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Subspace of P4

  1. Sep 26, 2012 #1
    1. Let V={(X^2+X+1)p(x) : p(x) [itex]\in[/itex] P2(x)}
    Show that V is a subspace of P4(x). Display a basis, with a proof. What is the dimension of V?



    2.



    3. I started to try to figure out how to prove that V is a subspace of P4, but I'm not sure how.

    To show that it is closed under addition:
    p(x)=x^2 is in P2(x)
    (x^2+x+1)(x^2)+(a^2+a+1)(a^2)
    x^4+x^3+x^2+a^4+a^3+a^2 is in P4, because no exponent is greater than 4.

    The zero vector is if p(x) = 0

    And then I'm not sure how to show that multiplication is closed
     
    Last edited by a moderator: Sep 26, 2012
  2. jcsd
  3. Sep 26, 2012 #2

    Mark44

    Staff: Mentor

    Please don't remove the template text. You are removing all of it except the [ b ] tags, which makes all of what you enter bold.
    Yes, but p(x) is supposed to represent any arbitrary function in P2(x), which could be a polynomial of degree 2 or degree 1, depending on how your text defines P2(x).

    In any case you need to look at (x2 + x + 1)*p(x), where p(x) is some arbitrary polynomial in P2.
    ???
    That would be multiplication by a scalar.
     
  4. Sep 26, 2012 #3
    So I can prove it is closed under scalar multiplication by saying k(x^2+x+1)(ax^2+bx+c) is still in P4. Also would the basis be {x^4, x^3, x^2, x, 1} of dim(5)?
     
  5. Sep 26, 2012 #4

    Zondrina

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    You have to prove three things :

    1. V≠∅ or similarily, 0[itex]\in[/itex]V.
    2. V is closed under addition.
    3. V is closed under scalar multiplication.

    Start by showing that 0 is in V because in most cases it will help you with part 2 and 3.
     
  6. Sep 26, 2012 #5
    Can't I just say that since 0 is in P2, then (x^2+x+1)(0)=0 is in P4?
    Does how I proved scalar multiplication before work?
    To prove scalar addition can't I say that (x^2+x+1)(ax^2+bx+c)+(x^2+x+1)(ex^2+fx+g) is still in P4? Thanks for your help. But I also just wanna double check with that basis and dim(5) that i stated earlier to see if that's right?
     
  7. Sep 26, 2012 #6

    Zondrina

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    Yes, dim(P4) = 5. In fact, you can say in general that dim(Pn) = n+1.

    Now as for your question. I want to make sure that I know what you're asking.

    V={(x2+x+1)p(x) | p(x) ∈ P2}
    Show that V is a subspace of P4. Display a basis, with a proof. What is the dimension of V?

    Fixed that up a bit so I could be more clear. Is this correct?
     
    Last edited: Sep 26, 2012
  8. Sep 26, 2012 #7
    yes that is is correct, thanks
     
  9. Sep 26, 2012 #8

    Zondrina

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    So V is the set containing all polynomials of the form (x2+x+1)p(x) where p(x) is a polynomial of degree of at most 2.

    So any general p(x) has the form ax2 + bx + c and any general element in V has the form (x2 + x + 1)(ax2 + bx + c).

    Indeed to show V≠∅, argue that the zero polynomial of P4.... ( I'll leave this to you ).

    To show additive closure, pick two arbitrary polynomials in V, (x2 + x + 1)(ax2 + bx + c) and (x2 + x + 1)(dx2 + ex + f) and show that their sum must also lie inside V.

    To show scalar closure, pick an arbitrary polynomial in V, (x2 + x + 1)(ax2 + bx + c) and some arbitrary scalar in ℂ and show that even after multiplying them together, the product must still be inside V.
     
  10. Sep 26, 2012 #9

    Dick

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    Multiply out (x^2+x+1)(ax^2+bx+c). The subspace CAN'T be five dimensional. It only has 3 free parameters! Write it as a*(...)+b*(...)+c*(...). Fill in the (...).
     
  11. Sep 27, 2012 #10
    How can it not be five dimensional if you multiply it out you get:

    a(x^4+x^3+x^2)+b(X^3+x^2+x)+c(x^2+x+1)

    Wait. so you can use that as a basis. I see, thanks a lot. Just out of curiosity though, how would you prove that the zero vector exists.
     
  12. Sep 27, 2012 #11

    HallsofIvy

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    Take a= b= c= 0 so that (x^2+x+1)p(x)= (x^2+ x+ 1)(0x^2+ 0x+ 0)= 0, of course.
     
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