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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 rowsQuote:
Originally Posted by eg37se
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: $\displaystyle c_0p_0(x) + c_1p_1(x) + \cdots + c_mp_m(x)=0.$
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.
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.
