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Subspace (question)

  1. May 7, 2005 #1
    I just wanted to know if my answer is acceptable.

    Q: S={(x,y,z) E [itex]\mathbb{R}^{3}[/itex] l x^2 + y^2 +z ^2 =0}
    Is it a subspace of [itex]\mathbb{R}^{3}[/itex]?


    My answer:

    It is a subspace if x=0, y =0, z= 0

    Let u=(0,0,0) u2=(0,0,0) and k be a scalar

    u + u2 = (0,0,0) Closed under addition

    ku = k(0,0,0) = (0,0,0) Closed under scalar multiplication



    Is my answer complete? I'm not sure if I'm allowed to assume that the values of x,y,z are 0.

    Thank you in advance
     
    Last edited: May 7, 2005
  2. jcsd
  3. May 7, 2005 #2

    quasar987

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    Actually, S={(x,y,z) E [itex]\mathbb{R}^{3}[/itex] l x^2 + y^2 +z ^2 =0} = {(0,0,0)}

    I.e. (0,0,0) is the only element in that set. It is made clear if you recall that x²+y²+z²=R² describe a sphere of radius R. Here R=0. The only point "on" a sphere of radius 0 is (0,0,0).

    So, with this in mind, instead of the tentative "It is a subspace if x=y=z=0", start the proof with: "The only element of S is (0,0,0)", and then go on to show that the 3 conditions for S to be a subspace of R^3 are satisfied (like you did).
     
    Last edited: May 7, 2005
  4. May 7, 2005 #3
    Ok thanks.

    What if I have a multiplication:

    S={(x,y,z) E R^3 l xy =0}.
    To prove that this is a subspace (or not a subspace), the only way I know to do so is by assuming a value for x, y and z before checking if it's closed under scalar mutiplication adn addition. Is there another way?
     
  5. May 7, 2005 #4
    You can use [tex]a=(a_1,a_2,a_3), b=(b_1,b_2,b_3)[/tex] and since either x or y must be 0 to satisfy your condition of xy=0, you can have the condition that either [tex]a_1[/tex] or [tex]a_2[/tex] must be zero, and similarly for [tex]b_1[/tex] and [tex]b_2[/tex]. Then show that a and b are closed under scalar addition and multiplication, and you have proofed it for all of the values in the subspace
     
  6. May 7, 2005 #5

    quasar987

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    You must never assume a particular value of (x,y,z). You must always work your ways through the 3 conditions while assuming the most general form possible for (x,y,z).

    First condition: Is (0,0,0) in S? Consider (x=0,y=0,z=0). Then xy=0, so (0,0,0) is in S. *check*

    Second condition: Consider two vectors [itex]\vec{u}_1 = (x_1,y_1,z_1)[/itex] and [itex]\vec{u}_2 = (x_2,y_2,z_2)[/itex] in S. Because they are in S, they have the property that [itex]x_1 y_1 = 0[/itex] and [itex]x_2 y_2 = 0[/itex]. According to the definition of addition in [itex]\mathbb{R}^3[/itex], we have [itex]\vec{u}_1+\vec{u}_2 = (x_1+x_2,y_1+y_2,z_1+z_2)[/itex]. Now the condition: Does [itex](x_1+x_2)(y_1+y_2)=0[/itex]?. Let's see: In view of the "axiom of distributivity" in [itex]\mathbb{R}[/itex], [itex](x_1+x_2)(y_1+y_2)=x_1y_1+x_1y_2+x_2y_1+x_2y_2[/itex]. We know by hypothesis that [itex]x_1y_1 = x_2y_2=0[/itex]. But what about the other two terms? It could be that [itex]x_1\neq 0, \ y_1 = 0, \ x_2 = 0, \ y_2 \neq 0[/itex]. In this case, [itex]x_1y_1 = 0[/itex] and [itex]x_2y_2 = 0[/itex] are indeed satisfied but [itex]x_1y_2 \neq 0[/itex]. Conclusion: [itex]\vec{u}_1+\vec{u}_2 \notin S \ \forall \vec{u}, \ \vec{u}_2 \ \Rightarrow[/itex] S is not a subspace of [itex]\mathbb{R}^3[/itex] because condition 2 (closed under addition) is not met.
     
    Last edited: May 7, 2005
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