# Proof Probem

• Sep 22nd 2009, 04:20 PM
eg37se
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.
• Sep 22nd 2009, 06:15 PM
NonCommAlg
Quote:

Originally Posted by eg37se
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 $\displaystyle p_i(x)=a_{i0} + a_{i1}x + \cdots + a_{im}x^m, \ \ 0 \leq i \leq m.$ let $\displaystyle A=\begin{pmatrix}a_{00} & a_{01} & . & . & . & a_{0m} \\ a_{10} & a_{11} & . & . & . & a_{1m} \\ . & . & . & . & . & . \\ . & . & . & . & . & . \\ . & . & . & . & . & . \\ a_{m0} & a_{m1} & . & . & . & a_{mm} \end{pmatrix}$ and $\displaystyle X=\begin{pmatrix}1 \\ 2 \\ . \\ . \\ . \\ 2^m \end{pmatrix}.$ then $\displaystyle AX=\bold{0}$ and so $\displaystyle \det A^T=\det A =0.$ thus the columns of $\displaystyle A^T$ and hence the rows
of $\displaystyle A$ are linearly dependent, i.e. there exist scalars $\displaystyle c_0, \cdots , c_m \in F,$ which are not all 0, such that $\displaystyle c_0R_1 + c_2R_2 + \cdots + c_mR_{m+1}=\bold{0},$ where $\displaystyle R_i$ is the $\displaystyle i$-th row of $\displaystyle A.$ this result is now equivalent

to what we're looking for: $\displaystyle c_0p_0(x) + c_1p_1(x) + \cdots + c_mp_m(x)=0.$
• Sep 22nd 2009, 06:42 PM
eg37se
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.
• Sep 22nd 2009, 07:23 PM
Bruno J.
Consider the set $\displaystyle S=\{p(x) \in P_m[x]\: : \: p(2)=0\}$. Clearly this is a subspace of $\displaystyle P_m[x]$, and every $\displaystyle p(x) \in S$ can be written as $\displaystyle (x-2)g(x)$. Consider the map

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

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

This is a surjective (i.e. onto) linear transformation, because to every $\displaystyle g(x) \in P_{m-1}[x]$, we have $\displaystyle T((x-2)g(x))=g(x)$. 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 $\displaystyle \mbox{dim }S = m$.

It follows that since you have $\displaystyle m+1$ elements of $\displaystyle S$ in your list, they cannot be linearly independent.