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Linear Algebra - Subspace Checks

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following sets form subspaces of R[itex]^{2}[/itex]

    A){([itex]x_{1}[/itex],[itex]x_{2}[/itex])[itex]^{T}[/itex] | [itex]x_{1}[/itex][itex]x_{2}[/itex]=0}
    B){([itex]x_{1}[/itex],[itex]x_{2}[/itex])[itex]^{T}[/itex] | [itex]x_{1}[/itex]=3[itex]x_{2}[/itex]}
    2. Relevant equations
    Does zero vector exist?
    Is the space closed under addition?
    Is the space closed under scalar multiplication?

    3. The attempt at a solution
    I know B is a subspace, but I'm not sure why. I can check the zero vector and the scalar, but I'm not 100% sure how to define closed under addition.

    Also, I know A is not a subspace. Again, I know how to check for the zero vector, but I'm lost on addition and scalar multiplication, at least as a general form.
    Last edited: Feb 7, 2013
  2. jcsd
  3. Feb 7, 2013 #2


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    Give me some examples of vectors in A.
  4. Feb 7, 2013 #3
    [1,0] [0,1] [n,0] [0,n].... one value has to be zero for x1*x2=0 I had the answers backwards. B is a subspace and A is not.
  5. Feb 7, 2013 #4


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    Good! So check addition [1,0]+[0,1]=[1,1]. Is [1,1] in A?
  6. Feb 7, 2013 #5


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    To prove closure under addition take [x,y] in B, i.e. x=3y and [u,v] in B, so u=3v. You want to show [x,y]+[u,v]=[x+u,y+v] is in B. Can you show that?
  7. Feb 8, 2013 #6


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    Not "does the zero vector exist" but rather "is the zero vector in the subspace"
    The zero vector, <0, 0>, clearly exists!

    If <x, y> and <u, v> are in A then we know that xy= 0 and uv= 0. Their sum is <x+u, y+ v> so we need to look at (x+ u)(y+ v)= xy+ xv+ uy+ uv= xv+ uy. Is that necessarily 0?
    If <x, y> and <u, v> are in B then we know that x= 3y and u= 3v. Their sum is <x+u, y+ v> so we need to compare x+ u and 3(y+ v). Are they necessarily equal?
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