1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Subspace (question)
  1. Subspace problem (Replies: 1)