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Proving that a subset is a subspace

  1. Oct 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine whether ##W = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : a_1 = 3a_3,~ a_3 = -a_2 \}## is a subspace of ##\mathbb{R}^3##.

    2. Relevant equations


    3. The attempt at a solution
    To show that a subset of vector space is a subspace we need to show three things: 1) That the zero vector of R^3 is in W. 2) That W is closed under vector addition. 3) That W is closed under scalar multiplication.

    1) if ##a_2 = 0## then ##a_1 = 0,~a_3 = 0##, so the zero vector is in W.
    2) I'm not exactly sure how to clearly show this one. Here is my attempt: ##(a_1, a_2, a_3)+ (b_1, b_2, b_3) = (a_1 + b_1, a_2 + b_2, a_3 + b_3) = (3(a_2 + b_2), a_2 + b_2, -(a_2 + b_2))##, which is of the form of the vector defined in W.
    3) Not exactly sure how to show this one either, but here is my attempt: ##c(a_1, a_2, a_3) = (ca_1, ca_2, ca_3) = (3(ca_2), ca_2, -(ca_2))##, which is of the form of the vectors in W.

    Thus, W is a subspace of R^3

    Is this a correct proof? Am I doing 2) and 3) right or is there a better way?
     
  2. jcsd
  3. Oct 9, 2016 #2

    Mark44

    Staff: Mentor

    Start with two vectors that clearly belong to W, such as ##u = <-3u_2, u_2, -u_2>## and ##v = <-3v_2, v_2, -v_2>##.

    Edit: From the problem statement, I determined that all three coordinates are directly or indirectly related to the second coordinate, so I wrote all three coordinates in terms of the second.
    Similar idea as above -- start with ##u = <-3u_2, u_2, -u_2>##.
     
    Last edited: Oct 9, 2016
  4. Oct 9, 2016 #3
    So what about for more complicated potential subspaces, such as ##W = \{(a_1, a_2, a_3) \in \mathbb{R}^3 : 5a_1^2 - 3a_2^2 + 6a_3^2 = 0 \}##? Would adding specific solutions to see if it is closed be better than solving for ##a_1## and putting that into the tuple and adding that to another tuple of the same form to see if that's closed?
     
  5. Oct 9, 2016 #4

    Mark44

    Staff: Mentor

    You could solve for a1 in terms of the other two variables.

    ##a_1 = \pm \sqrt{(3/5)a_2^2 - (6/5)a_3^2}##
    ##a_2 = a_2##
    ##a_3 = a_3##
    Now if you have two such vectors in this set, is their sum in the set? Is a scalar multiple of this vector in the set? I didn't check for a zero vector, since there is obviously such a vector in the set.
     
  6. Oct 9, 2016 #5
    Well, ##(\pm \sqrt{(3/5)a_2^2 - (6/5)a_3^2}) + (\pm \sqrt{(3/5)b_2^2 - (6/5)b_3^2}) \neq \pm \sqrt{(3/5)(a_2 + b_2)^2 - (6/5)(a_2 + b_2)^2}##, so is that enough to show that it is not closed under addition?
     
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