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Determining whether a set of vectors is a subspace of R^3?

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine whether the set of all vectors of the form (sin2t,sintcost,3sin2t) is a subspace of R^3 and if so, find a basis for it.


    2. Relevant equations

    I guess you just need to use the axioms where it is closed under scalar addition and multiplication.

    3. The attempt at a solution

    If I have two vectors u=(1,2,3) and v=(4,5,6) then u+v = (5,7,9). This gives us 5=sin2t, 7=sintcost, and 9=3sin2t. Am I right in saying there's no (real) value of t which will satisfy any of these equations, meaning (sin2t,sintcost,3sin2t) isn't closed under addition and thus not a subspace of R^3?
     
  2. jcsd
  3. Apr 18, 2013 #2

    HallsofIvy

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    Let v be such a "vector". Since a vector space must be closed under scalar multiplication, look at 1000v. Is that still in the set.

    (Hint: [itex]-1\le sin(x)\le 1[/itex] for all x.)
     
  4. Apr 18, 2013 #3

    mfb

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    This would require that u=(1,2,3) and v=(4,5,6) are elements of your set (they are not). You can use the approach posted by HallsofIvy, you just need some vector in the set to begin.
     
  5. Apr 18, 2013 #4
    Multiplying by 1000 would give

    [itex]-1\le sin(1000x)\le 1[/itex]

    right? And that statement remains true for all x?
     
    Last edited: Apr 18, 2013
  6. Apr 18, 2013 #5

    mfb

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    While that statement remains true for all x, this is not the point where a multiplication is useful.

    Start here, please. Can you find such a vector of your set?
     
  7. Apr 18, 2013 #6
    Not one with real components I don't think.
     
  8. Apr 18, 2013 #7

    mfb

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    You have vectors with real components. Actually, all vectors have real components, as you are in R^3. Just plug in some arbitrary value of t - I expect that t is real, so the vector components are real as well.
     
  9. Apr 18, 2013 #8
    t=1 say? This would give (0.91,0.45,2.72). Any number for t would give some values back in the form we want.
     
  10. Apr 18, 2013 #9

    mfb

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    Right. Now you can follow the advice of HallsofIvy and multiply this vector by 1000. Is the resulting vector (how does it look like) part of your set, too?
     
  11. Apr 18, 2013 #10

    HallsofIvy

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    No! 1000 sin(x) is NOT equal to sin(1000x).
     
  12. Apr 18, 2013 #11
    So that would mean you have 1000sin(x), which no longer oscillates between the same values as sin(x).
     
  13. Apr 19, 2013 #12

    mfb

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    There is no x here. You get (910,450,2720). Can this vector be part of your set? In other words, is there a t such that (sin2t,sintcost,3sin2t) = (910,450,2720)?
    If you can disprove this, you are done.
     
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