Results 1 to 4 of 4

Math Help - Proof Probem

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    23

    Proof Probem

    Problem: Suppose that po, p1, ...., pm are polynomials in Pm(F) such that pj(2)=0 for every j. Prove that (p0, p1, ...., pm) is not linearly independent in Pm(F).

    Not really sure where to start with this one guys. Any help would be much appreciated. THANKS.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by eg37se View Post
    Problem: Suppose that po, p1, ...., pm are polynomials in Pm(F) such that pj(2)=0 for every j. Prove that (p0, p1, ...., pm) is not linearly independent in Pm(F).

    Not really sure where to start with this one guys. Any help would be much appreciated. THANKS.
    let p_i(x)=a_{i0} + a_{i1}x + \cdots + a_{im}x^m, \ \ 0 \leq i \leq m. let A=\begin{pmatrix}a_{00} & a_{01} & . & . & . & a_{0m} \\ a_{10} & a_{11} & . & . & . & a_{1m} \\ . & . & . & . & . & . \\ . & . & . & . & . & . \\ . & . & . & . & . & . \\ a_{m0} & a_{m1} & . & . & . & a_{mm} \end{pmatrix} and X=\begin{pmatrix}1 \\ 2 \\ . \\ . \\ . \\ 2^m \end{pmatrix}. then AX=\bold{0} and so \det A^T=\det A =0. thus the columns of A^T and hence the rows
    of A are linearly dependent, i.e. there exist scalars c_0, \cdots , c_m \in F, which are not all 0, such that c_0R_1 + c_2R_2 + \cdots + c_mR_{m+1}=\bold{0}, where R_i is the i-th row of A. this result is now equivalent

    to what we're looking for: c_0p_0(x) + c_1p_1(x) + \cdots + c_mp_m(x)=0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    23
    I believe you that that works, but those are concepts we unfortunately have not covered yet. Therefore, I think I am going to try to use the Linear Dependence Lemma in my book which states that, (v1,...,vm) is linear dependent provided that:

    vj is an element of span (v1, ...., v(j-1))

    and

    if the jth term is removed from v1,..., vn the span of the remaining list equals span (v1,...., vn)


    Let me know if this looks like it could work. Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Consider the set S=\{p(x) \in P_m[x]\: : \: p(2)=0\}. Clearly this is a subspace of P_m[x], and every p(x) \in S can be written as (x-2)g(x). Consider the map

    T : S \rightarrow P_{m-1}[x]

    (x-2)g(x) \mapsto g(x)

    This is a surjective (i.e. onto) linear transformation, because to every g(x) \in P_{m-1}[x], we have T((x-2)g(x))=g(x). Moreover it is injective (i.e. one-to-one) because the inverse image of the zero polynomial is just the zero polynomial (check this). Hence \mbox{dim }S = m.

    It follows that since you have m+1 elements of S in your list, they cannot be linearly independent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probem Solcing (How Far can You See?)
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: October 9th 2010, 08:13 AM
  2. probem with function
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 10th 2009, 06:01 AM
  3. Solving this probem...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 7th 2008, 11:37 AM
  4. integration probem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 24th 2008, 03:42 AM
  5. story probem help
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 15th 2007, 07:20 PM

Search Tags


/mathhelpforum @mathhelpforum