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Subspace of Polynomials of degree 2

  1. Nov 8, 2009 #1
    I'm trying to show that a set W of polynomials in P2 such that p(1)=0 is a subspace of P2. Then find a basis for W and dim(W).

    I have already found that the set W is a subspace of P2 because it is closed under addition and scalar multiplication and have showed that. The thing i'm stuck on is finding a basis of W and the dim(W), but i think if i figure out how to find the basis i would be able to see what the dim(W) is. So my question is can anyone give me a hint on how to find the basis for W? Thanks
  2. jcsd
  3. Nov 8, 2009 #2


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    Welcome to PF!

    Hi Tomblue! Welcome to PF! :smile:
    Hint: what is a basis for the whole of P2? How large is it?

    So how large do you expect a basis for W to be?

    Now adjust the basis you got for P2. :wink:
  4. Nov 8, 2009 #3
    W=p(x)={a0 + a1x + a2x2}
    so i'm not sure if this is right but i'm trying to work through it, so since the set W is a set of polynomials in P2 then p(x)={a0+a1 +a2} and a0 would be zero since p(1)=0. Then therefore the basis for W would be = (0,-1,1) but i think the dim(W) is 2-dimensional but if that was correct then i should of had 2 free variables when finding my basis and it would be different. So i'm still not quite sure i did it right.
  5. Nov 8, 2009 #4
    I can find three linear independent solutions for [tex]\{a_0, a_1, a_2\}[/tex]; it looks like you stopped after you found one.

    You are jumping to conclusions when you say [tex]a_0[/tex] would be zero since p(1)=0.
    Last edited: Nov 8, 2009
  6. Nov 8, 2009 #5
    i found that a0=0, then a2=t, and a1=-t those are the three linearly independent solutions, so what i thought the basis would look like in vector form would bex € P2 such that x=t(0,-1,1)
  7. Nov 8, 2009 #6
    What if you let [tex]a_1=0[/tex] or [tex]a_2 = 0[/tex], first of all is it possible to let these variables take that value, and if so, what are then the corresponding values of the other two variables?
  8. Nov 8, 2009 #7
    If i let a2=0, then a0=-a1 and a1=-a0 but i dont see how that would help me
  9. Nov 8, 2009 #8
    It helps you because it gives you two more linear independent sets of coefficients [tex]\{a_0,a_1,a_2\}[/tex] that help to form a basis for W.
  10. Nov 9, 2009 #9
    Maybe it would help if you see it like this: (I'm no good with latex, sorry).


    \left[ \begin{array}{ccc} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & -1 \\ a_{31} & a_{32} & 1\end{array} \right] [ \begin{array}{ccc} a_{0} \\ a_{1} \\ a_{2}\end{array} ] = 0

  11. Nov 9, 2009 #10


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    Hi Tomblue! :smile:

    (just go up :zzz: …)
    I'm not convinced you understand what a basis is.

    (it would have helped if you'd actually answered my question)

    One basis for P2 would be the three elements 1 x and x2, which we can write as {1, x, x2}.

    Another basis would be {1, 1 + x, 2 + x2}.

    Any set of three independent polynomials will do.

    ok, now fiddle about with {1, x, x2} to get a (smaller) basis for W.
  12. Nov 12, 2009 #11
    Here's a different approach.

    W is the space of polynomials a0+a1x+a2x2
    this is really R3, or {a0 a1 a2}.

    Requiring that a0+a1+a2=0, is the same as requiring that the euclidean inner product of any element of W with {1 1 1} be zero.

    Hence the subspace W forms a plane in R3, namely the plane that is orthogonal to the vector {1 1 1}, and thus should have dimension 2.

    To find a basis simply take the cross product of {1 1 1} with any two other linearly independent vectors, for instance using {1 0 0} and {0 1 0} gives the basis {0 1 -1} and {-1 0 1}, which correspond to the polynomials x-x2 and -1+x2.

    Not so general, but gives a good mental picture.
  13. Nov 13, 2009 #12


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    Any polynomial in P2 can be written uniquely as [itex]a_0+ a_1x+ a_2x^2[/itex] (which is the same as saying that {1, x, x2} is a basis).

    Any polynomial, p, in P2 such that p(1)= 0 must satisfy [itex]a_0+ a_1(1)+ a_2(1^2)= a_0+ a_1+ a_2= 0[/itex]. We can solve for any one of those coefficients in terms of the other two. For example, we can say that [itex]a_0= -a_1- a_2[/itex]. That means that [itex]a_0+ a_1x+ a_2x^2= -a_1- a_2+ a_1x+ a_2x^2[/itex][itex]= (-1+x)a_1+ (-1+ x^2)a_2[/itex]. Now do you see what you can use as a basis? You should also be able to find two other bases for the same subspace by solving for [itex]a_1[/itex] in terms of [itex]a_2[/itex] and [itex]a_0[/itex] and by solving for [itex]a_2[/itex] in terms of [itex]a_0[/itex] and [itex]a_1[/itex].
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