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Checking if sets are subspaces of ##\mathbb{R}^{3}##

  1. Oct 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Is the set ##W## a subspace of ##\mathbb{R}^{3}##?
    ##W=\left \{ \begin{bmatrix}
    x\\
    y\\
    z
    \end{bmatrix}:x\leq y\leq z \right \}##
    2. Relevant equations


    3. The attempt at a solution
    I believe the set is indeed a subspace of ##\mathbb{R}^{3}##, since it looks like it will satisfy the three properties of a subspace. I'm wondering as to how one would go about explicitly proving this. Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Oct 8, 2015 #2

    Mark44

    Staff: Mentor

    What are the three properties you need to verify? You didn't list them in the 2nd section or anywhere else.

    Generally, you take one or two arbitrary members of your set and show that the set is closed under addition and scalar multiplication.
     
  4. Oct 8, 2015 #3
    So, the properties would be that the set needs to contain the zero vector, needs to be closed under scalar multiplication, and needs to be closed under addition. When you say take arbitrary members of the set to test the last two properties, what exactly do you mean?
     
  5. Oct 8, 2015 #4

    Mark44

    Staff: Mentor

    Let, say, ##\vec{u}## and ##\vec{v}## be elements of W, with ##\vec{u} = <u_1, u_2, u_3>## and ##\vec{v} = <v_1, v_2, v_3>##.
    What condition do these vectors need to satisfy in order to belong to set W? You are given this condition.

    Is 0 an element of W?
    If ##\vec{u}## and ##\vec{v}## are arbitrary members of W, is ##\vec{u} + \vec{v}## also in W?
    If ##\vec{u}## is an arbitrary member of W, and k is an arbitrary scalar, is ##k\vec{u}## also in W?

    If the answers to these questions are all "yes", then the set is a subspace of ##\mathbb{R}^3##.
     
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