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Which of the subsets of R^3 is a subspace of R^3.

  1. Nov 10, 2013 #1
    1. Which of the subsets of R3 is a subspace of R3.

    a) W = {(x,y,z): x + y + z = 0}

    b) W = {(x,y,z): x + y + z = 1}

    I was wondering if my answer for A is correct.

    2. Relevant equations



    3.

    A) W = {(x,y,z): x + y + z = 0}

    Since, x + y + z = 0. Then, the values for all the variables have to be zero. Therefore, the only vector in W is the zero vector. So, W is nonempty and a subset of R3.

    Furthermore, because W is closed under addition and scalar multiplication, it is a subspace of R3.

    Testing for closure under addition:

    Let a = (a1, a2, a3) and
    Let b = (b1, b2, b3)

    a + b = (a1, a2, a3) + (b1, b2, b3)

    = (a1 + b1, a2 + b2, a3 + b3)

    where x = a1 + b1, y = a2 + b2, z = a3 + b3.

    = (x,y,z) Closure under addition.


    Testing for closure under scalar multiplication:
    c is any real number

    c(x1, y2, z3)
    = (cx1, cy2, cz3) Where x1, y2, z3 = 0

    = (0,0,0)
     
  2. jcsd
  3. Nov 10, 2013 #2

    Simon Bridge

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    That just means (0,0,0) is one member of the set. Another member would be (1,1,-2) since 1+1-2=0 as well.
     
  4. Nov 10, 2013 #3

    LCKurtz

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    What about (-2,1,1)?

    You need to redo the closure arguments. The results of the operations have to satisfy x+y+z=0.
     
  5. Nov 10, 2013 #4
    If I wrote the first part like this:

    Since, x + y + z = 0. Then, the values for all the variables could equal to zero. Therefore, the zero vector is in W. So, W is nonempty and a subset of R3.

    would it work?

    and I might need some help on the closure arguments.
     
  6. Nov 10, 2013 #5

    Mark44

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    Restate the sentence that starts with "Then, the values ..." as If we let x = y = z = 0, then the zero vector is in this set, and W is nonempty.

    You don't need "and a subset of R3." This is given in the problem statement.
    Assume that u and v are two arbitrary vectors in your set. Show that u + v must also be contained in this set. The key here is arbitrary. You can't pick and choose specific vectors.

    Assume that k is an arbitrary scalar. Show that ku is in the set.
     
  7. Nov 10, 2013 #6

    HallsofIvy

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    Suppose (x, y, z) satisfies x+ y+ z= 0 and (a, b, c) satisfies a+ b+ c= 0. What can you say about (x+ a, y+ b, c+ d) and (ax, ay, az)?
     
  8. Nov 10, 2013 #7
    There closed under addition and scalar multiplication.
     
  9. Nov 10, 2013 #8
    Would this work?


    Let u = (u1, u2, u3) and
    Let v = (v1, v2, v3)

    u + v = (u1, u2, u3) + (v1, v2, v3)

    = (u1 + v1, u2 + v2, u3 + v3)

    where x = u1 + v1, y = u2 + v2, z = u3 + v3.

    = (x,y,z) Closure under addition.
     
  10. Nov 10, 2013 #9

    Mark44

    Staff: Mentor

    No, this won't work. Your u and v vectors are just arbitrary vectors in R3. You need to start with vectors in set W.

    Your OP gives a precise but terse definition of W. With this definition, one can distinguish between vectors in W and vectors in R3 that aren't in W.

    To get your intuition working, can you come up with specific vectors in W? Note that in proving that a set is a subspace, you can't use specific vectors, but this exercise might help you understand a little better what you need to do in your proof.
     
  11. Nov 10, 2013 #10
    (0,0,0)

    (1,1,-2)
     
  12. Nov 10, 2013 #11

    Simon Bridge

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    Back to basics.
    W is the set of all triples (x,y,z) which satisfy the condition C: x+y+z=0.

    You know that (0,0,0) is a member of W because it satisfies the condition that x+y+z=0
    Therefore W is not empty. So far so good...

    It should be easy to find other members ... how many members are there altogether?
    hint: more than 4 - see if you can come up with a few more, i.e. is (4,-1,7) a member of W?
    This is the exercise suggested by Mark44.

    If z=0, what possible values of x and y will work?
    What if z=1? z=-1?, z=1/2,1/3,10?
    If (x,y,z) were plotted on x,y,z axes, what geometric object would the members of W make?

    You cannot use specific vectors to prove the subspace...

    u=(a,b,c) and v=(e,f,g) are two representative members of W,
    this means that a+b+c=0 and e+f+g=0

    W is closed under addition if w=u+v is also a member of W.
    How would you go about showing this is the case?
    hint: find the components of w in terms of the components of u and of v, then see if they satisfy the condition C.
     
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