Proving Linear Dependence in Pm(F) Using Polynomials with p_j(2)=0

[gravenewworld]

Suppose that p_0,p_1,p_2...,p_m are polynomials in Pm(F) such that p_j(2)=0 for each j. Prove that (p_0,...,p_m) is not linearly independent in Pm(F).

So far I have, suppose that there is a polynomial in the list that is of degree 0, then that polynomial must be 0, hence the list is linearly dependent. If there is no polynomial of degree zero, there are at least two polynomials in the list that have the same degree. This where I get stuck, am I going in the right direction?

 

[e(ho0n3]

What in the world is Pm(F)?

 

[gravenewworld]

Sorry maybe I should have explained that. Is P subscript m (F). Its all polynomial s over a field F with degree at most m.

 

[gravenewworld]

Can anyone help me at all? This problem is driving me crazy

 

[arildno]

I think you may construct a weird induction argument on m.
here's my thoughts, anyway:
1. Any such set cannot contain a constant polynomial, sinze that would be the zero polynomial (and hence, the set would be linearly dependent)

2. Take the case of m=1:
We then have two linear polynomials,
$$p_{0}=a_{0}x+b_{0}$$
$$p_{1}=a_{1}x+b_{1}$$
(the a's distinct from zero)
But the condition $$p_{0}(2)=p_{1}(2)=0$$ implies that:
$$p_{0}=b_{0}(1-\frac{x}{2})$$
$$p_{1}=b_{1}(1-\frac{x}{2})$$
Hence, our polynomials are linearly dependent..
hope this helps a bit..

 

[vsage]

Sorry to resurrect old posts but my linear alg. professor came up with a viable way:

we're trying to prove a_0*P_0 + a_1*P_1 + ... + a_m*P_m = 0 for where anyone a_0 through a_m is nonzero. (Equation 1)

P_0, P_1, ..., P_m are m+1 vectors and you already proved that there can't be a p_i of degree 0 for 0<=i<=m . We also know that each p_i(x)%(x-2) = 0 for 0<=i<=m so dividing Equation 1 by (x-2) we have m+1 polynomials of varying degree i where 0<=i<=m-1. These polynomials have a maximum dimension of m, therefore. m+1 polynomials cannot be a basis of F^m because they're not linearly independent therefore the polynomials are linearly dependent. I hope I translated his thoughts correctly.