shown that the subspace [tex]S[/tex] spanned by [tex]\{(1,0,1), (0,1,0)\}[/tex] is contained in the plane [tex]P = \{(x, y, z) \in \mathbb{R}^3 : x = z \}[/tex]. You also need to show that [tex]S[/tex] contains [tex]P[/tex]. You could do this by direct calculation, or by an argument based on dimension.
     
  4. Jan 25, 2010 #3
    i see. So how would you calculate it then?
     
  5. Jan 25, 2010 #4
    You showed that [tex]S \subset P[/tex] by showing that if [tex]v \in S[/tex], that is, [tex]v[/tex] is a linear combination of [tex](1,0,1)[/tex] and [tex](0,1,0)[/tex], then [tex]v \in P[/tex], that is, the first and third coordinates of [tex]v[/tex] are equal.

    Show that [tex]S \supset P[/tex] by proving the reverse implication: if [tex]v \in P[/tex], that is, if the first and third coordinates of [tex]v[/tex] are equal, then [tex]v \in S[/tex], that is, [tex]v[/tex] is a linear combination of [tex](1,0,1)[/tex] and [tex](0,1,0)[/tex]. You should be able to exhibit explicitly the coefficients in this linear combination, using the components of [tex]v[/tex].
     
  6. Jan 25, 2010 #5
    got it, thanks!
     
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