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Determining if vectors in R3 are linear subspaces

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data
    U={(x1,x2,x3)[itex]\in[/itex]ℝ3 | x1+x2=0}
    Is this a linear subspace of ℝ3?


    2. Relevant equations
    x1+x2=0


    3. The attempt at a solution
    I know that in order to be a linear subspace, it must be closed under addition and scalar multiplication. I'm just not really sure how to incorporate the x1+x2=0. This is what I've done:
    x=(x1,x2,x3), y=(y1,y2,y3)
    x+y=(x1,x2,x3)+(y1,y2,y3)=(x1+y1, x2+y2, x3+y3)
    but how does this relate to x1+x2=0? there is no "x1+x2" to check... Confused please help!
    Edit:
    Do I simply just show this?
    (x1+y1)+(x2+y2)=
    (x1+x2)+(y1+y2)=0+0=0 (closed under addition)

    cx=c(x1,x2,x3)=(cx1,cx2,cx3)
    (cx1+cx2)=
    c(x1+x2)=c(0)=0 (closed under scalar multiplication)
    So it is a subspace?
     
    Last edited: Dec 14, 2011
  2. jcsd
  3. Dec 14, 2011 #2

    Mark44

    Staff: Mentor

    I'm sure this isn't the question. U is a set of vectors. The question is whether U is a linear subspace of R3.
    The key determinant of whether any vector <x1, x2, x3> is an element of U is whether its first two components of the vector add to 0.

    Start with your two vectors x and y, and assume that they are in U. What can you say for certain about these two vectors? What can you say about their sum?

    What can you say about kx, where k is a scalar?
     
  4. Dec 14, 2011 #3
    I think I got it...
    Do I simply just show this?
    (x1+y1)+(x2+y2)=
    (x1+x2)+(y1+y2)=0+0=0 (closed under addition)

    cx=c(x1,x2,x3)=(cx1,cx2,cx3)
    (cx1+cx2)=
    c(x1+x2)=c(0)=0 (closed under scalar multiplication)
    Thanks for your help!
     
  5. Dec 14, 2011 #4

    Mark44

    Staff: Mentor

    You should show a bit more, not just the calculations. For the first part, what you are doing is showing that if x and y are in U, then so is x + y. This shows that U is closed under vector addition.

    Similar for the other part.
     
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