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Homework Help: Find a basis of U, the subspace of P3

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Find a basis of U, the subspace of P3

    U = {p(x) in P3 | p(7) = 0, p(5) = 0}


    2. Relevant equations



    3. The attempt at a solution
    ax3+bx2+cx+d
    p(7)=343a+49b+7c+d=0
    p(5)=125a+25b+5c+d=0

    d=-343a-49b-7c
    d=-125a-25b-5c

    ax3+bx2+cx+{(d+d)/2} -->{(d+d)/2}=2d/2=d
    (-343a-49b-7c-125a-25b-5c)/2=-234a-37b-6c
    ax3+b2+cx-234a-37b-6c
    a(x3-234)+b(x2-37)+c(x-6)

    basis{x3-234,x2-37,x-6}
    dim=3
    please check if I m correct or not
    and is there a easier way to do it?

    also, if
    U = {p(x) in P3 | p(7) = 0, p(5) = 0,p(3) = 0,p(1) = 0}
    p(x) does not exist?
    thanks!
     
    Last edited: Oct 24, 2011
  2. jcsd
  3. Oct 25, 2011 #2

    Mark44

    Staff: Mentor

    No, that's not right.

    Your subspace of P3 is the set of 3rd degree polynomials p(x) such that p(7) = 0 and p(5) = 0. That means that any polynomials in this space must have factors of (x -7) and (x - 5). So any polynomial in this subspace must satisfy (x - 7)(x - 5)(ex + f) = 0, with e and f being arbitrary.

    There's more to do, but this will get you going in the right direction.
     
  4. Oct 25, 2011 #3
    so the basis would be {(x - 7)(x - 5)x,(x - 7)(x - 5)}? dim=2

    plz answer this one
    U = {p(x) in P3 | p(7) = 0, p(5) = 0,p(3) = 0,p(1) = 0}
    does p(x) exist?
    it has 4 degree(x - 7)(x - 5)(x - 3)(x - 1), but P3 is a set of third degree, so it does not exist
    is that right?
     
    Last edited: Oct 25, 2011
  5. Oct 25, 2011 #4

    Mark44

    Staff: Mentor

    For the first question: yes.
    For the second question: there is no 3rd degree polynomial with four zeroes.
     
  6. Oct 25, 2011 #5
    can i write does not exist?
    thanks!
     
  7. Oct 25, 2011 #6

    Mark44

    Staff: Mentor

    Sure. "There is no ..." and "does not exist" mean the same thing.
     
  8. Oct 25, 2011 #7

    Mark44

    Staff: Mentor

    Correction (and thanks to vela!): There is no 3rd degree polynomial with exactly four zeroes, but p(x) = 0x3 + 0x2 + 0x + 0 satisfies p(7) = p(5) = p(3) = p(1) = 0.
     
  9. Oct 25, 2011 #8
    does that mean the basis exist {1,x,x2,x3}?
     
  10. Oct 25, 2011 #9

    Mark44

    Staff: Mentor

    Let's get your questions straight.
    1) If U is as described in post #1, what is a basis for U? (Already answered)
    2) Is there a polynomial in P3 such that p(7) = p(5) = p(3) = p(1) = 0? Yes, the zero polynomial.
     
  11. Oct 25, 2011 #10
    Yes.
    But the question in my assignment is
    Find a basis of U, the subspace of P3
    d) U = {p(x) in P3 | p(7) = 0, p(5) = 0,p(3) = 0,p(1) = 0}
    and I wrote no basis
     
  12. Oct 25, 2011 #11

    Mark44

    Staff: Mentor

    If you give the reason, your answer will probably be OK.
     
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