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Subspaces & dimension

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Is the collection a subspace of the given vector space? If so what is the dimension?

    V={ax^2+bx+c: a=b+c} in P2


    2. Relevant equations



    3. The attempt at a solution
    The first part of the question is pretty straightforward. I just verified closure under addition and scalar multiplication to show that V is indeed a subspace of P2. But I am confused about the second part. I know that the dimension of P2 is 3, so dimV must be < or = 3. I want to say three, but that would mean that V=P2, and I'm not sure that that's true. Any ideas?
     
  2. jcsd
  3. May 16, 2009 #2
    You have the condition that a = b + c, so a is constrained. b and c completely specify a. Can you show the dimension explicitly from this?
     
  4. May 16, 2009 #3
    Hmmm...ok. I think that a basis for V would be {x^2+x, x^2+1}, and therefore the dimension is 2.
     
  5. May 17, 2009 #4

    HallsofIvy

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    Okay, why do you think that is a basis? Are x^2+ x and x^2+ 1 independent? Do they span V?
     
  6. May 17, 2009 #5
    Well, {x^2+x, x^2+1} is linearly independent, and I think they span V since it contains all of the different polynomial terms as well as satisfying the restrictions placed on the coefficients. Is that not right?
     
  7. May 17, 2009 #6

    matt grime

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    Thinking it is not the same as having a proof for it. Forget polynomials - it's just 3-tuples

    (a,b,c)

    such that a=b+c. Is (1,1,0) and (0,1,1) a basis? I.e. are the linearly independent, and do they span? Even more explicitly, if x(1,1,0)+y(0,1,1)=(0,0,0) what does that say about x and y? And given (a,b,c) in the subspace, can you write it as r(1,1,0)+s(0,1,1) for a suitable r and s?
     
  8. May 17, 2009 #7

    HallsofIvy

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    Of course, you can use x2, etc. if you want to!:tongue:

    To show they are independent, look at the equation [itex]\alpha(x^2+ x)+ \beta(x^2+ 1)= 0[/itex] (= 0 for all x). What must [itex]\alpha[/itex] and [itex]b[/itex] equal?

    Since V consists of all polynomials of the form [itex]ax^2+ bx+ c[/itex] with a= b+c or [itex](b+c)x^2+ bx+ c[/itex], to show that these span V, look at the equation [itex]\alpha(x^2+ x)+ \beta(x^2+ 1)= (b+c)x^2+ bc+ d[/itex]. Can that be solved for [itex]\alpha[/itex] and [itex]\beta[/itex] for all b and c?
     
  9. May 18, 2009 #8
    Yes, I've got the linear independence part, alpha and beta equal 0. So point #1 for a basis is satisfied by B= {x^2+x, x^2+1}

    Ok, so for any arbitrary polynomial in V, (b+c)x^2+bx+c, exists in the span of the basis vectors.

    so c1(x^2+x) + c2(x^2+1) = (b+c)x^2+bx+c
    =c1x^2 +c1x + c2x^2 + c2 = (b+c)x^2 + bx + c

    so the resulting system of equations is:
    c1+c2 = b + c
    c1=b
    c2=c and this system is consistent.

    So I have shown that my proposed B is indeed a basis for V and that V is a 2 dimensional subspace of P2.
     
  10. May 19, 2009 #9

    matt grime

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    You've just assumed the answer - you can't write (b+c)x^2+bx+c as an arbitrary vector in V when you're trying to show that such a decomposition exists.
     
  11. May 19, 2009 #10

    HallsofIvy

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    That's the wrong way- you have shown that (b+c)x^2+ bx+ c in of the form ax^2+ bx+c with a= b+ c which was obvious from the definition of the subset.

    I said before
    Do that!
     
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