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Homework Help: Determine whether Subsets are Subspaces

  1. Jul 8, 2009 #1
    Here we go....wheeeee

    1. The problem statement, all variables and given/known data
    For each of the following subsets of F3, determine whether it is a subspace of F3

    (a)[tex] {(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=0}[/tex]

    (b)[tex] {(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=4}[/tex]

    (c)[tex] {(x_1, x_2, x_3) \in \mathbf{F}^3: x_1x_2x_3=0}[/tex]

    (d)[tex] {(x_1, x_2, x_3) \in \mathbf{F}^3: x_1=5x_3}[/tex]

    My attempt

    additive ID:
    the additive ID exists in the subset since if (x1,x2,x3)=0 then the criteria that x_1+2x_2+3x_3=0 holds true.

    closure under addition:
    [tex] {(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=0}\text{ and }[/tex]
    [tex] {(y_1, y_2, y_3) \in \mathbf{F}^3: y_1+2y_2+3y_3=0}[/tex]


    [tex]x_1=-(2x_2+3x_3)\text{ and }y_1=-(2y_2+3y_3)[/tex]

    then [tex](x_1, x_2, x_3)+(y_1, y_2, y_3)=(x_1+y_1,x_2+y_2,x_3+y_3)[/tex]


    Applying the condition that [tex]x_1+2x_2+3x_3=0[/tex]

    We have:


    Closure under Scalar multiplication: given
    [tex] {a\in\mathbf{F}\text{ and }(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=0}[/tex]


    Applying the condition we have:


    Thus the subset is a subspace. Look okay?

    immediately we can see that if (x1,x2,x3)=0 than the condition that [itex]x_1+2x_2+3x_3=4[/itex] does not hold. Not a subspace.

    I am having a little trouble seeing this one. if (x1,x2,x3)=0 than the condition that [itex]x_12x_23x_3=0[/itex] is true confirming that the additive ID is included.

    Under addition we have

    [tex](x_1+y_1,x_2+y_2,x_3+y_3)[/tex] applying the condition that the product=0 we have


    I am not sure what to say about this??

    (d) .... working on it
  2. jcsd
  3. Jul 8, 2009 #2


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    Hi Saladsamurai! :smile:
    (no leaning over the side! :biggrin:)

    Yup, your (a) and (b) are fine. :smile:

    Hint for (c): what if x1 = y2 = 0?

    and try rewriting (d) as x1 - 5x3 = 0 :wink:
  4. Jul 8, 2009 #3


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    For (c) it may be clearer to look at a specific example. a=(1,1,0) and b=(0,1,1) are in your subset, right? What about a+b?
  5. Jul 8, 2009 #4
    Then a+b=(1+0, 1+1, 0+1) and the product of these elements [itex]\ne 0[/itex] therefore all a+b [itex]\not\in[/itex] U (where U is the subset in question)

    So c is not a subspace.
    Last edited: Jul 8, 2009
  6. Jul 8, 2009 #5
    For d : If U is the subset in question: The additive identity is on U since if x1=x2=x3=0 then x1=5*x3=0

    Closure under addition:
    Let a=(x1,x2,x3) where x1=5*x3 and b=(y1,y2,y3) where y1=5*y3
    => a+b = ((5*x3+5*y3), x2+y2, x3+y3)

    =(5(x3+y3), x2+y2, x3+y3) => closed.

    Closure under scalar multiplication: if r is a scalar and U= {(x1,x2,x3): x1=5*x3} then,

    r(x1,x2,x3)=(5r*x3,r*x2,r*x3) => closed under mult.

    So d is a subspace
  7. Jul 8, 2009 #6


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    Yup … your c and d are fine now! :biggrin:
  8. Jul 8, 2009 #7
    This isn't so bad

    :notices he has 10 more chapters to go: :uhh:
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