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Dimension of set S, subspace of R3?

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine whether set S = {2a,-4a+5b,4b| aε R ^ bε R} is a subspace of R3?

    If it is a subspace of R3, find the dimension?



    2. Relevant equations
    dimension= n if it forms the basis of Rn, meaning that its linear independent and span(S) = V



    3. The attempt at a solution
    I am confused on really where to start on this problem.
    What do I do with the a's and b's? I have only done problems like this with real values for the vectors. What is the first step in determining if this is a subspace?
     
  2. jcsd
  3. Apr 14, 2014 #2

    jbunniii

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    First step is to recall the definition of a subspace: it must be closed under addition and scalar multiplication.

    So start by taking two arbitrary elements of ##S##. An arbitrary element looks like ##v_1 = (2a_1, -4a_1 + 5b_1, 4b_1)##. A second one looks like ##v_2 = (2a_2, -4a_2 + 5b_2, 4b_2)##. What do you get when you add ##v_1## and ##v_2##?
     
  4. Apr 14, 2014 #3
    Adding v1 and v2

    well if you add them wouldn't you get a new vector
    that is
    (2a1 + 2a2, -4a1-4a2 + 5b1 +5b2, 4b1 + 4b2)
    Which is not in the same form as v1 and v2 thus it is not a subspace right?
    So you cannot give it a dimension? Is this correct?

    Thanks for the reply! Can you look at my other posts and help with those?
     
  5. Apr 14, 2014 #4

    jbunniii

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    What if I rearrange the expressions a bit: ##(2(a_1 + a_2), -4(a_1 + a_2) + 5(b_1 + b_2), 4(b_1 + b_2))##. Is this an element of ##S##?
     
  6. Apr 14, 2014 #5
    Okay gotcha so when you re-arrange like that them yes it is an element of S and thus is closed under addition. So it is a subspace?
     
  7. Apr 14, 2014 #6

    jbunniii

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    You also need to check that it is closed under scalar multiplication.
     
  8. Apr 14, 2014 #7

    HallsofIvy

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    As far as the dimension is concerned, notice that (2a,-4a+5b,4b)= (2a, -4a, 0)+ (0, 5b, 4b)= a(2, -4, 0)+ b(0, 5, 4).
     
  9. Apr 14, 2014 #8
    Multiplication

    So if I take v1 = (2a, -4a+5b , 4b)
    and multiply by the scalar k=3
    you would get
    v1 = (6a, -12a +15b, 12b)
    So it isn't closed under scalar multiplication or is it?
     
  10. Apr 14, 2014 #9

    jbunniii

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    We can rewrite that expression as ##(2(3a), -4(3a) + 5(3b), 4(3b))##. Is this in ##S##? What happens if you replace ##k=3## by something more general, i.e. just ##k##?
     
  11. Apr 14, 2014 #10
    Well I guess as long as k isn't zero then it is also closed under scalar multiplication right?
     
  12. Apr 14, 2014 #11
    Okay so it is subspace. Now how do I find the dimension?
     
  13. Apr 14, 2014 #12

    jbunniii

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    ##k=0## works too:
    $$0(2a, -4a + 5b, 4b) = (2(0), -4(0) + 5(0), 4(0)) = (2a', -4a' + 5b', 4b')$$
    where ##a' = b' = 0##.
     
  14. Apr 14, 2014 #13

    jbunniii

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    See HallsofIvy's note above.
     
  15. Apr 15, 2014 #14
    Okay so dimension is 2?
     
  16. Apr 15, 2014 #15

    jbunniii

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    That's right. As Halls noted,
    $$(2a,-4a+5b,4b)= (2a, -4a, 0)+ (0, 5b, 4b)= a(2, -4, 0)+ b(0, 5, 4)$$
    and this observation allows you to immediately see a basis for the subspace. Do you see why?
     
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